Exercise 1: One sign apart: a double eigenvalue decided by its eigenspace

An $n \times n$ matrix is diagonalizable when it has $n$ linearly independent eigenvectors. A repeated eigenvalue is the only place where that count can fall short, and it is exactly where students stop thinking and start guessing.

Consider $A = \begin{pmatrix} 3 & 0 & 0 \\ 1 & 2 & 1 \\ -1 & 1 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 0 & 0 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$ . The two matrices differ by a single sign, in the bottom left corner.

a) Compute the characteristic polynomial of $A$ by cofactor expansion along a well-chosen row, and give the eigenvalues with their algebraic multiplicities. Explain why $B$ has exactly the same polynomial.
b) For the repeated eigenvalue of $A$ , find the eigenspace: its dimension and a basis.
c) Do the same for $B$ .
d) Decide which of $A$ and $B$ is diagonalizable. For the one that is, give $P$ and $D$ with $P^{-1}AP = D$ .
e) Explain why the simple eigenvalue $\lambda = 1$ never needed a rank test, in either matrix.

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Answers

a) $\det(A - \lambda I) = (3-\lambda)^2(1-\lambda)$ for both: $\lambda = 3$ (algebraic multiplicity $2$ ), $\lambda = 1$ (multiplicity $1$ )
b) $\dim E_3 = 2$ , basis $(1,1,0)$ , $(-1,0,1)$
c) $\dim E_3 = 1$ , basis $(0,1,1)$
d) $A$ is diagonalizable, $P = \begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -1 \end{pmatrix}$ , $D = \text{diag}(3,3,1)$ ; $B$ is not
e) Algebraic multiplicity $1$ forces geometric multiplicity $1$

a) Expand along row $1$ , which holds two zeros, so only one cofactor survives: $\det(A - \lambda I) = (3-\lambda)\det\begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix} = (3-\lambda)\big((2-\lambda)^2 - 1\big)$ . The bracket is a difference of squares, $(2-\lambda-1)(2-\lambda+1) = (1-\lambda)(3-\lambda)$ , so the polynomial is $(3-\lambda)^2(1-\lambda)$ : $\lambda = 3$ with algebraic multiplicity $2$ and $\lambda = 1$ with multiplicity $1$ . For $B$ the same expansion along row $1$ deletes row $1$ and column $1$ , and what remains is the SAME $2 \times 2$ block: the entry that changed sign sits in column $1$ , which the minor never sees. Two quick checks: the trace $3 + 2 + 2 = 7 = 3 + 3 + 1$ , and $\det A = 3 \times (4 - 1) = 9 = 3 \times 3 \times 1$ .

b) $A - 3I = \begin{pmatrix} 0 & 0 & 0 \\ 1 & -1 & 1 \\ -1 & 1 & -1 \end{pmatrix}$ . The operation $R_3 \to R_3 + R_2$ turns row $3$ into a zero row, so one pivot remains: rank $1$ , and the nullity is $3 - 1 = 2$ . The single equation $x - y + z = 0$ leaves $y$ and $z$ free, $x = y - z$ , so $E_3 = \text{span}\{(1,1,0), (-1,0,1)\}$ . Check each one: $A(1,1,0) = (3,3,0)$ and $A(-1,0,1) = (-3,0,3)$ . The geometric multiplicity is $2$ , equal to the algebraic one: the double root has kept a whole PLANE of eigenvectors.

c) $B - 3I = \begin{pmatrix} 0 & 0 & 0 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix}$ . Now $R_3 \to R_3 - R_2$ gives $(0, 2, -2)$ , a second pivot: rank $2$ , nullity $1$ . The equations $x - y + z = 0$ and $y - z = 0$ give $y = z$ and $x = 0$ , so $E_3 = \text{span}\{(0,1,1)\}$ , and indeed $B(0,1,1) = (0,3,3)$ . The geometric multiplicity is $1$ , strictly less than $2$ : the eigenvalue is DEFECTIVE. Same polynomial, same eigenvalues, and one sign has cost a whole dimension of eigenvectors.

d) For $A$ , the eigenvalue $1$ : $A - I = \begin{pmatrix} 2 & 0 & 0 \\ 1 & 1 & 1 \\ -1 & 1 & 1 \end{pmatrix}$ gives $x = 0$ from $R_1$ , then $y + z = 0$ , so $E_1 = \text{span}\{(0,1,-1)\}$ . Together with the basis of $E_3$ that is $3$ eigenvectors, and they are independent: $\det P = -2 \neq 0$ for $P = \begin{pmatrix} 1 & -1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & -1 \end{pmatrix}$ . So $A$ is diagonalizable and $P^{-1}AP = D = \text{diag}(3, 3, 1)$ , with the eigenvalues in the order of the columns. For $B$ , the eigenspaces supply $1 + 1 = 2$ independent eigenvectors, one short of $3$ : no invertible $P$ can be built, and $B$ is NOT diagonalizable. The verdict came from one rank, never from the polynomial.

e) For every eigenvalue, $1 \le$ geometric multiplicity $\le$ algebraic multiplicity: an eigenvalue always has at least one eigenvector, and it can never have more independent ones than its multiplicity as a root. Algebraic multiplicity $1$ squeezes the geometric one to exactly $1$ , so a simple eigenvalue can never be the reason a matrix fails. On an exam, the rank test is spent on the repeated eigenvalues only; computing $E_1$ is still needed, but to fill a column of $P$ , not to decide anything.

Exercise 2: A parameter decides, and it is not the one you expect

Let $A = \begin{pmatrix} 2 & a & 1 \\ 0 & 2 & b \\ 0 & 0 & -1 \end{pmatrix}$ , where $a$ and $b$ are real numbers.

The matrix is triangular, so its eigenvalues cost nothing. The whole question is what happens inside the eigenspace of the repeated one, and which of the two parameters can reach it.

a) Give the eigenvalues of $A$ with their algebraic multiplicities, and justify in one line.
b) Find all values of $a$ and $b$ for which $A$ is diagonalizable.
c) Take $a = 0$ and $b = 3$ , so $A = \begin{pmatrix} 2 & 0 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & -1 \end{pmatrix}$ . Find an invertible $P$ and a diagonal $D$ with $P^{-1}AP = D$ .
d) For which real $k$ is $C = \begin{pmatrix} 1 & 1 \\ 0 & k \end{pmatrix}$ diagonalizable?
e) From d), a student concludes: "a matrix is diagonalizable exactly when its eigenvalues are distinct". Give the counterexample that kills the word exactly, and state the correct criterion.

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Answers

a) $2$ (algebraic multiplicity $2$ ) and $-1$ (multiplicity $1$ ), for all $a$ , $b$
b) Diagonalizable if and only if $a = 0$ , whatever $b$
c) $P = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -3 \\ 0 & 0 & 3 \end{pmatrix}$ , $D = \text{diag}(2, 2, -1)$
d) All $k \neq 1$
e) $I_2$ : repeated eigenvalue, already diagonal. Criterion: geometric $=$ algebraic multiplicity for every eigenvalue

a) $A - \lambda I$ is still upper triangular, and the determinant of a triangular matrix is the product of its diagonal: $\det(A - \lambda I) = (2-\lambda)^2(-1-\lambda)$ . So $\lambda = 2$ with algebraic multiplicity $2$ and $\lambda = -1$ with multiplicity $1$ , whatever $a$ and $b$ are. The parameters are invisible to the polynomial, which is exactly why the polynomial alone can never settle the question.

b) Only $\lambda = 2$ needs a test. $A - 2I = \begin{pmatrix} 0 & a & 1 \\ 0 & 0 & b \\ 0 & 0 & -3 \end{pmatrix}$ . Row $3$ is a multiple of $(0,0,1)$ and so is row $2$ : $R_2 \to R_2 + \frac{b}{3}R_3$ turns row $2$ into zeros for EVERY $b$ . What remains is $(0, a, 1)$ and $(0, 0, -3)$ . If $a \neq 0$ these two rows have pivots in columns $2$ and $3$ : rank $2$ , nullity $1 < 2$ , not diagonalizable. If $a = 0$ , $R_1 \to R_1 + \frac{1}{3}R_3$ kills row $1$ too: rank $1$ , nullity $2$ , diagonalizable. So $A$ is diagonalizable if and only if $a = 0$ , and $b$ plays no role. The reason is structural: $b$ couples the eigenvalue $2$ to the eigenvalue $-1$ , which are distinct, while $a$ couples the eigenvalue $2$ to ITSELF, and only that coupling can crush an eigenspace. Writing " $a = 0$ and $b = 0$ " costs the part.

c) $E_2$ : with $a = 0$ , $A - 2I$ reduces to the single equation $z = 0$ , so $E_2 = \text{span}\{(1,0,0), (0,1,0)\}$ . $E_{-1}$ : $A + I = \begin{pmatrix} 3 & 0 & 1 \\ 0 & 3 & 3 \\ 0 & 0 & 0 \end{pmatrix}$ gives $3x + z = 0$ and $3y + 3z = 0$ ; with $z = 3$ this is $x = -1$ , $y = -3$ , so the eigenvector is $(-1, -3, 3)$ . Check: $A(-1,-3,3) = (-2 + 3, -6 + 9, -3) = (1, 3, -3)$ , which is $-1$ times the vector. Hence $P = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -3 \\ 0 & 0 & 3 \end{pmatrix}$ and $D = \text{diag}(2, 2, -1)$ , the two columns for $\lambda = 2$ first because the two $2$ come first in $D$ . $P$ is triangular with $\det P = 3 \neq 0$ .

d) $C$ is triangular, with eigenvalues $1$ and $k$ . If $k \neq 1$ they are distinct, and two distinct eigenvalues give two independent eigenvectors: diagonalizable, no further work. If $k = 1$ , $C - I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has rank $1$ , so $\dim E_1 = 1 < 2$ and $C$ is not diagonalizable. Answer: every $k \neq 1$ .

e) The identity $I_2$ has the single eigenvalue $1$ , repeated, and it is diagonal already, with $P = I$ . The matrix of c) is another counterexample. Distinct eigenvalues are SUFFICIENT, never necessary. The correct criterion: $A$ is diagonalizable if and only if, for every eigenvalue, the geometric multiplicity equals the algebraic multiplicity; equivalently the dimensions of the eigenspaces add up to $n$ .

Exercise 3: Building P and D, and the order that ties them

Diagonalizing is a change of axes: along an eigenvector, $A$ does nothing but stretch. The matrix $P$ lists those axes as its COLUMNS, and $D$ lists the stretch factors in the SAME order. Both halves of that sentence cost marks when they are forgotten.

Let $A = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$ .

a) Find an invertible $P$ and a diagonal $D$ with $P^{-1}AP = D$ .
b) Check your answer without computing $P^{-1}$ , by comparing $AP$ with $PD$ column by column.
c) Swap the two columns of $P$ : what becomes of $D$ ? And if instead you multiply the first column of $P$ by $-5$ ?
d) A student writes the eigenvectors as the ROWS of a matrix $Q$ . Compute $Q^{-1}AQ$ and show it is not diagonal.
e) Compute $P^{-1}$ and verify that $PDP^{-1} = A$ .

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Answers

a) $P = \begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix}$ , $D = \text{diag}(4, -1)$
b) $AP = PD = \begin{pmatrix} 8 & -1 \\ 12 & 1 \end{pmatrix}$
c) Swapping gives $D = \text{diag}(-1, 4)$ ; scaling a column leaves $D$ unchanged
d) $Q^{-1}AQ = \frac{1}{5}\begin{pmatrix} 28 & 22 \\ -12 & -13 \end{pmatrix}$ , not diagonal
e) $P^{-1} = \frac{1}{5}\begin{pmatrix} 1 & 1 \\ 3 & -2 \end{pmatrix}$ and $PDP^{-1} = A$

a) $\det(A - \lambda I) = (1-\lambda)(2-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = (\lambda - 4)(\lambda + 1)$ . For $\lambda = 4$ : $A - 4I = \begin{pmatrix} -3 & 2 \\ 3 & -2 \end{pmatrix}$ , and $R_2 \to R_2 + R_1$ leaves $-3x + 2y = 0$ , so $(2, 3)$ . For $\lambda = -1$ : $A + I = \begin{pmatrix} 2 & 2 \\ 3 & 3 \end{pmatrix}$ , and $R_2 \to R_2 - \frac{3}{2}R_1$ leaves $x + y = 0$ , so $(1, -1)$ . Two distinct eigenvalues, so diagonalizability is automatic: $P = \begin{pmatrix} 2 & 1 \\ 3 & -1 \end{pmatrix}$ and $D = \text{diag}(4, -1)$ . The $4$ comes first in $D$ because $(2,3)$ comes first in $P$ .

b) $AP$ is $A$ applied to each column of $P$ : $A(2,3) = (2 + 6, 6 + 6) = (8, 12) = 4(2,3)$ and $A(1,-1) = (1 - 2, 3 - 2) = (-1, 1) = -1(1,-1)$ . And $PD$ multiplies each column of $P$ by the matching diagonal entry: $PD = \begin{pmatrix} 8 & -1 \\ 12 & 1 \end{pmatrix}$ . The two agree. Since $P$ is invertible ( $\det P = -2 - 3 = -5$ ), $AP = PD$ is the same statement as $P^{-1}AP = D$ , and it costs two matrix-vector products instead of an inverse. This is the check to write on every copy.

c) Swapping the columns, $P' = \begin{pmatrix} 1 & 2 \\ -1 & 3 \end{pmatrix}$ , forces $D' = \text{diag}(-1, 4)$ : the $i$ -th diagonal entry of $D$ is the eigenvalue of the $i$ -th column of $P$ , so the two lists move together. Keeping $D = \text{diag}(4, -1)$ with the swapped $P$ gives $AP' \neq P'D$ , and the answer is wrong although every number in it is right. Multiplying the first column by $-5$ gives $(-10, -15)$ , still an eigenvector for $4$ , so $D$ does not change. $P$ is never unique; $D$ is unique up to the order of its entries.

d) $Q = \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}$ , $\det Q = -5$ , $Q^{-1} = \frac{1}{5}\begin{pmatrix} 1 & 3 \\ 1 & -2 \end{pmatrix}$ . Then $AQ = \begin{pmatrix} 4 & 1 \\ 8 & 7 \end{pmatrix}$ and $Q^{-1}AQ = \frac{1}{5}\begin{pmatrix} 28 & 22 \\ -12 & -13 \end{pmatrix}$ , which is not diagonal. The reason is in the product itself: $AQ$ applies $A$ to the COLUMNS of $Q$ , here $(2,1)$ and $(3,-1)$ , and neither of those is an eigenvector. Rows of $P$ are never read by the product $AP$ .

e) For a $2 \times 2$ , swap the diagonal, negate the off-diagonal, divide by the determinant: $P^{-1} = \frac{1}{-5}\begin{pmatrix} -1 & -1 \\ -3 & 2 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 1 & 1 \\ 3 & -2 \end{pmatrix}$ . Then $PDP^{-1} = \frac{1}{5}\begin{pmatrix} 8 & -1 \\ 12 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 3 & -2 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 5 & 10 \\ 15 & 10 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix} = A$ . Note where $P$ sits: $A = PDP^{-1}$ , with $P$ on the LEFT. Multiplying $P^{-1}AP = D$ by $P$ on the left and by $P^{-1}$ on the right is the only way to get it back.

Exercise 4: Powers of a matrix, read in the eigenvector axes

Let $A = \begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$ . The figure draws its eigenvectors $v_1 = (1,1)$ , for $\lambda = 2$ , and $v_2 = (2,1)$ , for $\lambda = 3$ , as two new axes, and writes the standard vector $e_1 = (1,0)$ in those axes as the diagonal of a parallelogram.

In the eigenvector axes, $A$ multiplies the first coordinate by $2$ and the second by $3$ ; $A^n$ does it $n$ times. That is the whole reason diagonalization computes powers.

a) Check that $v_1$ and $v_2$ are eigenvectors, and write $P$ , $D$ and $P^{-1}$ .
b) Prove that $A^n = PD^nP^{-1}$ for every integer $n \ge 1$ .
c) Give $A^n$ entry by entry. Check the formula at $n = 0$ , at $n = 1$ , and against $A^3$ computed directly.
d) Read on the figure how $e_1$ is written in terms of $v_1$ and $v_2$ , and deduce $A^ne_1$ without any matrix product. Compare with c).
e) Put $n = -1$ in the formula of c) and check that it gives $A^{-1}$ .

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Answers

a) $Av_1 = 2v_1$ , $Av_2 = 3v_2$ ; $P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$ , $D = \text{diag}(2,3)$ , $P^{-1} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$
b) $P^{-1}P = I$ cancels between consecutive factors; induction on $n$
c) $A^n = \begin{pmatrix} 2 \cdot 3^n - 2^n & 2^{n+1} - 2 \cdot 3^n \\ 3^n - 2^n & 2^{n+1} - 3^n \end{pmatrix}$ , and $A^3 = \begin{pmatrix} 46 & -38 \\ 19 & -11 \end{pmatrix}$
d) $e_1 = -v_1 + v_2$ , so $A^ne_1 = -2^nv_1 + 3^nv_2 = (2 \cdot 3^n - 2^n, 3^n - 2^n)$
e) $A^{-1} = \begin{pmatrix} 1/6 & 1/3 \\ -1/6 & 2/3 \end{pmatrix}$

a) $Av_1 = (4 - 2, 1 + 1) = (2, 2) = 2v_1$ and $Av_2 = (8 - 2, 2 + 1) = (6, 3) = 3v_2$ . So $P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$ , with the columns in the order $v_1, v_2$ , and $D = \text{diag}(2, 3)$ in the same order. $\det P = 1 - 2 = -1$ , so $P^{-1} = \frac{1}{-1}\begin{pmatrix} 1 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$ . Check: $PP^{-1} = \begin{pmatrix} -1 + 2 & 2 - 2 \\ -1 + 1 & 2 - 1 \end{pmatrix} = I$ .

b) From $P^{-1}AP = D$ we get $A = PDP^{-1}$ . Then $A^2 = PDP^{-1}PDP^{-1} = PD(P^{-1}P)DP^{-1} = PD^2P^{-1}$ : the inner $P^{-1}P$ collapses to $I$ . By induction, if $A^n = PD^nP^{-1}$ then $A^{n+1} = A^nA = PD^nP^{-1}PDP^{-1} = PD^{n+1}P^{-1}$ . Finally $D^n = \text{diag}(2^n, 3^n)$ , because a product of diagonal matrices multiplies the diagonal entries. Only $D$ is raised entry by entry; $A^n$ is NOT the matrix of the $n$ -th powers of the entries of $A$ .

c) $PD^n = \begin{pmatrix} 2^n & 2 \cdot 3^n \\ 2^n & 3^n \end{pmatrix}$ , then multiply by $P^{-1}$ : $A^n = \begin{pmatrix} 2 \cdot 3^n - 2^n & 2^{n+1} - 2 \cdot 3^n \\ 3^n - 2^n & 2^{n+1} - 3^n \end{pmatrix}$ . At $n = 0$ : $\begin{pmatrix} 2 - 1 & 2 - 2 \\ 1 - 1 & 2 - 1 \end{pmatrix} = I$ . At $n = 1$ : $\begin{pmatrix} 6 - 2 & 4 - 6 \\ 3 - 2 & 4 - 3 \end{pmatrix} = A$ . At $n = 3$ : $\begin{pmatrix} 54 - 8 & 16 - 54 \\ 27 - 8 & 16 - 27 \end{pmatrix} = \begin{pmatrix} 46 & -38 \\ 19 & -11 \end{pmatrix}$ , and directly $A^2 = \begin{pmatrix} 14 & -10 \\ 5 & -1 \end{pmatrix}$ , $A^3 = A^2A = \begin{pmatrix} 56 - 10 & -28 - 10 \\ 20 - 1 & -10 - 1 \end{pmatrix}$ : the same. The two checks at $n = 0$ and $n = 1$ take ten seconds and catch a $P^{-1}$ on the wrong side every time.

d) Following the parallelogram, $e_1 = -v_1 + v_2$ : indeed $-(1,1) + (2,1) = (1, 0)$ . Its eigen-coordinates are $(-1, 1)$ , and $A^n$ acts on them one at a time: $A^ne_1 = -2^nv_1 + 3^nv_2 = (-2^n + 2 \cdot 3^n, -2^n + 3^n)$ . That is exactly the first column of c), as it must be, since $A^ne_1$ IS the first column of $A^n$ . The figure also says where things go: $3^n$ outgrows $2^n$ , so $A^ne_1$ turns toward the line of $v_2$ as $n$ grows.

e) With $n = -1$ : $A^{-1} = \begin{pmatrix} \frac{2}{3} - \frac{1}{2} & 1 - \frac{2}{3} \\ \frac{1}{3} - \frac{1}{2} & 1 - \frac{1}{3} \end{pmatrix} = \begin{pmatrix} 1/6 & 1/3 \\ -1/6 & 2/3 \end{pmatrix}$ . The $2 \times 2$ formula agrees: $\det A = 4 + 2 = 6$ and $A^{-1} = \frac{1}{6}\begin{pmatrix} 1 & 2 \\ -1 & 4 \end{pmatrix}$ . The formula is allowed at $n = -1$ because $D^{-1} = \text{diag}(\frac{1}{2}, \frac{1}{3})$ exists, that is because $0$ is not an eigenvalue.

Exercise 5: Similar matrices: what they share, and what sharing does not prove

Two $n \times n$ matrices $A$ and $B$ are similar when $B = P^{-1}AP$ for some invertible $P$ . They are then the same linear map written in two different bases, and everything that does not depend on the basis is shared: determinant, trace, rank, characteristic polynomial, eigenvalues.

The trap is the converse. Invariants can prove that two matrices are NOT similar; no list of invariants can prove that they are.

a) Prove that similar matrices have the same determinant and the same characteristic polynomial. Deduce that they have the same eigenvalues and the same trace.
b) Are $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ and $\begin{pmatrix} 5 & 0 \\ 1 & 0 \end{pmatrix}$ similar?
c) Show that $M = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ and $2I$ have the same characteristic polynomial, trace, determinant and rank, and yet are not similar.
d) Show that $M_1 = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix}$ and $M_2 = \begin{pmatrix} 3 & 0 \\ 5 & 1 \end{pmatrix}$ ARE similar, by going through a diagonal matrix, and give an $S$ with $M_2 = S^{-1}M_1S$ .
e) In Exercise 1, $A$ and $B$ had the same characteristic polynomial. Use an invariant to prove that they are not similar.

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Answers

a) $\det(P^{-1}AP) = \det A$ and $P^{-1}AP - \lambda I = P^{-1}(A - \lambda I)P$
b) No: $\det = -2$ against $0$ (the traces agree, which proves nothing)
c) Both $(2-\lambda)^2$ , trace $4$ , det $4$ , rank $2$ ; but $P^{-1}(2I)P = 2I \neq M$ for every $P$
d) Both are similar to $\text{diag}(1,3)$ ; $S = PQ^{-1} = \begin{pmatrix} -2 & 1 \\ 1/2 & 0 \end{pmatrix}$
e) $\text{rank}(A - 3I) = 1$ but $\text{rank}(B - 3I) = 2$ : not similar

a) $\det(P^{-1}AP) = \det(P^{-1})\det A\det P = \frac{1}{\det P}\det A\det P = \det A$ , by the product rule. For the polynomial, the key line is $P^{-1}AP - \lambda I = P^{-1}(A - \lambda I)P$ , true because $P^{-1}(\lambda I)P = \lambda P^{-1}P = \lambda I$ . Taking determinants, $\det(B - \lambda I) = \det(A - \lambda I)$ for every $\lambda$ : same polynomial, hence same eigenvalues with the same algebraic multiplicities. The trace is the sum of the eigenvalues (for a $2 \times 2$ it is the coefficient in $\lambda^2 - (\text{tr})\lambda + \det$ ), so it is shared too. Rank is shared because multiplying by invertible matrices never changes it.

b) The traces agree, $1 + 4 = 5 = 5 + 0$ , and a student who stops there concludes wrongly. The determinants are $4 - 6 = -2$ and $0$ . One invariant that differs is enough: NOT similar. The second matrix is singular and the first is not, and no change of basis can make a map lose invertibility.

c) $\det(M - \lambda I) = (2 - \lambda)^2 = \det(2I - \lambda I)$ ; trace $4$ and determinant $4$ for both; rank $2$ for both. Suppose $P^{-1}(2I)P = M$ . But $P^{-1}(2I)P = 2P^{-1}P = 2I$ , whatever $P$ is, and $2I \neq M$ . So the ONLY matrix similar to $2I$ is $2I$ itself. The four invariants were all equal and the conclusion is still no. What separates them is the eigenspace: $M - 2I$ has rank $1$ , so $M$ has a single line of eigenvectors, while every vector is an eigenvector of $2I$ .

d) $M_1$ is triangular with eigenvalues $1$ and $3$ , distinct, so $M_1 = PDP^{-1}$ with $D = \text{diag}(1, 3)$ : $E_1 = \text{span}\{(1,0)\}$ and $M_1 - 3I = \begin{pmatrix} -2 & 2 \\ 0 & 0 \end{pmatrix}$ gives $(1,1)$ , so $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ . $M_2$ is triangular with eigenvalues $3$ and $1$ : $M_2 - I = \begin{pmatrix} 2 & 0 \\ 5 & 0 \end{pmatrix}$ gives $(0,1)$ and $M_2 - 3I = \begin{pmatrix} 0 & 0 \\ 5 & -2 \end{pmatrix}$ gives $(2,5)$ ; ordering the columns to match the SAME $D$ , $Q = \begin{pmatrix} 0 & 2 \\ 1 & 5 \end{pmatrix}$ and $M_2 = QDQ^{-1}$ . Then $D = P^{-1}M_1P$ , so $M_2 = QP^{-1}M_1PQ^{-1} = S^{-1}M_1S$ with $S = PQ^{-1}$ . Here $Q^{-1} = \frac{1}{-2}\begin{pmatrix} 5 & -2 \\ -1 & 0 \end{pmatrix}$ and $S = \begin{pmatrix} -2 & 1 \\ 1/2 & 0 \end{pmatrix}$ . The route through the diagonal matrix is the only general way to PROVE similarity in this course.

e) If $B = S^{-1}AS$ , then $B - 3I = S^{-1}(A - 3I)S$ , so $A - 3I$ and $B - 3I$ would have the same rank. Exercise 1 found $\text{rank}(A - 3I) = 1$ and $\text{rank}(B - 3I) = 2$ . So $A$ and $B$ are not similar, although they share their polynomial $(3-\lambda)^2(1-\lambda)$ . Equivalently: a matrix similar to a diagonalizable one is diagonalizable, and $A$ is while $B$ is not. This is the $3 \times 3$ version of c).

Exercise 6: A population with two age classes, and its dominant eigenvalue

A bird population is counted each spring in two classes: juveniles $j_k$ , hatched the previous year, and adults $a_k$ . Each adult produces on average $3$ juveniles that survive to the next count, juveniles do not breed, half of the juveniles survive to become adults, and half of the adults survive another year. With $v_k = (j_k, a_k)$ this reads $v_{k+1} = Av_k$ with $A = \begin{pmatrix} 0 & 3 \\ 0.5 & 0.5 \end{pmatrix}$ .

The count starts at $v_0 = (100, 100)$ . The figure plots the first five counts in the $(j, a)$ plane, in hundreds of birds.

a) Diagonalize $A$ .
b) Write $v_0$ as a combination of the eigenvectors, and deduce a closed formula for $v_k$ .
c) Check the formula against the points $v_1$ and $v_2$ on the figure.
d) Describe the long-term behaviour: the growth rate per year and the eventual ratio of juveniles to adults. Explain why the negative eigenvalue does not matter in the long run.
e) Keep everything except the juvenile survival rate $s$ , the $0.5$ in the bottom left corner of $A$ . For which $s$ does the population neither grow nor shrink in the long run?

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Answers

a) $P = \begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}$ , $D = \text{diag}(1.5, -1)$
b) $v_0 = 80(2,1) + 20(-3,1)$ , so $v_k = 80(1.5)^k(2,1) + 20(-1)^k(-3,1)$
c) $v_1 = (300, 100)$ and $v_2 = (300, 200)$ , as plotted
d) Growth of $50$ per cent a year, juveniles to adults tending to $2 : 1$
e) $s = \frac{1}{6}$

a) $\det(A - \lambda I) = -\lambda(0.5 - \lambda) - 1.5 = \lambda^2 - 0.5\lambda - 1.5$ . Doubling clears the decimals: $2\lambda^2 - \lambda - 3 = (2\lambda - 3)(\lambda + 1)$ , so $\lambda_1 = 1.5$ and $\lambda_2 = -1$ . For $1.5$ : $A - 1.5I = \begin{pmatrix} -1.5 & 3 \\ 0.5 & -1 \end{pmatrix}$ , and $R_1 \to R_1 + 3R_2$ leaves $j = 2a$ , so $u_1 = (2, 1)$ . For $-1$ : $A + I = \begin{pmatrix} 1 & 3 \\ 0.5 & 1.5 \end{pmatrix}$ , and $R_2 \to R_2 - 0.5R_1$ leaves $j = -3a$ , so $u_2 = (-3, 1)$ . Distinct eigenvalues: $P = \begin{pmatrix} 2 & -3 \\ 1 & 1 \end{pmatrix}$ , $D = \text{diag}(1.5, -1)$ , $\det P = 5$ .

b) Solve $c_1(2, 1) + c_2(-3, 1) = (100, 100)$ : the second coordinate gives $c_2 = 100 - c_1$ , and the first $2c_1 - 300 + 3c_1 = 100$ , so $c_1 = 80$ and $c_2 = 20$ . Since $A^ku_i = \lambda_i^ku_i$ , $v_k = A^kv_0 = 80(1.5)^k(2, 1) + 20(-1)^k(-3, 1)$ . This is $A^k = PD^kP^{-1}$ applied to $v_0$ , without ever writing $P^{-1}$ : the coefficients $c_1, c_2$ ARE $P^{-1}v_0$ .

c) $k = 1$ : $(240, 120) + (60, -20) = (300, 100)$ , and directly $Av_0 = (0 + 300, 50 + 50) = (300, 100)$ . $k = 2$ : $80 \times 2.25 = 180$ , so $(360, 180) + (-60, 20) = (300, 200)$ , and directly $Av_1 = (300, 150 + 50) = (300, 200)$ . These are the points $(3, 1)$ and $(3, 2)$ on the figure, in hundreds.

d) Divide by $1.5^k$ : $\frac{v_k}{1.5^k} = 80(2, 1) + 20\left(\frac{-1}{1.5}\right)^k(-3, 1)$ , and the second term tends to $0$ . So in the long run $v_k \approx 80(1.5)^k(2, 1)$ : the population grows by $50$ per cent a year and the juveniles outnumber the adults two to one, whatever the starting count (as long as $c_1 \neq 0$ ). The eigenvalue $-1$ makes the second component flip sign every year, but its size stays $20$ while the first grows without bound, so its relative weight vanishes. On the figure the counts do not converge to a point, they run away; it is their DIRECTION that settles on the line of $u_1$ , the dashed line of the solution figure.

e) With $s$ in the corner, $\det(A - \lambda I) = \lambda^2 - 0.5\lambda - 3s$ . The population is stationary in the long run when the dominant eigenvalue is exactly $1$ : $1 - 0.5 - 3s = 0$ , so $s = \frac{1}{6}$ . Check that $1$ really is dominant: the two roots add up to $0.5$ , so the other one is $-0.5$ , smaller in absolute value. One juvenile in six must survive to adulthood; with one in two the colony grows, with fewer than one in six it dies out.

Exercise 7: A two-state Markov chain and its steady state

A bike-sharing network has two stations, Hilltop ( $H$ ) and Riverside ( $R$ ). Each bike is picked up once a day and returned to one of the two stations. Of the bikes taken at Hilltop, $80$ per cent come back to Hilltop and $20$ per cent end at Riverside; of those taken at Riverside, $30$ per cent end at Hilltop and $70$ per cent stay at Riverside. The figure is the transition diagram.

On the morning of day $0$ all $500$ bikes are at Hilltop. Let $x_k = (h_k, r_k)$ be the numbers of bikes at the two stations on the morning of day $k$ .

a) Write the transition matrix $T$ with $x_{k+1} = Tx_k$ . What do its columns add up to, and why?
b) Show that $1$ is an eigenvalue of $T$ without solving the characteristic equation, then get the other eigenvalue from the trace.
c) Diagonalize $T$ and give $x_k$ as an explicit formula.
d) Give the steady state and the long-run share of each station. Show that it does not depend on how the $500$ bikes were placed on day $0$ .
e) From which day on does Riverside hold at least $190$ bikes?

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Answers

a) $T = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}$ ; each column adds up to $1$
b) $T - I$ has rows adding to zero, so $\lambda = 1$ ; then $\lambda_2 = 1.5 - 1 = 0.5$
c) $P = \begin{pmatrix} 3 & 1 \\ 2 & -1 \end{pmatrix}$ , $D = \text{diag}(1, 0.5)$ , $x_k = (300 + 200(0.5)^k, 200 - 200(0.5)^k)$
d) $(300, 200)$ : $60$ per cent at Hilltop, $40$ per cent at Riverside, for any start with $500$ bikes
e) From day $5$ on

a) Column $j$ says where the bikes that start the day at station $j$ end up: $T = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}$ , so that $h_{k+1} = 0.8h_k + 0.3r_k$ and $r_{k+1} = 0.2h_k + 0.7r_k$ . Each column adds up to $1$ because every bike ends the day SOMEWHERE: no bike is created or lost. Writing the probabilities along the rows instead gives the transpose, and $x_{k+1} = Tx_k$ then produces nonsense; check on day $1$ : $T(500, 0) = (400, 100)$ , which adds up to $500$ as it must.

b) $T - I = \begin{pmatrix} -0.2 & 0.3 \\ 0.2 & -0.3 \end{pmatrix}$ : its second row is minus its first, because each column of $T - I$ adds up to $1 - 1 = 0$ . So $T - I$ is singular and $1$ is an eigenvalue, for EVERY matrix whose columns add up to $1$ . The trace is the sum of the eigenvalues: $0.8 + 0.7 = 1.5 = 1 + \lambda_2$ , so $\lambda_2 = 0.5$ . Check with the determinant: $0.56 - 0.06 = 0.5 = 1 \times 0.5$ .

c) For $1$ : $-0.2h + 0.3r = 0$ , so $u_1 = (3, 2)$ . For $0.5$ : $T - 0.5I = \begin{pmatrix} 0.3 & 0.3 \\ 0.2 & 0.2 \end{pmatrix}$ gives $h + r = 0$ , so $u_2 = (1, -1)$ . Hence $P = \begin{pmatrix} 3 & 1 \\ 2 & -1 \end{pmatrix}$ and $D = \text{diag}(1, 0.5)$ . Decompose $x_0$ : $c_1(3, 2) + c_2(1, -1) = (500, 0)$ gives, by adding the two coordinates, $5c_1 = 500$ , so $c_1 = 100$ and $c_2 = 200$ . Then $x_k = 100(3, 2) + 200(0.5)^k(1, -1) = (300 + 200(0.5)^k, 200 - 200(0.5)^k)$ . Check: $x_1 = (400, 100)$ and $x_2 = (350, 150)$ , while directly $T(400, 100) = (320 + 30, 80 + 70) = (350, 150)$ .

d) $(0.5)^k \to 0$ , so $x_k \to (300, 200)$ : in the long run $60$ per cent of the bikes are at Hilltop and $40$ per cent at Riverside. The steady-state vector is the eigenvector for $1$ SCALED to the total, $(0.6, 0.4)$ as proportions; quoting $(3, 2)$ as the answer is not a distribution. It does not depend on the start: the eigenvector $(1, -1)$ adds up to $0$ , so the total of $x_0$ equals $5c_1$ , and a total of $500$ forces $c_1 = 100$ whatever the split. Only the transient term, which dies out, remembers where the bikes started.

e) $r_k \ge 190$ means $200 - 200(0.5)^k \ge 190$ , that is $(0.5)^k \le 0.05$ , that is $2^k \ge 20$ . Since $2^4 = 16$ and $2^5 = 32$ , this holds from $k = 5$ on. On day $4$ Riverside has $200 - 12.5 = 187.5$ bikes on average, on day $5$ it has $200 - 6.25 = 193.75$ . No calculator and no logarithm: the powers of $2$ do the work.

Exercise 8: Five statements to correct

Each statement below was written by a student revising diagonalization for the MATH 133 final. Each is false as written. Say what is wrong, give the smallest counterexample that settles it, and write the correct statement.

a) "A matrix with a repeated eigenvalue cannot be diagonalized."
b) "A diagonalizable matrix is invertible, and an invertible matrix is diagonalizable."
c) "If $P^{-1}AP = D$ , then $A = P^{-1}DP$ ."
d) "The long-run behaviour of $v_{k+1} = Av_k$ depends only on the largest eigenvalue, never on $v_0$ ."
e) "If $A^2$ is diagonalizable, then so is $A$ ."

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Answers

a) False: $I_2$ has the double eigenvalue $1$ and is diagonal
b) False both ways: $\text{diag}(1, 0)$ is diagonalizable and singular; $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ is invertible and not diagonalizable
c) False: $A = PDP^{-1}$ ; with $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , $D = \text{diag}(1,2)$ the two products differ
d) False: with $A = \text{diag}(2, 0.5)$ and $v_0 = (0, 1)$ , $v_k \to 0$
e) False: $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has $N^2 = 0$ diagonal, but $N$ is not diagonalizable

a) FALSE. The identity $I_2$ has the single eigenvalue $1$ , repeated twice, and it is already diagonal; so is the matrix $A$ of Exercise 1, with its double eigenvalue $3$ . A repeated eigenvalue is where diagonalizability CAN fail, not where it does. Correct statement: $A$ is diagonalizable if and only if, for each eigenvalue, $\dim E_\lambda$ equals its algebraic multiplicity; only the repeated eigenvalues need that test.

b) FALSE in both directions, because the two properties answer different questions. $\text{diag}(1, 0)$ is diagonal, so diagonalizable, and singular. $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has determinant $1$ , so it is invertible, and its double eigenvalue $1$ has $\text{rank}(A - I) = 1$ , so $\dim E_1 = 1 < 2$ : not diagonalizable. Correct: invertibility asks whether $0$ is an eigenvalue; diagonalizability asks whether there are enough eigenvectors. Neither implies the other.

c) FALSE: the $P$ and the $P^{-1}$ have changed sides. Multiply $P^{-1}AP = D$ on the left by $P$ and on the right by $P^{-1}$ : $A = PDP^{-1}$ . With $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ and $D = \text{diag}(1, 2)$ : $PDP^{-1} = \begin{pmatrix} 1 & 1 \\ 0 & 2 \end{pmatrix}$ while $P^{-1}DP = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}$ . They differ, and only the first one has $(1,1)$ , the second column of $P$ , as an eigenvector for $2$ . Correct: $A = PDP^{-1}$ , with the eigenvector matrix on the LEFT.

d) FALSE. Writing $v_0 = c_1u_1 + c_2u_2$ , the dominant eigenvalue wins only if $c_1 \neq 0$ . With $A = \text{diag}(2, 0.5)$ and $v_0 = (0, 1)$ , $c_1 = 0$ and $v_k = (0, 0.5^k) \to 0$ , although the largest eigenvalue is $2$ . Correct: if $|\lambda_1|$ is strictly the largest and the component of $v_0$ along $u_1$ is non-zero, then $v_k$ behaves like $c_1\lambda_1^ku_1$ . In real models the condition $c_1 \neq 0$ is almost always true, which is why it is forgotten.

e) FALSE. $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ has $N^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ , which is diagonal. But $N$ has the double eigenvalue $0$ and $\text{rank}(N - 0I) = 1$ , so $\dim E_0 = 1 < 2$ : not diagonalizable. The TRUE direction is the other one: if $A = PDP^{-1}$ then $A^2 = PD^2P^{-1}$ with the same $P$ , so every power of a diagonalizable matrix is diagonalizable.

Exercise 9: Counting pavings: a recurrence solved by diagonalization

A garden path $2$ units wide and $n$ units long is paved with two kinds of slabs: $1 \times 2$ rectangles, laid either across the path or along it, and $2 \times 2$ squares. Let $t_n$ be the number of different pavings, with $t_0 = 1$ (the empty path) and $t_1 = 1$ (one slab across).

The figure shows the three pavings of a path of length $2$ . The recurrence behind this count is a cousin of Fibonacci's, and diagonalization turns it into an exact formula.

a) Explain why $t_n = t_{n-1} + 2t_{n-2}$ for $n \ge 2$ , and compute $t_2$ to $t_5$ .
b) With $V_n = (t_n, t_{n+1})$ , find the matrix $A$ such that $V_{n+1} = AV_n$ .
c) Diagonalize $A$ .
d) Deduce that $t_n = \frac{2^{n+1} + (-1)^n}{3}$ , and check it for $n = 0$ to $5$ .
e) Compute $t_{10}$ , and explain why $\frac{t_{n+1}}{t_n}$ tends to $2$ .

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Answers

a) Last slab across, or a $2 \times 2$ end block in two ways; $t_2 = 3$ , $t_3 = 5$ , $t_4 = 11$ , $t_5 = 21$
b) $A = \begin{pmatrix} 0 & 1 \\ 2 & 1 \end{pmatrix}$
c) $P = \begin{pmatrix} 1 & 1 \\ 2 & -1 \end{pmatrix}$ , $D = \text{diag}(2, -1)$
d) $V_0 = \frac{2}{3}(1,2) + \frac{1}{3}(1,-1)$ , first coordinate $t_n = \frac{2^{n+1} + (-1)^n}{3}$
e) $t_{10} = 683$ ; the ratio tends to the dominant eigenvalue $2$

a) Look at the right end of the path. Either the last column is ONE slab laid across: what is left is a path of length $n - 1$ , paved in $t_{n-1}$ ways. Or it is not: then the last column is covered by slabs that also cover the column before it, so the last $2 \times 2$ block is filled either by two slabs along the path or by one square, two ways, and what is left has length $n - 2$ . The cases are disjoint and cover everything, so $t_n = t_{n-1} + 2t_{n-2}$ . Then $t_2 = 1 + 2 = 3$ , the three pictures of the figure, $t_3 = 3 + 2 = 5$ , $t_4 = 5 + 6 = 11$ , $t_5 = 11 + 10 = 21$ .

b) $V_{n+1} = (t_{n+1}, t_{n+2}) = (t_{n+1}, 2t_n + t_{n+1})$ , so the first new coordinate copies the old second one, and $A = \begin{pmatrix} 0 & 1 \\ 2 & 1 \end{pmatrix}$ . Turning a recurrence of order two into a first-order system on a PAIR of consecutive terms is the whole trick of Nicholson 3.4.

c) $\det(A - \lambda I) = -\lambda(1 - \lambda) - 2 = \lambda^2 - \lambda - 2 = (\lambda - 2)(\lambda + 1)$ . For $2$ : $A - 2I = \begin{pmatrix} -2 & 1 \\ 2 & -1 \end{pmatrix}$ , and $R_2 \to R_2 + R_1$ leaves $y = 2x$ , so $(1, 2)$ . For $-1$ : $A + I = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$ , and $R_2 \to R_2 - 2R_1$ leaves $x + y = 0$ , so $(1, -1)$ . Hence $P = \begin{pmatrix} 1 & 1 \\ 2 & -1 \end{pmatrix}$ , $D = \text{diag}(2, -1)$ , $\det P = -3$ .

d) $V_0 = (1, 1) = c_1(1, 2) + c_2(1, -1)$ : adding the two equations $c_1 + c_2 = 1$ and $2c_1 - c_2 = 1$ gives $3c_1 = 2$ , so $c_1 = \frac{2}{3}$ , $c_2 = \frac{1}{3}$ . Then $V_n = A^nV_0 = \frac{2}{3}2^n(1, 2) + \frac{1}{3}(-1)^n(1, -1)$ , whose first coordinate is $t_n = \frac{2^{n+1} + (-1)^n}{3}$ . Check: $\frac{2+1}{3} = 1$ , $\frac{4-1}{3} = 1$ , $\frac{8+1}{3} = 3$ , $\frac{16-1}{3} = 5$ , $\frac{32+1}{3} = 11$ , $\frac{64-1}{3} = 21$ . A formula for an integer sequence with a $3$ in the denominator is believable only after this check: the numerator must always be a multiple of $3$ , and it is.

e) $t_{10} = \frac{2^{11} + 1}{3} = \frac{2049}{3} = 683$ , a number nobody would enumerate by hand. In the ratio $\frac{t_{n+1}}{t_n} = \frac{2^{n+2} - (-1)^n}{2^{n+1} + (-1)^n}$ , the terms $\pm 1$ stay bounded while the powers of $2$ explode, so the ratio tends to $2$ : the DOMINANT eigenvalue. The solution figure shows it: the points hug the curve $\frac{2}{3}2^n$ , alternately $\frac{1}{3}$ above and below it, and that alternating gap is the eigenvalue $-1$ at work.

Exercise 10: Foxes and hares: where the dominant eigenvalue crosses 1

In a predator and prey model, $F_k$ is the number of foxes and $H_k$ the number of hares, in hundreds, in year $k$ . Without hares the foxes would halve every year; without foxes the hares would grow by $10$ per cent a year. With $x_k = (F_k, H_k)$ the model is $x_{k+1} = Ax_k$ , where $A = \begin{pmatrix} 0.5 & 0.4 \\ -p & 1.1 \end{pmatrix}$ and $p > 0$ measures how many hares a fox eats.

Every characteristic polynomial below factors by hand. This is a final-exam question in its usual shape: diagonalize, then read the long run off the eigenvalues.

a) Take $p = 0.2$ . Diagonalize $A$ and describe what happens to both populations in the long run.
b) Find the value of $p$ for which $1$ is an eigenvalue of $A$ .
c) Take $p = 0.125$ and $x_0 = (40, 30)$ . Diagonalize $A$ and write $x_k$ explicitly.
d) Give the limit of $x_k$ and interpret it. Would another starting point lead to the same limit?
e) Without computing any eigenvector, explain what happens when $p$ is slightly below $0.125$ .

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Answers

a) $\lambda = 0.9$ , $(1,1)$ and $\lambda = 0.7$ , $(2,1)$ : both populations die out, about $100$ hares per fox
b) $p = 0.125$
c) $P = \begin{pmatrix} 4 & 4 \\ 5 & 1 \end{pmatrix}$ , $D = \text{diag}(1, 0.6)$ , $x_k = (20 + 20(0.6)^k, 25 + 5(0.6)^k)$
d) $(20, 25)$ : $20$ foxes and $2500$ hares; another start gives another point of the line $H = 1.25F$
e) The dominant eigenvalue exceeds $1$ : both populations eventually grow

a) $\det(A - \lambda I) = (0.5 - \lambda)(1.1 - \lambda) + 0.4p = \lambda^2 - 1.6\lambda + 0.55 + 0.4p$ . With $p = 0.2$ : $\lambda^2 - 1.6\lambda + 0.63 = (\lambda - 0.9)(\lambda - 0.7)$ , since $0.9 + 0.7 = 1.6$ and $0.9 \times 0.7 = 0.63$ . For $0.9$ : $A - 0.9I = \begin{pmatrix} -0.4 & 0.4 \\ -0.2 & 0.2 \end{pmatrix}$ gives $F = H$ , so $(1, 1)$ . For $0.7$ : $A - 0.7I = \begin{pmatrix} -0.2 & 0.4 \\ -0.2 & 0.4 \end{pmatrix}$ gives $F = 2H$ , so $(2, 1)$ . $P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$ , $D = \text{diag}(0.9, 0.7)$ . Then $x_k = c_10.9^k(1, 1) + c_20.7^k(2, 1) \to (0, 0)$ : both species die out. The slower factor $0.9$ dominates, so the decline settles at $10$ per cent a year with $F_k \approx H_k$ , one fox per hundred hares, all the way down.

b) $1$ is an eigenvalue exactly when $A - I$ is singular: $\det(A - I) = (-0.5)(0.1) + 0.4p = 0$ , so $0.4p = 0.05$ and $p = 0.125$ . No need to solve the quadratic: the definition of an eigenvalue IS the singularity test.

c) $p = 0.125$ : $\lambda^2 - 1.6\lambda + 0.6 = (\lambda - 1)(\lambda - 0.6)$ . For $1$ : $A - I = \begin{pmatrix} -0.5 & 0.4 \\ -0.125 & 0.1 \end{pmatrix}$ gives $5F = 4H$ , so $(4, 5)$ . For $0.6$ : $A - 0.6I = \begin{pmatrix} -0.1 & 0.4 \\ -0.125 & 0.5 \end{pmatrix}$ gives $F = 4H$ , so $(4, 1)$ . $P = \begin{pmatrix} 4 & 4 \\ 5 & 1 \end{pmatrix}$ , $D = \text{diag}(1, 0.6)$ . Decompose $x_0$ : $c_1(4, 5) + c_2(4, 1) = (40, 30)$ ; the first coordinate gives $c_1 + c_2 = 10$ , the second $5c_1 + c_2 = 30$ , so $c_1 = 5$ and $c_2 = 5$ . Hence $x_k = 5(4, 5) + 5(0.6)^k(4, 1) = (20 + 20(0.6)^k, 25 + 5(0.6)^k)$ . Check $k = 1$ : the formula gives $(32, 28)$ and directly $(0.5 \times 40 + 0.4 \times 30, -0.125 \times 40 + 1.1 \times 30) = (32, 28)$ .

d) $(0.6)^k \to 0$ , so $x_k \to (20, 25)$ : $20$ foxes and $2500$ hares, a stable coexistence. The solution figure shows the approach: the points slide toward the limit along the direction $(4, 1)$ of the eigenvalue $0.6$ , while the dashed line $H = 1.25F$ is the line of eigenvalue $1$ , every point of which is a fixed point. So another starting point gives another limit, $c_1(4, 5)$ , with the SAME ratio of $125$ hares per fox. Compare with a Markov chain, where the total is conserved and fixes $c_1$ ; nothing is conserved here.

e) Write $q(\lambda) = \lambda^2 - 1.6\lambda + 0.55 + 0.4p$ . Then $q(1) = -0.05 + 0.4p$ , which is NEGATIVE when $p < 0.125$ . The parabola $q$ opens upward and is positive for large $\lambda$ , so it crosses zero somewhere beyond $1$ : the dominant eigenvalue is greater than $1$ . Then $c_1\lambda_1^k$ grows, and both populations eventually increase together in the proportions of the dominant eigenvector. The value $p = 0.125$ is the knife edge between extinction and growth, and the whole analysis came from the sign of one number.

Corrected exercises: diagonalization and its applications (MATH 133)

Course recap

Part A: the basics (/50)

Exercise 1: One sign apart: a double eigenvalue decided by its eigenspace

Exercise 2: A parameter decides, and it is not the one you expect

Exercise 3: Building P and D, and the order that ties them

Exercise 4: Powers of a matrix, read in the eigenvector axes

Exercise 5: Similar matrices: what they share, and what sharing does not prove

Part B: problems and reasoning (/50)

Exercise 6: A population with two age classes, and its dominant eigenvalue

Exercise 7: A two-state Markov chain and its steady state

Exercise 8: Five statements to correct

Exercise 9: Counting pavings: a recurrence solved by diagonalization

Exercise 10: Foxes and hares: where the dominant eigenvalue crosses 1

See also

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