Corrected exercises: eigenvalues and eigenvectors (MATH 133)

a) $Au = (1 \cdot 1 + 2 \cdot (-1),\ 3 \cdot 1 + 2 \cdot (-1)) = (-1, 1)$. Compare with $u$: $(-1, 1) = -1 \cdot (1, -1)$, so $Au = -u$. The vector $u$ is an eigenvector and its eigenvalue is $\lambda = -1$. Now $Av = (1, 3)$, the first column of $A$. For $v$ to be an eigenvector we would need $(1, 3) = \lambda (1, 0) = (\lambda, 0)$, and the second entry $3 \neq 0$ makes that impossible: $v$ is NOT an eigenvector. The method in both cases is the same two steps, compute $Ax$, then ask whether it is a multiple of $x$ by comparing entry by entry. Writing $\lambda$ down before checking that the SAME number works in every entry is the classic slip: $(4, 6)$ is not a multiple of $(1, 3)$ even though $4/1$ looks like a candidate.

b) $Aw = (2 + 6,\ 6 + 6) = (8, 12) = 4(2, 3)$, so $w$ is an eigenvector with $\lambda = 4$. Then $A(-3, 3) = (-3 + 6,\ -9 + 6) = (3, -3) = -1 \cdot (-3, 3)$: an eigenvector again, with $\lambda = -1$. That is no coincidence: $(-3, 3) = -3u$, and $A(cu) = cAu = c(\lambda u) = \lambda (cu)$. Every non-zero multiple of an eigenvector is an eigenvector for the SAME eigenvalue. An eigenvector is really a whole line of them, which is why the next exercises ask for a basis of the eigenspace rather than for one vector.

c) If $x = 0$ were allowed, then $A0 = 0 = \lambda 0$ would hold for EVERY real $\lambda$, and every number would be an eigenvalue of every matrix. The notion would carry no information, so the definition requires $x \neq 0$. The eigenvalue itself has no such restriction: $\lambda = 0$ is perfectly legal and says that some non-zero $x$ satisfies $Ax = 0$, that is the homogeneous system has a non-trivial solution, that is $A$ is NOT invertible. Two different zeros, two different statements: the zero VECTOR is never an eigenvector, the zero NUMBER is an eigenvalue exactly when $A$ is singular.

d) $\lambda$ is an eigenvalue exactly when $(\lambda I - A)x = 0$ has a non-zero solution, that is when $\lambda I - A$ is singular. Here $3I - A = \begin{pmatrix} 2 & -2 \\ -3 & 1 \end{pmatrix}$, whose determinant is $2 \cdot 1 - (-2)(-3) = 2 - 6 = -4 \neq 0$. The matrix is invertible, the only solution is $x = 0$, and $3$ is NOT an eigenvalue. Next $4I - A = \begin{pmatrix} 3 & -2 \\ -3 & 2 \end{pmatrix}$, with determinant $6 - 6 = 0$: singular, so $4$ IS an eigenvalue, which part b) had already shown with $w$. Testing one candidate costs one $2 \times 2$ determinant, far less than the full polynomial.

e) On the left of the figure, $Au$ points exactly OPPOSITE to $u$, yet $u$ is an eigenvector: the eigenvalue $-1$ reverses it. So keeps its direction is false as soon as an eigenvalue is negative, and students who believe it throw away half of the eigenvectors of reflections. On the right, $Av$ leaves the dashed line of $v$ altogether: that is what failing looks like. The correct sentence: $x \neq 0$ is an eigenvector of $A$ when $Ax$ lies on the LINE through the origin spanned by $x$; the eigenvalue says whether it is stretched ($\lambda > 1$), shrunk ($0 < \lambda < 1$), reversed ($\lambda < 0$) or sent to $0$ ($\lambda = 0$).

a) $xI - A = \begin{pmatrix} x - 1 & -2 \\ 2 & x - 6 \end{pmatrix}$. Every entry of $A$ changes sign except on the diagonal, where $x$ is added: the off-diagonal $2$ becomes $-2$ and the $-2$ becomes $2$. Then $c_A(x) = (x - 1)(x - 6) - (-2)(2) = x^2 - 7x + 6 + 4 = x^2 - 7x + 10 = (x - 2)(x - 5)$. The eigenvalues are $2$ and $5$. The sign of the product of the off-diagonal entries is where marks are lost: $(-2)(2) = -4$ is SUBTRACTED, which adds $4$. Writing $x^2 - 7x + 2$ gives a polynomial with no rational root, and in a course without calculators that alone is a signal that something is wrong.

b) For $\lambda = 2$: $2I - A = \begin{pmatrix} 1 & -2 \\ 2 & -4 \end{pmatrix}$. $R_2 \to R_2 - 2R_1$ gives $\begin{pmatrix} 1 & -2 \\ 0 & 0 \end{pmatrix}$. One pivot, one free variable: $x_2 = t$, $x_1 = 2t$, so $x = t(2, 1)$ and the basic eigenvector is $(2, 1)$. For $\lambda = 5$: $5I - A = \begin{pmatrix} 4 & -2 \\ 2 & -1 \end{pmatrix}$. $R_1 \leftrightarrow R_2$, then $R_2 \to R_2 - 2R_1$ gives $\begin{pmatrix} 2 & -1 \\ 0 & 0 \end{pmatrix}$, so $2x_1 = x_2$; with $x_1 = t$, $x = t(1, 2)$ and the basic eigenvector is $(1, 2)$. The zero row is not a nuisance, it is the GUARANTEE: $\lambda I - A$ is singular by construction, so its reduction must leave at least one free variable. The eigenspaces are $E_2 = \text{span}\{(2, 1)\}$ and $E_5 = \text{span}\{(1, 2)\}$, two lines through the origin, drawn on the figure of the solution.

c) $A(2, 1) = (1 \cdot 2 + 2 \cdot 1,\ -2 \cdot 2 + 6 \cdot 1) = (4, 2) = 2(2, 1)$, and $A(1, 2) = (1 + 4,\ -2 + 12) = (5, 10) = 5(1, 2)$. Both hold. This check costs ten seconds and catches both kinds of error at once, a wrong eigenvalue and a wrong vector. A correct eigenvalue with a vector from the wrong null space fails it immediately, which is exactly what happens when a student solves $(\lambda I - A)x = 0$ with the rows of $A$ instead of those of $\lambda I - A$.

d) If $4I - A$ reduces to $I$, then $(4I - A)x = 0$ has only the solution $x = 0$, and $4$ is NOT an eigenvalue. Indeed $4I - A = \begin{pmatrix} 3 & -2 \\ 2 & -2 \end{pmatrix}$ has determinant $-6 + 4 = -2$, and $c_A(4) = 16 - 28 + 10 = -2$: the same number, as it must be, since $c_A(4)$ IS $\det(4I - A)$. What she must NOT do is declare the eigenspace to be $\{0\}$ or report $x = 0$ as the eigenvector: that answer is always wrong. The reduction is telling her the $\lambda$ is wrong, so she goes back to the polynomial, finds the arithmetic slip, and starts again with $2$ and $5$.

e) $A - xI = -(xI - A)$, and multiplying an $n \times n$ matrix by $-1$ multiplies its determinant by $(-1)^n$. So $\det(A - xI) = (-1)^n c_A(x)$. For this $2 \times 2$ the two polynomials are equal. For a $3 \times 3$ they are opposite: $\det(A - xI) = -c_A(x)$, with leading term $-x^3$. The ROOTS never change, and neither do the eigenspaces, since $(\lambda I - A)x = 0$ and $(A - \lambda I)x = 0$ have the same solutions. What does change is any statement read off the coefficients, such as $c_A(0)$: keep one convention for the whole computation, and in MATH 133 that convention is $\det(xI - A)$, monic, leading coefficient $1$.

a) The second row of $xI - A$ is $(0,\ x - 5,\ 0)$: two zeros, so only one cofactor survives. The entry sits in position $(2, 2)$ and its sign is $(-1)^{2+2} = +1$. Deleting row 2 and column 2 leaves $\begin{pmatrix} x - 4 & 2 \\ -1 & x - 1 \end{pmatrix}$, so $c_A(x) = (x - 5)\left[(x - 4)(x - 1) - (2)(-1)\right] = (x - 5)(x^2 - 5x + 4 + 2) = (x - 5)(x^2 - 5x + 6)$. The quadratic factors as $(x - 2)(x - 3)$, since $2 + 3 = 5$ and $2 \cdot 3 = 6$. Final answer: $c_A(x) = (x - 2)(x - 3)(x - 5)$. The factor $(x - 5)$ came out without ever being looked for: that is what the choice of row buys, and it is the method the marker expects to see named.

b) The roots of $c_A$: $\lambda = 2$, $\lambda = 3$ and $\lambda = 5$, three distinct real eigenvalues.

c) For $\lambda = 2$: $2I - A = \begin{pmatrix} -2 & -1 & 2 \\ 0 & -3 & 0 \\ -1 & -1 & 1 \end{pmatrix}$. $R_1 \leftrightarrow R_3$, then $R_3 \to R_3 - 2R_1$ turns the new third row into $(0, 1, 0)$; $R_2 \to -\frac{1}{3}R_2$ gives $(0, 1, 0)$, $R_3 \to R_3 - R_2$ gives a zero row, $R_1 \to -R_1$ then $R_1 \to R_1 - R_2$ gives $(1, 0, -1)$. So $x_2 = 0$, $x_1 = x_3 = t$: $E_2 = \text{span}\{(1, 0, 1)\}$. For $\lambda = 3$: $3I - A = \begin{pmatrix} -1 & -1 & 2 \\ 0 & -2 & 0 \\ -1 & -1 & 2 \end{pmatrix}$. $R_3 \to R_3 - R_1$ gives a zero row, $R_2 \to -\frac{1}{2}R_2$ gives $(0, 1, 0)$, $R_1 \to -R_1$ then $R_1 \to R_1 - R_2$ gives $(1, 0, -2)$. So $x_2 = 0$, $x_1 = 2x_3$: $E_3 = \text{span}\{(2, 0, 1)\}$. For $\lambda = 5$: $5I - A = \begin{pmatrix} 1 & -1 & 2 \\ 0 & 0 & 0 \\ -1 & -1 & 4 \end{pmatrix}$. $R_3 \to R_3 + R_1$ gives $(0, -2, 6)$, $R_2 \leftrightarrow R_3$, $R_2 \to -\frac{1}{2}R_2$ gives $(0, 1, -3)$, $R_1 \to R_1 + R_2$ gives $(1, 0, -1)$. So $x_2 = 3x_3$, $x_1 = x_3$: $E_5 = \text{span}\{(1, 3, 1)\}$. Each reduction left exactly one free variable, as it must for a singular $\lambda I - A$.

d) $A(1, 0, 1) = (4 - 2,\ 0,\ 1 + 1) = (2, 0, 2)$, twice the vector. $A(2, 0, 1) = (8 - 2,\ 0,\ 2 + 1) = (6, 0, 3)$, three times. $A(1, 3, 1) = (4 + 3 - 2,\ 15,\ 1 + 3 + 1) = (5, 15, 5)$, five times. The trace of $A$ is $4 + 5 + 1 = 10$ and $2 + 3 + 5 = 10$: the sum of the eigenvalues matches, a check that takes three seconds and catches most sign slips in the polynomial.

e) Expanding the factored form: $(x - 5)(x^2 - 5x + 6) = x^3 - 5x^2 + 6x - 5x^2 + 25x - 30 = x^3 - 10x^2 + 31x - 30$, the same polynomial. His route is correct but expensive twice over: three $2 \times 2$ minors instead of one, and then a cubic whose roots must be found by testing the divisors of $30$ ($c(1) = -8$, $c(2) = 0$, then a division) before the quadratic appears. On an exam without a calculator this is where time runs out and sign errors multiply. The rule: before expanding $xI - A$, scan it for the row or column with the most zeros, and expand along that one.

a) A reflection leaves every point of its mirror where it is: any vector along $L$ is sent to itself, so it is an eigenvector with eigenvalue $1$. The line $y = \frac{x}{2}$ is spanned by $d = (2, 1)$. A vector PERPENDICULAR to the mirror is sent to its opposite: eigenvalue $-1$, with eigenvector $n = (1, -2)$, since $d \cdot n = 2 - 2 = 0$. Every other vector, like $v$ on the figure, is sent to its mirror image, which is on another line: not an eigenvector. So the eigenvalues are $1$ and $-1$, and a $2 \times 2$ has at most two, so the list is complete.

b) $Qd = \frac{1}{5}(3 \cdot 2 + 4 \cdot 1,\ 4 \cdot 2 - 3 \cdot 1) = \frac{1}{5}(10, 5) = (2, 1) = d$. $Qn = \frac{1}{5}(3 - 8,\ 4 + 6) = \frac{1}{5}(-5, 10) = (-1, 2) = -n$. Both confirmed. The figure of the solution draws them: $d$ untouched, $n$ reversed on its own line (drawn there at half length). Notice that the reversed vector is a full eigenvector: the eigenvalue $-1$ is exactly the case the sentence keeps its direction would throw away.

c) Projecting onto $L$ keeps every vector of $L$: eigenvalue $1$, $E_1(P) = \text{span}\{(2, 1)\}$. It crushes every vector perpendicular to $L$ to the zero vector: $Pn = 0 = 0 \cdot n$, eigenvalue $0$, $E_0(P) = \text{span}\{(1, -2)\}$. Check: $P(2, 1) = \frac{1}{5}(8 + 2,\ 4 + 1) = (2, 1)$ and $P(1, -2) = \frac{1}{5}(4 - 4,\ 2 - 2) = (0, 0)$. Since $0$ is an eigenvalue, $Px = 0$ has a non-zero solution, so $P$ is NOT invertible: geometrically, two different vectors such as $v$ and $v + n$ have the same shadow, and no matrix can undo that. Note that the eigenvector for $\lambda = 0$ is $n$, a non-zero vector; the zero vector $Pn$ is its image, not the eigenvector.

d) A rotation by $90^\circ$ turns every non-zero vector a quarter turn: $a$ and $Ra$, $b$ and $Rb$ are perpendicular on the figure, so $Rx$ is never on the line of $x$, and no real $\lambda$ can satisfy $Rx = \lambda x$ with $x \neq 0$. Algebraically, $xI - R = \begin{pmatrix} x & 1 \\ -1 & x \end{pmatrix}$, so $c_R(x) = x \cdot x - (1)(-1) = x^2 + 1$. For every real $x$, $x^2 + 1 \ge 1 > 0$: no real root, no real eigenvalue, no eigenvector in $\mathbb{R}^2$. In MATH 133 this is the complete answer. The polynomial does have roots in the complex numbers, but eigenvalues here are real numbers attached to vectors of $\mathbb{R}^n$, and computing with complex ones is outside the course.

e) For $Q$: trace $\frac{3 - 3}{5} = 0$ and $1 + (-1) = 0$; determinant $\frac{3 \cdot (-3) - 4 \cdot 4}{25} = \frac{-25}{25} = -1$ and $1 \cdot (-1) = -1$. For $P$: trace $\frac{4 + 1}{5} = 1$ and $1 + 0 = 1$; determinant $\frac{4 \cdot 1 - 2 \cdot 2}{25} = 0$ and $1 \cdot 0 = 0$. The determinant $-1$ of the reflection is its reversal of orientation, and the determinant $0$ of the projection is its collapse onto a line: the two numbers that the geometry announces are exactly the products of the eigenvalues that the geometry showed.

a) $xI - A = \begin{pmatrix} xI - B & -C \\ 0 & xI - D \end{pmatrix}$ is block upper triangular, with the zero block below the diagonal, so its determinant is the product of the determinants of the diagonal blocks: $c_A(x) = \det(xI - B)\det(xI - D) = c_B(x)\,c_D(x)$. The block $C$ plays no part. Now $c_B(x) = \det\begin{pmatrix} x - 4 & 3 \\ -1 & x \end{pmatrix} = x(x - 4) + 3 = x^2 - 4x + 3 = (x - 1)(x - 3)$ and $c_D(x) = \det\begin{pmatrix} x - 4 & -2 \\ -1 & x - 3 \end{pmatrix} = (x - 4)(x - 3) - 2 = x^2 - 7x + 10 = (x - 2)(x - 5)$. So $c_A(x) = (x - 1)(x - 3)(x - 2)(x - 5)$ and the eigenvalues are $1$, $3$, $2$ and $5$.

b) The block product gives $A(u, w) = (Bu + Cw,\ Dw)$. With $w = 0$: $A(u, 0) = (Bu, 0)$. So if $Bu = \lambda u$ with $u \neq 0$, then $A(u, 0) = (\lambda u, 0) = \lambda(u, 0)$, and $(u, 0)$ is an eigenvector of $A$. For $\lambda = 1$: $I - B = \begin{pmatrix} -3 & 3 \\ -1 & 1 \end{pmatrix}$, $R_1 \to -\frac{1}{3}R_1$ then $R_2 \to R_2 + R_1$ leaves $u_1 = u_2$, so $u = (1, 1)$ and $(1, 1, 0, 0)$ is an eigenvector. For $\lambda = 3$: $3I - B = \begin{pmatrix} -1 & 3 \\ -1 & 3 \end{pmatrix}$, $R_2 \to R_2 - R_1$ leaves $u_1 = 3u_2$, so $(3, 1, 0, 0)$. Checks: $A(1, 1, 0, 0) = (4 - 3,\ 1,\ 0,\ 0) = (1, 1, 0, 0)$ and $A(3, 1, 0, 0) = (12 - 3,\ 3,\ 0,\ 0) = (9, 3, 0, 0)$. Padding works here because the zero block kills everything below.

c) $x = (u, w)$ is an eigenvector for $5$ when $Bu + Cw = 5u$ and $Dw = 5w$. If $w$ were $0$, the first equation would say $Bu = 5u$ with $u \neq 0$, impossible since $5$ is not an eigenvalue of $B$. So $w \neq 0$ and $Dw = 5w$: $w$ is an eigenvector of $D$. From $5I - D = \begin{pmatrix} 1 & -2 \\ -1 & 2 \end{pmatrix}$ and $R_2 \to R_2 + R_1$: $w_1 = 2w_2$, so $w = t(2, 1)$. Take $w = (2, 1)$. Then $(5I - B)u = Cw = (2 - 1,\ 0 - 1) = (1, -1)$, with $5I - B = \begin{pmatrix} 1 & 3 \\ -1 & 5 \end{pmatrix}$. $R_2 \to R_2 + R_1$ gives $8u_2 = 0$, so $u_2 = 0$ and $u_1 = 1$. Hence $E_5(A) = \text{span}\{(1, 0, 2, 1)\}$. Check: $A(1, 0, 2, 1) = (4 + 2 - 1,\ 1 - 1,\ 8 + 2,\ 2 + 3) = (5, 0, 10, 5)$, five times the vector.