A real matrix can have a characteristic polynomial with no real roots, and then it has no real eigenvectors at all. The roots are still there; they're complex. This page reads them. A pair of complex eigenvalues $a \pm bi$ is a turn and a stretch hiding inside the matrix: $|a + bi|$ says how much it stretches, the angle of $a + bi$ says how far it turns, and the real and imaginary parts of an eigenvector say in which slanted grid. The scalars stay real throughout. Complex numbers only show up as roots and as eigenvector entries.
Some matrices turn every arrow, so no arrow stays on its own line and there are no eigenvectors to find. The characteristic polynomial still has roots; they just aren't on the real number line. They're complex numbers like $1 + i$. This page shows that those complex roots aren't nonsense: the point $1 + i$ sits at a $45^\circ$ angle, at distance $\sqrt{2}$ from $0$, and the matrix turns every arrow by $45^\circ$ and stretches it by $\sqrt{2}$. The complex eigenvalue is a picture of the turn.
The quarter turn $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ had characteristic polynomial $t^2 + 1$ and no real eigenvalues. Its roots are $\pm i$, and $i$ is exactly a quarter of the way around the unit circle.
You've met complex numbers $a + bi$, with $i^2 = -1$, in algebra. Picture $a + bi$ as the point $(a, b)$ in a plane, the complex plane: real part across, imaginary part up. Two more words come with them.
For a complex number $z = a + bi$, its complex conjugate is $\overline{z} = a - bi$, its mirror image across the real axis. Its modulus is $|z| = \sqrt{a^2 + b^2}$, its distance from $0$. For a vector $v$ with complex entries, $\overline{v}$ conjugates every entry.
The modulus is the Pythagorean theorem on the point $(a, b)$, and multiplying a number by its conjugate gives the modulus squared: $(a + bi)(a - bi) = a^2 - b^2i^2 = a^2 + b^2$. Now a matrix whose characteristic polynomial won't factor over the real numbers.
Find the eigenvalues of $A = \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix}$.
The trace is $4$ and the determinant is $3 + 2 = 5$, so the characteristic polynomial is $f(t) = t^2 - 4t + 5$. The quadratic formula gives
$$t = \frac{4 \pm \sqrt{16 - 20}}{2} = \frac{4 \pm \sqrt{-4}}{2} = \frac{4 \pm 2i}{2} = 2 \pm i.$$The eigenvalues are $2 + i$ and $2 - i$: a conjugate pair. (Check: they add up to $4$, the trace, and multiply to $4 + 1 = 5$, the determinant.)
In the picture, every arrow is knocked off its line, and the two roots sit above and below the real axis like reflections.
The $\pm$ in the quadratic formula is why they came as a pair. Watch what happens to the two roots as a matrix's characteristic polynomial rises off the axis.
That pairing isn't special to $2 \times 2$ matrices.
Let $A$ be a real $n \times n$ matrix. If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, then $\overline{\lambda}$ is an eigenvalue of $A$ with eigenvector $\overline{v}$. So the eigenvalues that aren't real come in conjugate pairs $a \pm bi$.
Conjugation respects sums and products: $\overline{z + w} = \overline{z} + \overline{w}$ and $\overline{zw} = \overline{z}\,\overline{w}$, and a real number is its own conjugate. Conjugate every entry of $Av = \lambda v$. Each entry of $Av$ is a sum of products $A_{ij}v_j$ with $A_{ij}$ real, so $\overline{Av} = A\overline{v}$, and $\overline{\lambda v} = \overline{\lambda}\,\overline{v}$. So $A\overline{v} = \overline{\lambda}\,\overline{v}$, and $\overline{v} \neq 0$ because $v \neq 0$. $\blacksquare$
A real $2 \times 2$ matrix has either two real eigenvalues (maybe equal) or a pair $a \pm bi$ with $b \neq 0$. A real $3 \times 3$ matrix has either three real eigenvalues or one real eigenvalue and a pair $a \pm bi$.
This is a visual argument, not a formal proof. For a $2 \times 2$ the roots are $\frac{\tr(A) \pm \sqrt{\text{discriminant}}}{2}$. When the discriminant is negative, its square root is $ci$ for a real $c$, and the $\pm$ hands you $a + bi$ and $a - bi$ together, as in the animation above. For a $3 \times 3$ the polynomial is a cubic, which always crosses the axis. Either it crosses three times, or it crosses once and the cubic splits as $(\text{real root} - t)$ times a quadratic that never touches the axis, whose two roots are a mirror pair.
The full proof is in the n version.
Finding them is the same computation as always, with the quadratic formula allowed to take the square root of a negative number. In free response, type the pair like 2 ± i (or 2+i, 2-i).
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Eigenvectors work exactly as on the last page: solve $(A - \lambda I)v = 0$. The only change is that the arithmetic has $i$'s in it, and the answer does too.
Find an eigenvector of $A = \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix}$ for $\lambda = 2 - i$.
$A - \lambda I = \begin{pmatrix} 1 - (2 - i) & -2 \\ 1 & 3 - (2 - i) \end{pmatrix} = \begin{pmatrix} -1 + i & -2 \\ 1 & 1 + i \end{pmatrix}$.
Its first row says $(-1 + i)v_1 - 2v_2 = 0$. Take $v_1 = 2$; then $2v_2 = 2(-1 + i)$, so $v_2 = -1 + i$, and $v = (2, -1 + i)$.
Check with row 2: $1 \cdot 2 + (1 + i)(-1 + i) = 2 + (-1 + i - i + i^2) = 2 - 2 = 0$. Good.
By the conjugate pairs theorem, $\overline{v} = (2, -1 - i)$ is then an eigenvector for $2 + i$, for free.
Notice you never needed row 2 except as a check.
Since $\det(A - \lambda I) = 0$, the two rows of a $2 \times 2$ matrix $A - \lambda I$ are multiples of each other (by a complex number). So one row is enough: if row 1 is $(p, q)$ with $q \neq 0$, then $v = (q, -p)$ works. Don't try to row reduce with complex pivots; it's the same answer with more ways to slip.
Your turn. Any nonzero complex multiple of the right vector counts; in free response, type the entries separated by a comma, like 5, 3 - i.
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The simplest matrices with complex eigenvalues have a shape you can spot on sight.
A rotation-scaling matrix is a $2 \times 2$ matrix of the form $C = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$, with $a$ and $b$ real and not both $0$.
The name says what it does. Its first column, where $e_1$ lands, is the point $(a, b)$; read that point the way you'd read $a + bi$ in the complex plane.
Let $C = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$, let $r = \sqrt{a^2 + b^2}$, and let $\theta$ be the angle of the point $(a, b)$, so $a = r\cos\theta$ and $b = r\sin\theta$. Then
$$C = r\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}:$$$C$ turns every vector by $\theta$ and stretches it by $r$. Its eigenvalues are $a \pm bi$, and $|a \pm bi| = r$.
Substituting $a = r\cos\theta$, $b = r\sin\theta$ and factoring out $r$ gives the displayed form, and $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ is the turn by $\theta$. Its trace is $2a$ and its determinant $a^2 + b^2$, so $f(t) = t^2 - 2at + a^2 + b^2 = (t - a)^2 + b^2$, which is $0$ exactly when $t - a = \pm bi$. Finally $|a \pm bi| = \sqrt{a^2 + b^2} = r$. $\blacksquare$
This is a visual argument, not a formal proof. The columns of $C$ are $(a, b)$ and $(-b, a)$. The second is the first turned by $90^\circ$ (swap the entries and flip a sign), so the two have the same length and meet at a square corner. That means the unit square lands on a square: same shape, just turned by $\theta$ and grown by the length $r$ of $(a, b)$. And that length is the hypotenuse of a right triangle with legs $a$ and $b$.
The full proof is in the n version.
So for a rotation-scaling matrix, the eigenvalue $a + bi$ is literally the picture of the move: its angle is the turn and its modulus is the stretch.
Describe $C = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$ and find $C^8$.
Here $a = b = 1$. The point $(1, 1)$ has length $r = \sqrt{2}$ and angle $\theta = 45^\circ$, so $C$ turns every arrow by $45^\circ$ and stretches it by $\sqrt{2}$. Its eigenvalues are $1 \pm i$.
Doing $C$ eight times turns by $8 \cdot 45^\circ = 360^\circ$, which is no turn at all, and stretches by $(\sqrt{2})^8 = 16$. So $C^8 = 16I$.
Follow one point: each hop turns it by $45^\circ$ and pushes it $\sqrt{2}$ times farther out, so the hops trace a spiral.
Build your own. Drag the blue arrow, the first column $(a, b)$; the rest of the matrix follows, and so do its eigenvalues on the right.
The eigenvalue dot on the right and the blue arrow on the left are always in the same spot. Now read a turn and a stretch off a matrix. In free response, type $r$ and then $\theta$ in degrees, like 2, 120 (sqrt(2) works for $\sqrt{2}$).
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So multiplying by $a + bi$ in the complex plane is the same as multiplying by $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ in the real one. Which means complex numbers were $2 \times 2$ matrices in a trench coat all along. Somebody should have told algebra class.
Most matrices with complex eigenvalues don't have the rotation-scaling shape: $\begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix}$ certainly doesn't. But every one of them is a rotation-scaling matrix seen through a slanted grid, the same way a change of coordinates makes a reflection look messy. The grid comes from an eigenvector.
Let $A$ be a real $2 \times 2$ matrix with an eigenvalue $\lambda = a - bi$, $b \neq 0$, and an eigenvector $v$ for it. Write $v = \operatorname{Re} v + i\operatorname{Im} v$ with $\operatorname{Re} v$ and $\operatorname{Im} v$ real, and let $Q = (\operatorname{Re} v \ \ \operatorname{Im} v)$ be the matrix with those columns. Then $Q$ is invertible and
$$A = QCQ^{-1}, \qquad C = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}.$$Write $x = \operatorname{Re} v$, $y = \operatorname{Im} v$. Then $Av = Ax + iAy$, while $\lambda v = (a - bi)(x + iy) = (ax + by) + i(ay - bx)$. Both $Ax, Ay$ and $ax + by, ay - bx$ are real, so matching real and imaginary parts gives
$$Ax = ax + by, \qquad Ay = -bx + ay.$$Those say $AQ = (Ax \ \ Ay) = Q\begin{pmatrix} a & -b \\ b & a \end{pmatrix} = QC$, column by column. If $Q$ were not invertible, $x$ and $y$ would lie on one line through $0$, so $v = zw$ for a real vector $w \neq 0$ and a complex number $z \neq 0$. Then $Aw = \lambda w$ with $Aw$ real but $\lambda w$ not (it has an entry $\lambda w_j$ with $w_j \neq 0$ and $\lambda$ not real), which is impossible. So $Q$ is invertible and $A = QCQ^{-1}$. $\blacksquare$
This is a visual argument, not a formal proof. The columns of $Q$ draw a slanted grid. Name a point by how many steps along each column it takes to reach it, and in those slanted coordinates $A$ does exactly what $C$ does in the ordinary ones: it turns and stretches. A circle drawn in the slanted grid looks like an oval, so on the right the ring of dots is an oval, but dot for dot it moves just like the ring on the left. Here it is for the example below.
The full proof is in the n version.
Back to the running example, which now unmasks completely.
Write $A = \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix}$ as $QCQ^{-1}$ with $C$ a rotation-scaling matrix.
We found the eigenvector $v = (2, -1 + i)$ for $\lambda = 2 - i$, so $a = 2$, $b = 1$. Splitting $v$: $\operatorname{Re} v = (2, -1)$ and $\operatorname{Im} v = (0, 1)$. So
$$Q = \begin{pmatrix} 2 & 0 \\ -1 & 1 \end{pmatrix}, \qquad C = \begin{pmatrix} 2 & -1 \\ 1 & 2 \end{pmatrix}, \qquad Q^{-1} = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 1 & 2 \end{pmatrix}.$$Check: $QC = \begin{pmatrix} 4 & -2 \\ -1 & 3 \end{pmatrix}$ and $QCQ^{-1} = \frac12\begin{pmatrix} 4 - 2 & -4 \\ -1 + 3 & 6 \end{pmatrix} = \begin{pmatrix} 1 & -2 \\ 1 & 3 \end{pmatrix}$.
So in the slanted grid of $(2, -1)$ and $(0, 1)$, $A$ turns by the angle of $(2, 1)$, about $26.6^\circ$, and stretches by $|2 - i| = \sqrt{5}$.
Which eigenvalue you start from matters.
Use the eigenvalue with the minus sign, $a - bi$, to get $C = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$. Starting from $a + bi$ with the same recipe gives $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ instead: the same stretch, turning the other way, because the slanted grid comes out mirrored. Both factorizations are correct; just don't mix the recipe of one with the eigenvalue of the other.
Here's a matrix that looks like nothing in particular.
Find the eigenvalues of $A = \begin{pmatrix} -2 & -1 \\ 5 & 2 \end{pmatrix}$, and compute $A^4$.
The trace is $0$ and the determinant is $-4 + 5 = 1$, so $f(t) = t^2 + 1$ and the eigenvalues are $\pm i$: modulus $1$, angle $90^\circ$. For $\lambda = -i$, row 1 of $A - \lambda I$ is $(-2 + i, \ -1)$, so $v = (1, -2 + i)$, $Q = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix}$, and $C = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$: a quarter turn.
So $A$ is a quarter turn in disguise, and four quarter turns are no turn: $A^4 = QC^4Q^{-1} = QIQ^{-1} = I$. (Check by multiplying: $A^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$, so $A^4 = I$.)
Watch a point make the trip: four hops around one oval and it's home.
Apply a matrix over and over, $x, Ax, A^2x, \ldots$, and the complex eigenvalues decide the shape of the path. Since $A^k = QC^kQ^{-1}$, every step turns by $\theta$ and stretches by $r$ in the slanted grid, so only one number matters for "in or out": the modulus.
Let $A$ be a real $2 \times 2$ matrix with eigenvalues $a \pm bi$, $b \neq 0$, and let $r = |a \pm bi| = \sqrt{a^2 + b^2} = \sqrt{\det(A)}$. For any $x \neq 0$, the points $x, Ax, A^2x, \ldots$
By the two theorems above, $A = QCQ^{-1}$ with $C = rR_\theta$, where $R_\theta$ is the turn by $\theta$. Then $A^k = QC^kQ^{-1} = r^kQR_{k\theta}Q^{-1}$. Put $y = Q^{-1}x \neq 0$; the turn keeps lengths, so $u_k = R_{k\theta}y$ lies on the circle of radius $\|y\|$, and $A^kx = r^kQu_k$. The points $Qu_k$ lie on the oval $Q(\text{circle})$, whose distance from $0$ stays between two positive numbers $m$ and $M$ because $Q$ is invertible. So $r^km \le \|A^kx\| \le r^kM$: if $r = 1$ the points stay on the oval, if $r < 1$ they go to $0$, and if $r > 1$ they grow without bound. Finally $\det(A) = \det(C) = a^2 + b^2$. $\blacksquare$
This is a visual argument, not a formal proof. Measure in the slanted grid of $Q$. There every step turns the point by $\theta$ and multiplies its distance from $0$ by $r$. Turning never changes the distance, so the distance after $k$ steps is $r \cdot r \cdots r = r^k$ times the start. Shrinking factors pile up toward $0$, growing ones pile up forever, and $r = 1$ keeps the point on one circle of the slanted grid, which on paper is an oval. The animation below shows all three.
The full proof is in the n version.
Here are the three cases side by side, with everything else the same.
The determinant gives $r$ without solving anything, which makes "in or out" a one-line question.
Don't judge by the real part alone. $\begin{pmatrix} 0.5 & -1 \\ 1 & 0.5 \end{pmatrix}$ has $a = 0.5 < 1$, which looks like shrinking, but $r = \sqrt{0.25 + 1} \approx 1.12 > 1$: it spirals out. And check that the eigenvalues really are complex first, with $\tr(A)^2 - 4\det(A) < 0$; a matrix with real eigenvalues doesn't spiral at all.
Try it. Pick a matrix, drag the starting point, and slide $r$ through $1$.
Your turn: no drawing, just the matrix.
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Those spirals are matrix powers $A^kx$. The next page, Matrix Powers & Recurrences, takes powers seriously: when the eigenvalues are real, the biggest one wins in the long run.
In bigger matrices, each complex pair still owns a plane.
A real $3 \times 3$ matrix always has at least one real eigenvalue. If the other two are a complex pair, the matrix does two things at once: it stretches along a line, and it turns-and-stretches a plane.
Let $A$ be a real $n \times n$ matrix with an eigenvalue $\lambda = a - bi$, $b \neq 0$, and an eigenvector $v$ for it. Then the plane $\span\{\operatorname{Re} v, \operatorname{Im} v\}$ is carried into itself by $A$, and in the basis $\operatorname{Re} v, \operatorname{Im} v$ of that plane, $A$ acts as $C = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$.
The proof of the $2 \times 2$ theorem above never used the size of $A$: with $x = \operatorname{Re} v$ and $y = \operatorname{Im} v$ it gives $Ax = ax + by$ and $Ay = -bx + ay$, both in $\span\{x, y\}$, and the same argument shows $x, y$ are independent, so they span a plane. Those two equations say the coordinates of $Ax$ and $Ay$ in the basis $x, y$ are the columns of $C$. $\blacksquare$
Let $A$ be a real $3 \times 3$ matrix with eigenvalues $\mu$ (real) and $a \pm bi$ ($b \neq 0$). Then $A$ stretches a line through $0$ by $\mu$, and carries a plane through $0$ into itself, turning and stretching it like $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ in that plane's slanted grid. The line is not in the plane, but it need not be perpendicular to it.
This is a visual argument, not a formal proof. The line is the real eigenvalue's eigenspace. For the plane, take an eigenvector $v$ for $a - bi$: the same matching of real and imaginary parts as in the $2 \times 2$ case shows that $A$ sends $\operatorname{Re} v$ and $\operatorname{Im} v$ back into the plane they span, acting there like $C$. The line can't lie in that plane, because inside the plane $A$ turns everything and keeps no line. That's the swirl in the animation below: the plane is purple, the axis green.
The full proof is in the n version.
Here is one such matrix applied over and over: a turn of $50^\circ$ with a shrink of $0.9$ on a tilted plane, and a stretch of $1.08$ along an axis that leans away from it. Three starting points show the three things that can happen.
A matrix from the last page is the cleanest example.
Find all the eigenvalues of $R = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$, which sends $e_1 \to e_2 \to e_3 \to e_1$, and read off how far it turns.
Expanding along the first row, $f(t) = \det\begin{pmatrix} -t & 0 & 1 \\ 1 & -t & 0 \\ 0 & 1 & -t \end{pmatrix} = -t(t^2) + 1 \cdot 1 = 1 - t^3$, which factors as $-(t - 1)(t^2 + t + 1)$.
The real eigenvalue is $1$, with eigenvector $(1, 1, 1)$: the axis. The quadratic gives $t = \frac{-1 \pm \sqrt{1 - 4}}{2} = -\frac12 \pm \frac{\sqrt{3}}{2}i$. Its modulus is $\sqrt{\frac14 + \frac34} = 1$, so there's no stretch, and $\cos\theta = -\frac12$, $\sin\theta = \frac{\sqrt3}{2}$ puts it at $\theta = 120^\circ$. So $R$ turns the plane $x + y + z = 0$ by $120^\circ$ about the axis, and three turns bring everything home: $R^3 = I$.
That example read the angle off a complex pair. Adding up eigenvalues gives a faster route, and it works with complex ones too.
Let $A$ be an $n \times n$ matrix, and list its eigenvalues $\lambda_1, \ldots, \lambda_n$, complex ones included, each repeated as many times as it is a root of $f(t)$ (its algebraic multiplicity). Then
$$\lambda_1 + \lambda_2 + \cdots + \lambda_n = \tr(A), \qquad \lambda_1\lambda_2\cdots\lambda_n = \det(A).$$Over the complex numbers every polynomial factors into linear factors (the fundamental theorem of algebra), so with the leading coefficient $(-1)^n$ from the degree theorem, $f(t) = (-1)^n(t - \lambda_1)\cdots(t - \lambda_n)$. Setting $t = 0$: $\det(A) = f(0) = (-1)^n(-1)^n\lambda_1\cdots\lambda_n = \lambda_1\cdots\lambda_n$. For the trace, compare the coefficients of $t^{n-1}$. In the factored form it is $(-1)^n \cdot (-(\lambda_1 + \cdots + \lambda_n)) = (-1)^{n-1}(\lambda_1 + \cdots + \lambda_n)$. In $\det(A - tI)$, expanded as in the degree theorem's proof, every term except the product of the diagonal entries $(A_{11} - t)\cdots(A_{nn} - t)$ misses at least two diagonal entries, so has degree at most $n - 2$; the diagonal product's $t^{n-1}$ coefficient is $(-1)^{n-1}(A_{11} + \cdots + A_{nn}) = (-1)^{n-1}\tr(A)$. $\blacksquare$
For a $2 \times 2$ or $3 \times 3$ matrix, the eigenvalues, complex ones included and a repeated one counted each time it repeats, add up to the trace and multiply to the determinant.
For $2 \times 2$ this is the trace and determinant pattern, and the $\pm bi$ parts cancel in the sum: $(a + bi) + (a - bi) = 2a$. For $3 \times 3$, write $f(t) = -(t - \lambda_1)(t - \lambda_2)(t - \lambda_3)$. Setting $t = 0$ gives $\det(A) = \lambda_1\lambda_2\lambda_3$. And the $t^2$ term of $\det(A - tI)$ can only come from multiplying the three diagonal entries $(A_{11} - t)(A_{22} - t)(A_{33} - t)$, which gives $\tr(A)\,t^2$; the factored form gives $(\lambda_1 + \lambda_2 + \lambda_3)\,t^2$.
The full proof is in the n version.
Check it on $R$: $1 + \left(-\frac12 + \frac{\sqrt3}{2}i\right) + \left(-\frac12 - \frac{\sqrt3}{2}i\right) = 0 = \tr(R)$, and the product is $1 \cdot 1 = 1 = \det(R)$.
A spacecraft's attitude, a robot arm's wrist, a 3D game camera: each gets turned by a rotation matrix $R$, and engineers want two things from it, the axis and the angle. The axis is the eigenspace for $\lambda = 1$. The angle comes from the complex pair. If $R$ turns space by $\theta$ about an axis, it turns the plane perpendicular to the axis by $\theta$, exactly like $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$, so its eigenvalues are $1$ and $\cos\theta \pm i\sin\theta$. Adding them up,
$$\tr(R) = 1 + 2\cos\theta.$$No eigenvector needed: the diagonal alone tells you how far it turns.
The matrix $R = \frac13\begin{pmatrix} 2 & -1 & 2 \\ 2 & 2 & -1 \\ -1 & 2 & 2 \end{pmatrix}$ is a rotation. Find its axis and its angle.
Angle: $\tr(R) = \frac{2 + 2 + 2}{3} = 2 = 1 + 2\cos\theta$, so $\cos\theta = \frac12$ and $\theta = 60^\circ$.
Axis: $R - I = \frac13\begin{pmatrix} -1 & -1 & 2 \\ 2 & -1 & -1 \\ -1 & 2 & -1 \end{pmatrix}$. Each row adds up to $0$, so $(1, 1, 1)$ solves $(R - I)v = 0$, and the first two rows are independent, so that's the whole eigenspace: the axis is the line through $(1, 1, 1)$.
Here is that turn played on the unit cube, with the trace read off as it lands.
The trace gives $\cos\theta$, which can't tell $60^\circ$ from $-60^\circ$; that's fair, since turning one way about $(1, 1, 1)$ is turning the other way about $(-1, -1, -1)$.
The toolkit for this page: non-real eigenvalues of a real matrix come in conjugate pairs $a \pm bi$, with conjugate eigenvectors. The rotation-scaling matrix $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ turns by the angle of $a + bi$ and stretches by $r = |a + bi|$. Every real $2 \times 2$ with eigenvalue $a - bi$ is $QCQ^{-1}$ with $Q = (\operatorname{Re} v \ \ \operatorname{Im} v)$. Its powers spiral in, circle an oval, or spiral out as $r = \sqrt{\det(A)}$ is less than, equal to, or more than $1$. A real $3 \times 3$ with a complex pair stretches an axis and swirls a plane, and the eigenvalues always add to the trace.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Complex answers are multiple choice here.
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Made it this far? Time for something with a little heat: a high power of a matrix, without multiplying it out over and over. In free response, type the four entries row by row.
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