Most vectors get knocked off course by a matrix: $v$ points one way and $Av$ points somewhere else. A few special vectors come out pointing along the same line they went in on, only stretched, shrunk, or flipped. Along those directions an $n \times n$ matrix is as simple as it can possibly be, multiplication by a number. This page finds them: first by looking, then with the determinant.
Apply a matrix to a fan of arrows and nearly all of them get turned. A few don't: they land on the same line through $0$ they started on, just stretched, shrunk, or flipped. Those arrows are the matrix's favorite directions, and the stretch factors are its favorite numbers. This page hunts for them, first by eye and then with the determinant.
The reflection across $y = 2x$ got an easy matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ because $(1, 2)$ stayed put and $(-2, 1)$ flipped. Those two arrows were eigenvectors. Now we find them on purpose.
The arrows that came out on their own lines in that figure get a name.
Let $A$ be an $n \times n$ matrix. A nonzero vector $v \in \R^n$ is an eigenvector of $A$ if $Av = \lambda v$ for some number $\lambda$. That number $\lambda$ is the eigenvalue of $A$ belonging to $v$.
Let $A$ be a $2 \times 2$ or $3 \times 3$ matrix. An arrow $v \neq 0$ is an eigenvector of $A$ if $Av$ lands on the line through $0$ and $v$, that is, $Av = \lambda v$ for some number $\lambda$. That stretch factor $\lambda$ is the eigenvalue belonging to $v$.
The same words work for any linear map $T\colon \V \to \V$ from a space to itself: $v \neq 0$ with $T(v) = \lambda v$. For the map $x \mapsto Ax$ that's exactly the matrix version, so the rest of the page talks about matrices.
Checking whether a given vector is an eigenvector takes one multiplication.
Let $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$. Is $v = (1, 1)$ an eigenvector of $A$? Is $u = (1, 2)$?
$Av = (1 + 2, \ 2 + 1) = (3, 3) = 3(1, 1)$. Yes: $v$ is an eigenvector with eigenvalue $\lambda = 3$.
$Au = (1 + 4, \ 2 + 2) = (5, 4)$. A multiple of $(1, 2)$ would have its second entry twice its first, and $4 \neq 2 \cdot 5$. So $u$ is not an eigenvector.
In the picture, $Av$ runs along $v$'s own dashed line and $Au$ wanders off $u$'s.
Two things in the definition are easy to get backwards.
Eigenvectors are nonzero by definition: $A0 = \lambda 0$ for every $\lambda$, so if $0$ counted, every number would be an eigenvalue. Eigenvalues, on the other hand, can be $0$. That just means $Av = 0$ for some nonzero $v$: the matrix squashes that direction flat.
Also notice that if $v$ is an eigenvector, so is $2v$, or $-5v$, or any nonzero multiple, with the same $\lambda$: $A(cv) = cAv = \lambda(cv)$. Eigenvectors come in whole lines. Your turn: one multiplication, then read off the factor.
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For some maps you don't need to multiply at all: you can see which lines they keep.
Find the eigenvectors and eigenvalues of (a) the projection of $\R^3$ onto the floor, $(x, y, z) \mapsto (x, y, 0)$, and (b) the rotation of the plane by $90^\circ$.
(a) An arrow in the floor, like $(2, 1, 0)$, doesn't move: it's an eigenvector with $\lambda = 1$. An arrow straight up, like $(0, 0, 2)$, goes to $0 = 0 \cdot (0, 0, 2)$: an eigenvector with $\lambda = 0$. A leaning arrow like $(1, 2, 2)$ lands on $(1, 2, 0)$, a different direction, so it isn't one. And $2$ is not an eigenvalue: doubling would need $x = 2x$, $y = 2y$, $0 = 2z$, which only $0$ solves.
(b) A quarter turn moves every nonzero arrow off its line. No eigenvectors at all, so no (real) eigenvalues.
For a general matrix, go hunting. Drag $v$ around the circle; $Av$ follows. Find every direction where $Av$ lands on the line through $v$.
The shear in Mission 2 keeps only one line, and the rotation keeps none. Now read eigenvalues straight off a move, no matrix in sight.
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"Eigen" is German for "own." So an eigenvector is the matrix's own vector. Its favorite. The one it doesn't push around. Every matrix has a favorite child and it's not subtle about it.
Hunting by eye works for a picture. Here's how to find eigenvalues exactly. $Av = \lambda v$ is the same as $Av - \lambda v = 0$, which is $(A - \lambda I)v = 0$. So $\lambda$ is an eigenvalue exactly when the matrix $A - \lambda I$ sends some nonzero vector to $0$, and the determinant knows when that happens.
A number $\lambda$ is an eigenvalue of a square matrix $A$ if and only if $\det(A - \lambda I) = 0$.
$\lambda$ is an eigenvalue exactly when $Av = \lambda v$ for some $v \neq 0$, that is, when $(A - \lambda I)v = 0$ has a solution other than $v = 0$. By the Invertible Matrix Theorem, items (6) and (2), that happens exactly when $\det(A - \lambda I) = 0$. $\blacksquare$
This is a visual argument, not a formal proof. $\det(A - \lambda I)$ is the signed area of the parallelogram on the columns of $A - \lambda I$. It's $0$ exactly when that parallelogram is flat, which is exactly when $A - \lambda I$ flattens something: some arrow $v \neq 0$ goes to $0$. And $(A - \lambda I)v = 0$ says $Av = \lambda v$. Watch the parallelogram for $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ go flat twice as $t$ slides.
The full proof is in the n version.
Replace $\lambda$ by a variable $t$ and $\det(A - tI)$ becomes a polynomial in $t$. Its roots are the eigenvalues.
The characteristic polynomial of a square matrix $A$ is $f(t) = \det(A - tI)$.
Computing one straight from the definition is a $2 \times 2$ determinant with $t$'s in it.
Find the characteristic polynomial and the eigenvalues of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$.
The eigenvalues are $2$ and $5$. (This is the second matrix in the widget below.)
How big is that polynomial in general? Expanding the determinant answers it.
For an $n \times n$ matrix $A$, the characteristic polynomial has degree $n$, leading coefficient $(-1)^n$, and constant term $\det(A)$. So $A$ has at most $n$ different eigenvalues.
By induction on $n$, expanding $\det(A - tI)$ along the first row: every entry of $A - tI$ is a number or a number minus $t$, and $t$ appears only on the diagonal. The term $(A_{11} - t)\det(\tilde{B}_{11})$, where $\tilde{B}_{11}$ is $A - tI$ with row 1 and column 1 deleted, has degree $1 + (n - 1) = n$ with leading coefficient $(-1)(-1)^{n-1} = (-1)^n$ by induction; every other term multiplies a number by a minor that is missing two diagonal entries, so it has degree at most $n - 2$. Setting $t = 0$ gives $f(0) = \det(A)$. A polynomial of degree $n$ has at most $n$ roots. $\blacksquare$
The characteristic polynomial of a $2 \times 2$ matrix is a quadratic, and of a $3 \times 3$ matrix a cubic. So a $2 \times 2$ matrix has at most $2$ different eigenvalues and a $3 \times 3$ matrix at most $3$.
The variable $t$ only sits on the diagonal of $A - tI$. Expanding the determinant, the most $t$'s any one product can collect is one per diagonal entry: two for a $2 \times 2$, three for a $3 \times 3$. A quadratic crosses $0$ at most twice, a cubic at most three times.
The full proof is in the n version.
For $2 \times 2$ matrices the polynomial has a pattern worth memorizing.
For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the characteristic polynomial is
$$f(t) = t^2 - \tr(A)\,t + \det(A) = t^2 - (a + d)t + (ad - bc).$$So the two eigenvalues (when they're real) add up to $\tr(A)$ and multiply to $\det(A)$.
$\det\begin{pmatrix} a - t & b \\ c & d - t \end{pmatrix} = (a - t)(d - t) - bc = t^2 - (a + d)t + (ad - bc)$. If its roots are $\lambda_1$ and $\lambda_2$, then $f(t) = (t - \lambda_1)(t - \lambda_2) = t^2 - (\lambda_1 + \lambda_2)t + \lambda_1\lambda_2$; compare coefficients. $\blacksquare$
That's the fastest route for any $2 \times 2$.
Find the eigenvalues of $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$.
The trace is $1 + 1 = 2$ and the determinant is $1 - 4 = -3$, so
$$f(t) = t^2 - 2t - 3 = (t - 3)(t + 1).$$The eigenvalues are $\lambda = 3$ and $\lambda = -1$. Check: $3 + (-1) = 2$, the trace, and $3 \cdot (-1) = -3$, the determinant.
Here is why that check works. On the parallelogram built from the two eigenvectors, $A$ stretches by $3$ one way and flips the other way, so it multiplies area by $3 \cdot (-1)$: that's the determinant.
The usual slip is a sign.
The middle term of $t^2 - \tr(A)\,t + \det(A)$ has a minus sign. Drop it and both eigenvalues come out with the wrong sign. The trace and determinant check catches it in two seconds.
See the polynomial in action. Slide $t$ and watch $\det(A - tI)$; it hits $0$ exactly where the columns of $A - tI$ line up.
For a triangular matrix there's nothing to solve at all.
The eigenvalues of a triangular matrix are its diagonal entries.
If $A$ is triangular, so is $A - tI$, with diagonal entries $A_{ii} - t$. A triangular determinant is the product of its diagonal, so $f(t) = (A_{11} - t)(A_{22} - t)\cdots$, whose roots are exactly $A_{11}, A_{22}, \ldots$. $\blacksquare$
Here's a $3 \times 3$ where that saves real work.
Find the eigenvalues of $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{pmatrix}$.
$A - tI$ is upper triangular, so its determinant is the product of its diagonal entries:
$$f(t) = (2 - t)(3 - t)(-1 - t).$$The eigenvalues are $2$, $3$, and $-1$: just the diagonal. (The eigenvectors take more work. Solving $(A - \lambda I)v = 0$ for each, the way the last section of this page shows, gives $(1, 0, 0)$, $(1, 1, 0)$, and $(1, -3, 3)$.)
The shortcut has a strict condition.
Don't read eigenvalues off the diagonal of a matrix that isn't triangular. For $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$ the diagonal says $1, 1$, and the eigenvalues are $3$ and $-1$.
Your turn. In free response, type the eigenvalues separated by commas, like 3, -1.
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The rotation by $90^\circ$, $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, has $f(t) = t^2 - 0 \cdot t + 1 = t^2 + 1$. That's never $0$ for a real $t$, which matches the picture: a quarter turn keeps no line. So a real $2 \times 2$ matrix can have no real eigenvalues. Odd sizes can't dodge like that.
If $n$ is odd, every real $n \times n$ matrix has at least one real eigenvalue.
The characteristic polynomial has odd degree $n$ and leading coefficient $(-1)^n = -1$, so $f(t) \to -\infty$ as $t \to \infty$ and $f(t) \to \infty$ as $t \to -\infty$. A polynomial is continuous, so by the intermediate value theorem it equals $0$ somewhere in between. That root is a real eigenvalue. $\blacksquare$
Every real $3 \times 3$ matrix has at least one real eigenvalue, so it keeps at least one line through $0$.
This is a visual argument, not a formal proof. The characteristic polynomial of a $3 \times 3$ matrix is a cubic, $-t^3 + \cdots$. Far to the left it's huge and positive, far to the right huge and negative, and its graph is one unbroken curve, so it has to cross $0$ on the way. In space you can see the consequence: a turn has an axis that stays put. This one is special to odd sizes; it's the reason a $2 \times 2$ rotation can escape and a $3 \times 3$ one can't.
The full proof is in the n version.
So every real $3 \times 3$ matrix keeps at least one line. Go hunting for them in space. The sliders aim $v$: turn spins it around the $z$-axis, tilt lifts it off the floor.
$t^2 + 1 = 0$ has no real roots, but it has two complex ones, $t = \pm i$. A later page, Complex Eigenvalues, takes those seriously and finds the turning hiding inside them.
Once you have an eigenvalue $\lambda$, the eigenvectors for it are the nonzero solutions of $(A - \lambda I)v = 0$: a null space.
Let $\lambda$ be an eigenvalue of $A$. A vector $v$ is an eigenvector of $A$ for $\lambda$ if and only if $v \neq 0$ and $v \in \N(A - \lambda I)$.
$Av = \lambda v$ if and only if $Av - \lambda Iv = 0$, that is, $(A - \lambda I)v = 0$, which says $v \in \N(A - \lambda I)$. An eigenvector is a nonzero $v$ with $Av = \lambda v$. $\blacksquare$
So the eigenvectors for $\lambda$, with $0$ added back in, form a null space. That set gets its own name.
For an eigenvalue $\lambda$ of $A$, the eigenspace of $\lambda$ is $E_\lambda = \N(A - \lambda I) = \{v : Av = \lambda v\}$.
Back to the running example, finished off: its eigenspaces are two null spaces.
Find the eigenspaces of $A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$.
For $\lambda = 3$: $A - 3I = \begin{pmatrix} -2 & 2 \\ 2 & -2 \end{pmatrix}$. Both rows say $-2v_1 + 2v_2 = 0$, so $v_2 = v_1$, and $E_3 = \span\{(1, 1)\}$.
For $\lambda = -1$: $A + I = \begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}$. Both rows say $v_1 + v_2 = 0$, so $E_{-1} = \span\{(1, -1)\}$.
Check one: $A(1, -1) = (1 - 2, \ 2 - 1) = (-1, 1) = -1 \cdot (1, -1)$. Good.
Knowing both eigenspaces tells you the whole map: stretch by $3$ along one line, flip along the other.
Each eigenspace there is a line through $0$, which is a subspace. That always happens: a null space is closed under adding and scaling, so $E_\lambda$ is too.
Every eigenspace $E_\lambda$ of $A$ is a subspace, and it contains a nonzero vector.
$E_\lambda = \N(A - \lambda I)$ is the null space of the map $x \mapsto (A - \lambda I)x$, and null spaces are subspaces. It contains an eigenvector because $\lambda$ is an eigenvalue. $\blacksquare$
This is a visual argument, not a formal proof. If $A$ stretches $u$ by $\lambda$ and $v$ by $\lambda$, it stretches the whole parallelogram on them by $\lambda$, corner $u + v$ included; and it stretches $cu$ by $\lambda$ too. So adding and scaling never leave $E_\lambda$. In $\R^3$ that leaves four possible shapes, the subspaces: a line through $0$, a plane through $0$, or all of space (never just $\{0\}$, since an eigenvalue has an eigenvector). Here is a plane eigenspace, from the projection onto the floor.
The full proof is in the n version.
A $3 \times 3$ works the same way, with one surprise at the end.
Find the eigenvalues and eigenspaces of $A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 0 & -2 \\ -1 & 2 & 3 \end{pmatrix}$.
Expanding $\det(A - tI)$ along the first row and collecting terms gives
$$f(t) = -t^3 + 4t^2 - 4t = -t(t - 2)^2,$$so the eigenvalues are $0$ and $2$.
For $\lambda = 0$: $Av = 0$. Row $2$ says $2v_1 - 2v_3 = 0$, so $v_3 = v_1$; then row $1$ says $v_1 + 2v_2 + v_1 = 0$, so $v_2 = -v_1$. $E_0 = \span\{(1, -1, 1)\}$.
For $\lambda = 2$: $A - 2I = \begin{pmatrix} -1 & 2 & 1 \\ 2 & -2 & -2 \\ -1 & 2 & 1 \end{pmatrix}$. Row $2$ gives $v_1 = v_2 + v_3$; putting that into row $1$ gives $v_2 = 0$, so $v_1 = v_3$. $E_2 = \span\{(1, 0, 1)\}$.
The surprise: $(t - 2)$ appears twice in $f(t)$, but $E_2$ is only a line.
That doesn't always happen. Sometimes a repeated root does come with a bigger eigenspace.
Find the eigenspaces of the projection onto the floor, $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$.
$A$ is triangular, so $f(t) = (1 - t)^2(0 - t)$ and the eigenvalues are $1$ (twice) and $0$. $A - I = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ only asks for $v_3 = 0$, so $E_1$ is the whole floor, $\span\{(1, 0, 0), (0, 1, 0)\}$: a plane. $A$ itself only kills vectors with $v_1 = v_2 = 0$, so $E_0 = \span\{(0, 0, 1)\}$.
It's the plane eigenspace pictured in the subspace theorem's "Why it's true" above.
One habit catches most arithmetic slips.
If $A - \lambda I$ row reduces with no free variable, you made a mistake: a real eigenvalue always leaves at least one, because $\det(A - \lambda I) = 0$ means the rank dropped. Go back and recheck $\lambda$.
For $\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$, the eigenvectors $(1, 1)$ and $(1, -1)$ form a basis, and in that basis the matrix is diagonal. The $3 \times 3$ above, with only two lines of eigenvectors, can't do that. Counting roots against eigenspaces is the next page.
Your turn. In free response, type any eigenvector, like 2, 1; every nonzero multiple counts.
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A satellite, a drone, or a camera in a 3D game gets turned around by a rotation matrix, and a natural question is: around which axis? The axis is the line the rotation doesn't move, so it's the eigenspace for $\lambda = 1$. By the odd-size theorem above, every rotation of space has one.
The rotation $R = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$ sends $e_1 \to e_2 \to e_3 \to e_1$. Find its axis.
Solve $(R - I)v = 0$: $\begin{pmatrix} -1 & 0 & 1 \\ 1 & -1 & 0 \\ 0 & 1 & -1 \end{pmatrix}v = 0$ says $v_3 = v_1$, $v_1 = v_2$, $v_2 = v_3$. So $E_1 = \span\{(1, 1, 1)\}$: the rotation turns space about the cube's long diagonal. (Three turns bring every $e_i$ home, so it's a turn by $120^\circ$.)
Here is that turn, played out on the unit cube.
The same computation, one null space, finds the axis of any rotation matrix, however messy its entries.
The toolkit for this page: an eigenvector is $v \neq 0$ with $Av = \lambda v$. The eigenvalues are the roots of $f(t) = \det(A - tI)$ (for $2 \times 2$: $t^2 - \tr(A)t + \det(A)$; for triangular: the diagonal). For each $\lambda$, the eigenvectors are the nonzero vectors of $E_\lambda = \N(A - \lambda I)$. A real $3 \times 3$ always has a real eigenvalue; a $2 \times 2$ rotation can have none.
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. Type a number or a vector when it asks you to.
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Made it this far? Time for something with a little heat: pick the number $k$ that makes a given $\lambda$ an eigenvalue. In free response, type one number.
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