Corrected exercises: diagonalization and its applications (MATH 133)

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 $\det A = 3 \times (4 - 1) = 9 = 3 \times 3 \times 1$.

$A - 3I = \begin{pmatrix} 0 & 0 & 0 \\ 1 & -1 & 1 \\ -1 & 1 & -1 \end{pmatrix}$. 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.

$B - 3I = \begin{pmatrix} 0 & 0 & 0 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{pmatrix}$. 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.

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.

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.

$A - \lambda I$ 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.

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.

$E_2$: 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 $\det P = 3 \neq 0$.

$C$ 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 $k \neq 1$.

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 $n$.

$\det(A - \lambda I) = (1-\lambda)(2-\lambda) - 6 = \lambda^2 - 3\lambda - 4 = (\lambda - 4)(\lambda + 1)$. 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 $P$.

$AP$ 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.

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.

$Q = \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}$, $\det Q = -5$, $Q^{-1} = \frac{1}{5}\begin{pmatrix} 1 & 3 \\ 1 & -2 \end{pmatrix}$. 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 $AP$.

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.

$Av_1 = (4 - 2, 1 + 1) = (2, 2) = 2v_1$ 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: $PP^{-1} = \begin{pmatrix} -1 + 2 & 2 - 2 \\ -1 + 1 & 2 - 1 \end{pmatrix} = I$.

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 $A$.

$PD^n = \begin{pmatrix} 2^n & 2 \cdot 3^n \\ 2^n & 3^n \end{pmatrix}$, 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.

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.

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.