A nonzero vector $\mathbf{x}$ with $A\mathbf{x}=\lambda\mathbf{x}$ keeps its direction: the matrix only scales it by $\lambda$. We find every such $\lambda$ from $\det(A-\lambda I)=0$ and the vectors from $\operatorname{Nul}(A-\lambda I)$.
A matrix can do complicated things to the plane โ stretch it, shear it, flip it, rotate it. But look closely and you will usually find a few lines through the origin that the matrix maps onto themselves. A vector on such a line is not turned at all; it is only made longer, shorter, or reversed. Along those lines the matrix behaves like ordinary multiplication by a number, and that is as simple as linear algebra ever gets.
Finding these special directions is one of the most useful things you can do with a square matrix. They tell an engineer how a structure will vibrate, a statistician which combination of variables carries the most information, and an ecologist whether a population will boom or crash. In this section we define eigenvalues and eigenvectors, learn how to compute them with the characteristic equation, and collect the properties that make them so powerful.
Vectors That Keep Their Direction
๐ Definition 7.1 โ Eigenvalue and eigenvector
Let $A$ be an $n\times n$ matrix. A scalar $\lambda$ is an eigenvalue of $A$ if there is a nonzero vector $\mathbf{x}$ in $\mathbb{R}^n$ such that
$$A\mathbf{x}=\lambda\mathbf{x}.$$
Such a vector $\mathbf{x}$ is called an eigenvector of $A$ corresponding to (or belonging to) $\lambda$.
Two remarks are worth making straight away. First, the zero vector is never an eigenvector โ $A\mathbf{0}=\lambda\mathbf{0}$ for every $\lambda$, so allowing it would make every number an eigenvalue. The number $\lambda=0$, on the other hand, is perfectly allowed as an eigenvalue. Second, only square matrices have eigenvalues: $A\mathbf{x}$ and $\lambda\mathbf{x}$ must live in the same space.
Geometric meaning. If $\mathbf{x}$ is an eigenvector, $A\mathbf{x}$ lies on the line through $\mathbf{0}$ and $\mathbf{x}$. The eigenvalue says what happens along that line:
$\lambda\gt1$vectors on the line are stretched
$0\lt\lambda\lt1$vectors are shrunk towards $\mathbf{0}$
$\lambda\lt0$vectors are reversed (and scaled by $|\lambda|$)
$\lambda=0$the whole line is crushed to $\mathbf{0}$
If $\mathbf{x}$ is an eigenvector, so is every nonzero multiple $c\mathbf{x}$, with the same eigenvalue: $A(c\mathbf{x})=cA\mathbf{x}=c\lambda\mathbf{x}=\lambda(c\mathbf{x})$. So an eigenvector really describes a whole direction, and we are free to scale it to make its entries nice.
โ๏ธ Example 1 โ Checking eigenvectors
Let $A=\begin{bmatrix}4&-2\\1&1\end{bmatrix}$. Decide whether $\mathbf{u}=\begin{bmatrix}1\\1\end{bmatrix}$, $\mathbf{v}=\begin{bmatrix}2\\1\end{bmatrix}$ and $\mathbf{w}=\begin{bmatrix}1\\0\end{bmatrix}$ are eigenvectors of $A$, and give the eigenvalue when they are.
๐ก Show solutionโถ
Method Multiply, then ask whether the answer is a scalar multiple of the vector you started with.
$\mathbf{u}$ $A\mathbf{u}=\begin{bmatrix}4-2\\1+1\end{bmatrix}=\begin{bmatrix}2\\2\end{bmatrix}=2\begin{bmatrix}1\\1\end{bmatrix}$. So $\mathbf{u}$ is an eigenvector with eigenvalue $\lambda=2$.
$\mathbf{v}$ $A\mathbf{v}=\begin{bmatrix}8-2\\2+1\end{bmatrix}=\begin{bmatrix}6\\3\end{bmatrix}=3\begin{bmatrix}2\\1\end{bmatrix}$. So $\mathbf{v}$ is an eigenvector with eigenvalue $\lambda=3$.
$\mathbf{w}$ $A\mathbf{w}=\begin{bmatrix}4\\1\end{bmatrix}$. If this were $\lambda\begin{bmatrix}1\\0\end{bmatrix}$ the second entry would be $0$; it is $1$. So $\mathbf{w}$ is not an eigenvector โ the matrix turns it.
Picture Choose this matrix in Interactive 1: the rotating unit vector lines up with its image exactly when it points along $(1,1)$ or $(2,1)$.
Interactive 1 โ The Eigen-Circle: Hunting for Directions That Line Up
The blue unit vector $\mathbf{x}$ travels round the circle; the red arrow is its image $A\mathbf{x}$, which traces out the faint ellipse. Usually the two arrows point in different directions. Whenever they fall on the same line, $\mathbf{x}$ is an eigenvector: the animation pauses, marks the direction in green, and reads off $\lambda$ as the signed ratio of lengths.
$\mathbf{x}$ (angle)โ
$A\mathbf{x}$โ
angle between $\mathbf{x}$ and $A\mathbf{x}$โ
$\lambda$ (when lined up)โ
The Characteristic Equation
Guessing eigenvectors is not a method. To find them, rewrite $A\mathbf{x}=\lambda\mathbf{x}$ as $A\mathbf{x}-\lambda I\mathbf{x}=\mathbf{0}$, that is,
$$(A-\lambda I)\mathbf{x}=\mathbf{0}.$$
(We need $\lambda I$, not just $\lambda$, because we can only subtract a matrix from a matrix.) So $\lambda$ is an eigenvalue exactly when this homogeneous system has a nonzero solution โ that is, when the square matrix $A-\lambda I$ is not invertible. By the determinant test from Chapter 3, that happens exactly when its determinant is zero.
๐ Theorem 7.1 โ The characteristic equation
Let $A$ be an $n\times n$ matrix. A scalar $\lambda$ is an eigenvalue of $A$ if and only if
$$\det(A-\lambda I)=0 .$$
The eigenvectors belonging to $\lambda$ are the nonzero solutions of the homogeneous system $(A-\lambda I)\mathbf{x}=\mathbf{0}$.
For an $n\times n$ matrix $A$, $p(\lambda)=\det(A-\lambda I)$ is a polynomial of degree $n$ in $\lambda$, called the characteristic polynomial of $A$; the equation $p(\lambda)=0$ is the characteristic equation. Hence an $n\times n$ matrix has at most $n$ eigenvalues.
If $\lambda$ is an eigenvalue of $A$, the set of all solutions of $(A-\lambda I)\mathbf{x}=\mathbf{0}$,
$$E_\lambda=\operatorname{Nul}(A-\lambda I),$$
is called the eigenspace of $A$ for $\lambda$. It consists of all eigenvectors for $\lambda$ together with $\mathbf{0}$, and, being a null space, it is a subspace of $\mathbb{R}^n$.
For a $2\times2$ matrix the characteristic polynomial has a memorable shape. With $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$,
$$\det(A-\lambda I)=\begin{vmatrix}a-\lambda&b\\c&d-\lambda\end{vmatrix}=(a-\lambda)(d-\lambda)-bc=\lambda^2-(\operatorname{tr}A)\,\lambda+\det A,$$
where $\operatorname{tr}A=a+d$ is the trace (sum of the diagonal entries).
๐งญ Problem-solving strategy โ finding eigenvalues and eigenvectors
Write down $A-\lambda I$ (subtract $\lambda$ from each diagonal entry only).
Compute $p(\lambda)=\det(A-\lambda I)$ and solve $p(\lambda)=0$. Keep the polynomial factorised whenever you can โ for $3\times3$ matrices look for a common factor before expanding everything.
For each eigenvalue $\lambda$, row reduce $A-\lambda I$ (it must have a free variable) and write the general solution of $(A-\lambda I)\mathbf{x}=\mathbf{0}$ in parametric vector form. The vectors multiplying the free parameters form a basis for $E_\lambda$.
Check: multiply $A$ by each basis vector and confirm you get $\lambda$ times it.
โ๏ธ Example 2 โ Eigenvalues and eigenspaces of a $2\times2$ matrix
Find the eigenvalues of $A=\begin{bmatrix}1&6\\5&2\end{bmatrix}$ and a basis for each eigenspace.
๐ก Show solutionโถ
Characteristic equation $$\det(A-\lambda I)=\begin{vmatrix}1-\lambda&6\\5&2-\lambda\end{vmatrix}=(1-\lambda)(2-\lambda)-30=\lambda^2-3\lambda-28=(\lambda-7)(\lambda+4).$$ The eigenvalues are $\lambda_1=7$ and $\lambda_2=-4$.
$\lambda=7$ $$A-7I=\begin{bmatrix}-6&6\\5&-5\end{bmatrix}\ \xrightarrow{R_1\to-\frac16R_1,\ R_2\to R_2-5R_1}\ \begin{bmatrix}1&-1\\0&0\end{bmatrix}.$$ So $x_1=x_2$ with $x_2=t$ free: $\mathbf{x}=t\begin{bmatrix}1\\1\end{bmatrix}$. A basis for $E_7$ is $\left\{\begin{bmatrix}1\\1\end{bmatrix}\right\}$.
$\lambda=-4$ $$A+4I=\begin{bmatrix}5&6\\5&6\end{bmatrix}\ \xrightarrow{R_2\to R_2-R_1}\ \begin{bmatrix}5&6\\0&0\end{bmatrix}.$$ So $5x_1+6x_2=0$; taking $x_2=-5t$ gives $x_1=6t$. A basis for $E_{-4}$ is $\left\{\begin{bmatrix}6\\-5\end{bmatrix}\right\}$.
Check $A\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}7\\7\end{bmatrix}=7\begin{bmatrix}1\\1\end{bmatrix}$ โ and $A\begin{bmatrix}6\\-5\end{bmatrix}=\begin{bmatrix}6-30\\30-10\end{bmatrix}=\begin{bmatrix}-24\\20\end{bmatrix}=-4\begin{bmatrix}6\\-5\end{bmatrix}$ โ.
โ๏ธ Example 3 โ A $3\times3$ matrix with a two-dimensional eigenspace
Find the eigenvalues of $A=\begin{bmatrix}3&-1&1\\-1&3&-1\\1&-1&3\end{bmatrix}$ and a basis for each eigenspace.
๐ก Show solutionโถ
Expand Expanding along the first row,
$$\begin{aligned}\det(A-\lambda I)&=\begin{vmatrix}3-\lambda&-1&1\\-1&3-\lambda&-1\\1&-1&3-\lambda\end{vmatrix}\\&=(3-\lambda)\left[(3-\lambda)^2-1\right]+1\cdot\left[-(3-\lambda)+1\right]+1\cdot\left[1-(3-\lambda)\right].\end{aligned}$$
Factorise Put $\mu=3-\lambda$: $\ \mu(\mu^2-1)+2(1-\mu)=(\mu-1)\left[\mu(\mu+1)-2\right]=(\mu-1)(\mu+2)(\mu-1)$. Since $\mu-1=2-\lambda$ and $\mu+2=5-\lambda$,
$$\det(A-\lambda I)=(2-\lambda)^2(5-\lambda).$$ The eigenvalues are $\lambda=2$ (a double root) and $\lambda=5$.
$\lambda=2$ $$A-2I=\begin{bmatrix}1&-1&1\\-1&1&-1\\1&-1&1\end{bmatrix}\ \xrightarrow{R_2\to R_2+R_1,\ R_3\to R_3-R_1}\ \begin{bmatrix}1&-1&1\\0&0&0\\0&0&0\end{bmatrix}.$$ One equation $x_1=x_2-x_3$, two free variables $x_2=s$, $x_3=t$: $$\mathbf{x}=s\begin{bmatrix}1\\1\\0\end{bmatrix}+t\begin{bmatrix}-1\\0\\1\end{bmatrix}.$$ Basis for $E_2$: $\left\{(1,1,0),\ (-1,0,1)\right\}$ โ a plane of eigenvectors.
$\lambda=5$ $$A-5I=\begin{bmatrix}-2&-1&1\\-1&-2&-1\\1&-1&-2\end{bmatrix}\ \xrightarrow{R_1\leftrightarrow R_3}\ \begin{bmatrix}1&-1&-2\\-1&-2&-1\\-2&-1&1\end{bmatrix}\ \xrightarrow{R_2\to R_2+R_1,\ R_3\to R_3+2R_1}\ \begin{bmatrix}1&-1&-2\\0&-3&-3\\0&-3&-3\end{bmatrix}$$ $$\xrightarrow{R_3\to R_3-R_2,\ R_2\to-\frac13R_2,\ R_1\to R_1+R_2}\ \begin{bmatrix}1&0&-1\\0&1&1\\0&0&0\end{bmatrix}.$$ So $x_1=x_3$, $x_2=-x_3$: basis for $E_5$ is $\{(1,-1,1)\}$.
Check $A(1,1,0)=(2,2,0)$ โ, $A(-1,0,1)=(-2,0,2)$ โ, $A(1,-1,1)=(5,-5,5)$ โ. Also $2+2+5=9=\operatorname{tr}A$ โ see Theorem 7.4 below.
Interactive 2 โ The Characteristic Polynomial and Its Roots
The curve is $p(\lambda)=\det(A-\lambda I)$. Slide $\lambda$ along the axis: wherever the curve crosses or touches zero, $A-\lambda I$ stops being invertible, its rank drops, and $\lambda$ is an eigenvalue (green dots). The readouts compare the sum and product of the roots with $\operatorname{tr}A$ and $\det A$, and show the dimension of each eigenspace.
$\lambda$โ
$p(\lambda)=\det(A-\lambda I)$โ
$\operatorname{rank}(A-\lambda I)$โ
real roots (sum ยท product)โ
$\operatorname{tr}A$ ยท $\det A$โ
Triangular Matrices and Multiplicity
For a triangular matrix the characteristic polynomial comes for free: $A-\lambda I$ is still triangular, and the determinant of a triangular matrix is the product of its diagonal entries.
๐ Theorem 7.2 โ Eigenvalues of a triangular matrix
If $A$ is an upper or lower triangular $n\times n$ matrix (in particular, a diagonal matrix), then $$\det(A-\lambda I)=(a_{11}-\lambda)(a_{22}-\lambda)\cdots(a_{nn}-\lambda),$$ so the eigenvalues of $A$ are its diagonal entries $a_{11},a_{22},\dots,a_{nn}$.
When a root is repeated we need to count two different things: how many times it appears as a root, and how many independent eigenvectors it actually produces.
๐ Definition 7.3 โ Algebraic and geometric multiplicity
Let $\lambda_0$ be an eigenvalue of the $n\times n$ matrix $A$.
The algebraic multiplicity of $\lambda_0$ is the number of times $(\lambda_0-\lambda)$ occurs as a factor of $\det(A-\lambda I)$.
The geometric multiplicity of $\lambda_0$ is $\dim E_{\lambda_0}=\operatorname{nullity}(A-\lambda_0I)$, which equals $n-\operatorname{rank}(A-\lambda_0I)$.
๐ Theorem 7.3 โ Geometric multiplicity is at most algebraic multiplicity
For every eigenvalue $\lambda_0$ of a square matrix $A$,
$$1\le\text{geometric multiplicity of }\lambda_0\le\text{algebraic multiplicity of }\lambda_0 .$$
In particular, an eigenvalue that is a simple (non-repeated) root always has a one-dimensional eigenspace.
โ๏ธ Example 4 โ A triangular matrix with too few eigenvectors
Find the eigenvalues of $A=\begin{bmatrix}2&1&3\\0&2&-1\\0&0&5\end{bmatrix}$, their algebraic and geometric multiplicities, and a basis for each eigenspace. Compare with Example 3.
๐ก Show solutionโถ
Eigenvalues $A$ is upper triangular, so by Theorem 7.2, $\det(A-\lambda I)=(2-\lambda)^2(5-\lambda)$: $\lambda=2$ with algebraic multiplicity $2$, and $\lambda=5$ with algebraic multiplicity $1$.
$\lambda=2$ $$A-2I=\begin{bmatrix}0&1&3\\0&0&-1\\0&0&3\end{bmatrix}\ \xrightarrow{R_3\to R_3+3R_2,\ R_2\to-R_2,\ R_1\to R_1-3R_2}\ \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}.$$ So $x_2=0$, $x_3=0$, and only $x_1=t$ is free: $E_2=\operatorname{span}\{(1,0,0)\}$. Geometric multiplicity $1$.
$\lambda=5$ $$A-5I=\begin{bmatrix}-3&1&3\\0&-3&-1\\0&0&0\end{bmatrix}.$$ With $x_3=9t$: the second row gives $-3x_2-9t=0$, so $x_2=-3t$; the first gives $-3x_1-3t+27t=0$, so $x_1=8t$. Basis for $E_5$: $\{(8,-3,9)\}$. Check: $A(8,-3,9)=(16-3+27,\,-6-9,\,45)=(40,-15,45)=5(8,-3,9)$ โ.
Compare Examples 3 and 4 have the same characteristic polynomial $(2-\lambda)^2(5-\lambda)$. But in Example 3 the double eigenvalue $2$ has a whole plane of eigenvectors (geometric multiplicity $2$), while here it has only a line (geometric multiplicity $1$). Section 7.2 shows that this difference decides whether the matrix can be diagonalised. You can see it in Interactive 2: switch between these two matrices and compare the rank at $\lambda=2$.
โ ๏ธ Common mistakes with eigenvalues
1. Row reducing $A$ instead of $A-\lambda I$. Row operations change the eigenvalues of a matrix, so you cannot simplify $A$ first. Row reduce $A-\lambda I$, and only after you know $\lambda$.
2. โThe only solution is $\mathbf{x}=\mathbf{0}$.โ If $\lambda$ really is an eigenvalue, $A-\lambda I$ is singular and must produce a free variable. Getting only the zero solution means there is an arithmetic slip in $\lambda$ or in the row reduction.
3. Subtracting $\lambda$ from every entry. $A-\lambda I$ changes the diagonal only.
4. Eigenvalues are not the pivots. The pivots of an echelon form of $A$ have nothing to do with its eigenvalues (except for a matrix that is already triangular).
Properties of Eigenvalues
๐ Theorem 7.4 โ Eigenvalues, invertibility, powers, trace and determinant
Let $A$ be an $n\times n$ matrix.
$0$ is an eigenvalue of $A$ if and only if $A$ is not invertible. (So โ$0$ is not an eigenvalue of $A$โ joins the Invertible Matrix Theorem of Sections 2.2, 3.2 and 5.4: it is true exactly when $A$ is invertible, $\det A\ne0$, $\operatorname{rank}A=n$, and all the other conditions on that list hold.)
If $A\mathbf{x}=\lambda\mathbf{x}$, then $A^k\mathbf{x}=\lambda^k\mathbf{x}$ for every positive integer $k$.
If $A$ is invertible and $A\mathbf{x}=\lambda\mathbf{x}$, then $\lambda\ne0$ and $A^{-1}\mathbf{x}=\dfrac1\lambda\,\mathbf{x}$.
If $A\mathbf{x}=\lambda\mathbf{x}$ and $c$ is a scalar, then $(A+cI)\mathbf{x}=(\lambda+c)\mathbf{x}$.
If $\lambda_1,\dots,\lambda_n$ are the $n$ roots of the characteristic polynomial, listed with their algebraic multiplicities and including any complex roots, then $$\lambda_1+\lambda_2+\dots+\lambda_n=\operatorname{tr}A\qquad\text{and}\qquad\lambda_1\lambda_2\cdots\lambda_n=\det A .$$
Why these hold. (1) $0$ is an eigenvalue $\iff\det(A-0I)=\det A=0$. (2) $A^2\mathbf{x}=A(\lambda\mathbf{x})=\lambda A\mathbf{x}=\lambda^2\mathbf{x}$, and so on. (3) Multiply $A\mathbf{x}=\lambda\mathbf{x}$ by $A^{-1}$: $\mathbf{x}=\lambda A^{-1}\mathbf{x}$; $\lambda\ne0$ by (1), so divide by $\lambda$. (4) $(A+cI)\mathbf{x}=A\mathbf{x}+c\mathbf{x}=(\lambda+c)\mathbf{x}$. (5) For a $2\times2$ matrix, $\lambda^2-(\operatorname{tr}A)\lambda+\det A=(\lambda-\lambda_1)(\lambda-\lambda_2)$, and the right side expands to $\lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2$; compare coefficients. The $n\times n$ case works the same way. Notice that in (2)โ(4) the eigenvectors do not change โ only the eigenvalues do.
โ๏ธ Example 5 โ Using the properties
For $A=\begin{bmatrix}1&6\\5&2\end{bmatrix}$ (Example 2, eigenvalues $7$ and $-4$): (a) check the trace and determinant rules; (b) is $A$ invertible? (c) find the eigenvalues of $A^3$, $A^{-1}$ and $A-3I$; (d) compute $A^3\begin{bmatrix}6\\-5\end{bmatrix}$ without multiplying matrices.
๐ก Show solutionโถ
(a) $7+(-4)=3=1+2=\operatorname{tr}A$ โ and $7\cdot(-4)=-28=1\cdot2-6\cdot5=\det A$ โ.
(b) $0$ is not an eigenvalue, so $A$ is invertible (indeed $\det A=-28\ne0$).
(c) $A^3$: $7^3=343$ and $(-4)^3=-64$. $A^{-1}$: $\tfrac17$ and $-\tfrac14$. $A-3I$: $7-3=4$ and $-4-3=-7$. The eigenvectors $(1,1)$ and $(6,-5)$ are the same for all three matrices.
(d) $(6,-5)$ is an eigenvector for $\lambda=-4$, so $A^3\begin{bmatrix}6\\-5\end{bmatrix}=(-4)^3\begin{bmatrix}6\\-5\end{bmatrix}=\begin{bmatrix}-384\\320\end{bmatrix}$.
Check $A^{-1}=\dfrac{1}{-28}\begin{bmatrix}2&-6\\-5&1\end{bmatrix}$, and $A^{-1}\begin{bmatrix}1\\1\end{bmatrix}=\dfrac{1}{-28}\begin{bmatrix}-4\\-4\end{bmatrix}=\dfrac17\begin{bmatrix}1\\1\end{bmatrix}$ โ.
A first look at complex eigenvalues
The characteristic polynomial of a real matrix can have complex roots. Geometrically this happens when the matrix turns every direction, as a rotation does. We will not develop complex eigenvectors in depth, but you should recognise the situation.
โ๏ธ Example 6 โ A rotation has no real eigenvectors
Let $R=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$, the rotation of the plane through $90^\circ$ anticlockwise. Find its eigenvalues, and an eigenvector in $\mathbb{C}^2$ for one of them.
๐ก Show solutionโถ
Geometry A $90^\circ$ rotation moves every nonzero vector off its line, so we expect no real eigenvalues.
Algebra $\det(R-\lambda I)=\begin{vmatrix}-\lambda&-1\\1&-\lambda\end{vmatrix}=\lambda^2+1$, which has no real roots. Over the complex numbers, $\lambda=\pm i$.
Eigenvector For $\lambda=i$: $R-iI=\begin{bmatrix}-i&-1\\1&-i\end{bmatrix}$; the first row gives $-ix_1-x_2=0$, so $x_2=-ix_1$ and $\mathbf{x}=\begin{bmatrix}1\\-i\end{bmatrix}$. Check: $R\mathbf{x}=\begin{bmatrix}i\\1\end{bmatrix}=i\begin{bmatrix}1\\-i\end{bmatrix}$ โ (since $i\cdot(-i)=1$).
In general The rotation through angle $\theta$, $\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$, has $\det(R-\lambda I)=\lambda^2-2\cos\theta\,\lambda+1$ and eigenvalues $\cos\theta\pm i\sin\theta$. They are real only when $\sin\theta=0$: $\theta=0$ ($\lambda=1$, the identity) or $\theta=\pi$ ($\lambda=-1$, a half-turn). The trace rule still works: $(\cos\theta+i\sin\theta)+(\cos\theta-i\sin\theta)=2\cos\theta=\operatorname{tr}R$.
Exercises
I
Checking eigenvectors and eigenvalues. In parts 1โ4, $A=\begin{bmatrix}3&2\\3&-2\end{bmatrix}$; decide whether the vector is an eigenvector of $A$ and, if so, give its eigenvalue.
1 $\begin{bmatrix}2\\1\end{bmatrix}$
2 $\begin{bmatrix}1\\-3\end{bmatrix}$
3 $\begin{bmatrix}1\\1\end{bmatrix}$
4 $\begin{bmatrix}0\\0\end{bmatrix}$
5 Is $\lambda=2$ an eigenvalue of $B=\begin{bmatrix}3&1\\1&3\end{bmatrix}$? If so, find an eigenvector.
6 Is $\lambda=3$ an eigenvalue of $C=\begin{bmatrix}1&2&0\\0&3&0\\2&-4&2\end{bmatrix}$? If so, find an eigenvector.
7 Is $\lambda=4$ an eigenvalue of the matrix $C$ of part 6?
8 Is $(1,0,-1)$ an eigenvector of $M=\begin{bmatrix}2&1&0\\1&2&1\\0&1&2\end{bmatrix}$? If so, for which eigenvalue?
9 Is $(1,1,1)$ an eigenvector of the matrix $M$ of part 8?
3 $A\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}5\\1\end{bmatrix}$, not a multiple of $(1,1)$: no.
4 No โ the zero vector is never an eigenvector, by definition (even though $A\mathbf{0}=\lambda\mathbf{0}$ for every $\lambda$).
5 $B-2I=\begin{bmatrix}1&1\\1&1\end{bmatrix}$ has determinant $0$, so yes. $x_1+x_2=0$ gives the eigenvector $(1,-1)$; check $B(1,-1)=(2,-2)$ โ.
6 $C-3I=\begin{bmatrix}-2&2&0\\0&0&0\\2&-4&-1\end{bmatrix}$ has a zero row, so $\det(C-3I)=0$: yes. From row 1, $x_1=x_2$; from row 3, $2x_1-4x_2-x_3=0$, so $x_3=-2x_2$. Eigenvector $(1,1,-2)$; check $C(1,1,-2)=(3,3,-6)$ โ.
7 $C-4I=\begin{bmatrix}-3&2&0\\0&-1&0\\2&-4&-2\end{bmatrix}$. Expanding along row 2: $\det(C-4I)=(-1)\left[(-3)(-2)-0\cdot2\right]=-6\ne0$. So $4$ is not an eigenvalue. (In fact $\det(C-\lambda I)=(1-\lambda)(3-\lambda)(2-\lambda)$, so the eigenvalues are $1,2,3$.)
9 $M(1,1,1)=(3,4,3)$, not a multiple of $(1,1,1)$: no.
II
For each matrix find the characteristic polynomial, the real eigenvalues, and a basis for each eigenspace. State the algebraic and geometric multiplicity of any repeated eigenvalue.
5 $p(\lambda)=(4-\lambda)^2$. $A-4I=O$, so every vector is an eigenvector: $E_4=\mathbb{R}^2$ with basis $\{(1,0),(0,1)\}$. Algebraic and geometric multiplicity are both $2$.
6 $p(\lambda)=(1-\lambda)^2+4=\lambda^2-2\lambda+5$, with discriminant $4-20=-16\lt0$: no real eigenvalues. (Over $\mathbb{C}$, $\lambda=1\pm2i$; for $1+2i$ an eigenvector is $(1,i)$.) Geometrically the matrix is a rotation combined with scaling by $\sqrt5$.
7 Lower triangular, so $p(\lambda)=(2-\lambda)(3-\lambda)(-1-\lambda)$ and the eigenvalues are $2,3,-1$. $E_2$: $A-2I=\begin{bmatrix}0&0&0\\-1&1&0\\4&5&-3\end{bmatrix}$ gives $x_1=x_2$, $9x_2=3x_3$: basis $\{(1,1,3)\}$. $E_3$: $A-3I=\begin{bmatrix}-1&0&0\\-1&0&0\\4&5&-4\end{bmatrix}$ gives $x_1=0$, $5x_2=4x_3$: basis $\{(0,4,5)\}$. $E_{-1}$: $A+I=\begin{bmatrix}3&0&0\\-1&4&0\\4&5&0\end{bmatrix}$ gives $x_1=x_2=0$: basis $\{(0,0,1)\}$.
8 Expanding along row 2: $p(\lambda)=(1-\lambda)\left[(1-\lambda)^2-1\right]=(1-\lambda)(-\lambda)(2-\lambda)$. Eigenvalues $0,1,2$. $E_0$: $x_1+x_3=0$, $x_2=0$: basis $\{(1,0,-1)\}$. $E_1$: $x_1=x_3=0$: basis $\{(0,1,0)\}$. $E_2$: $x_1=x_3$, $x_2=0$: basis $\{(1,0,1)\}$. (The eigenvalue $0$ tells us the matrix is singular โ two columns are equal.)
9 $p(\lambda)=-(\lambda-2)^2(\lambda-8)$ (subtract row 2 from rows 1 and 3, or note that $A-2I$ is the all-twos matrix). $E_2$: $A-2I$ reduces to $x_1+x_2+x_3=0$, basis $\{(-1,1,0),(-1,0,1)\}$ โ algebraic and geometric multiplicity $2$. $E_8$: basis $\{(1,1,1)\}$. Check: $2+2+8=12=\operatorname{tr}A$ โ, $2\cdot2\cdot8=32=\det A$ โ.
10 Triangular: $p(\lambda)=(3-\lambda)^3$, so $\lambda=3$ with algebraic multiplicity $3$. $A-3I=\begin{bmatrix}0&0&0\\1&0&0\\0&0&0\end{bmatrix}$ has rank $1$, giving $x_1=0$ with $x_2$, $x_3$ free: basis $\{(0,1,0),(0,0,1)\}$, geometric multiplicity $2$.
III
Properties and theory.
1 A $3\times3$ matrix $A$ has eigenvalues $2$, $-1$ and $3$. Find $\det A$ and $\operatorname{tr}A$.
2 For the matrix of part 1, find the eigenvalues of $A^2$ and of $A^{-1}$.
3 For the matrix of part 1, find the eigenvalues of $A+4I$ and of $5A$.
4 For the matrix of part 1, is $A-3I$ invertible? Explain without computing anything.
5 Show that $0$ is an eigenvalue of $\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$ and find a corresponding eigenvector.
6 Show that $A$ and $A^{T}$ have the same characteristic polynomial. Then use $A=\begin{bmatrix}1&1\\0&2\end{bmatrix}$ to show that they need not have the same eigenvectors.
7 Suppose $A^2=O$. Show that $0$ is the only possible eigenvalue of $A$.
8 Suppose $A^2=A$ (a projection). Show that every eigenvalue of $A$ is $0$ or $1$.
9 For which real $k$ does $\begin{bmatrix}2&k\\1&2\end{bmatrix}$ have (a) two different real eigenvalues, (b) one repeated eigenvalue, (c) no real eigenvalues?
10True or false? โThe eigenvalues of $A+B$ are the sums of the eigenvalues of $A$ and of $B$.โ Justify your answer.
๐ก Show solutionsโถ
1 $\det A=2\cdot(-1)\cdot3=-6$ and $\operatorname{tr}A=2-1+3=4$ (Theorem 7.4(5)).
2 $A^2$: $4,1,9$. $A^{-1}$ exists because $0$ is not an eigenvalue; its eigenvalues are $\tfrac12,-1,\tfrac13$.
3 $A+4I$: $6,3,7$. $5A$: if $A\mathbf{x}=\lambda\mathbf{x}$ then $(5A)\mathbf{x}=5\lambda\mathbf{x}$, so $10,-5,15$.
4 No. The eigenvalues of $A-3I$ are $2-3$, $-1-3$, $3-3$, i.e. $-1,-4,0$. Since $0$ is an eigenvalue, $A-3I$ is singular.
5 Column 2 is the average of columns 1 and 3, so the columns are dependent, $\det A=0$, and $0$ is an eigenvalue. Solving $A\mathbf{x}=\mathbf{0}$: the rows reduce to $x_1-x_3=0$, $x_2+2x_3=0$, so $\mathbf{x}=(1,-2,1)$. Check: $1-4+3=0$, $4-10+6=0$, $7-16+9=0$ โ.
6 $\det(A^{T}-\lambda I)=\det\left((A-\lambda I)^{T}\right)=\det(A-\lambda I)$, since $I^{T}=I$ and a matrix and its transpose have the same determinant. For $A=\begin{bmatrix}1&1\\0&2\end{bmatrix}$: $A(1,1)=(2,2)$, so $(1,1)$ is an eigenvector of $A$ for $\lambda=2$. But $A^{T}=\begin{bmatrix}1&0\\1&2\end{bmatrix}$ has $A^{T}(1,1)=(1,3)$; its eigenvector for $\lambda=2$ is $(0,1)$.
7 If $A\mathbf{x}=\lambda\mathbf{x}$ with $\mathbf{x}\ne\mathbf{0}$, then $\mathbf{0}=A^2\mathbf{x}=\lambda^2\mathbf{x}$, so $\lambda^2=0$ and $\lambda=0$.
8 If $A\mathbf{x}=\lambda\mathbf{x}$, $\mathbf{x}\ne\mathbf{0}$: $\lambda\mathbf{x}=A\mathbf{x}=A^2\mathbf{x}=\lambda^2\mathbf{x}$, so $(\lambda^2-\lambda)\mathbf{x}=\mathbf{0}$, hence $\lambda(\lambda-1)=0$.
10 False. Take $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $B=\begin{bmatrix}0&0\\1&0\end{bmatrix}$: each has only the eigenvalue $0$, but $A+B=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ has eigenvalues $1$ and $-1$. (Eigenvalues do add when $A$ and $B$ share an eigenvector, as in Theorem 7.4(4) with $B=cI$.)
IV
Geometry and applications.
1 Without any algebra, find the eigenvalues and eigenvectors of the reflection in the line $y=x$, $\begin{bmatrix}0&1\\1&0\end{bmatrix}$. Then confirm with the characteristic equation.
2 Do the same for the projection onto the $x$-axis, $\begin{bmatrix}1&0\\0&0\end{bmatrix}$.
3 Find the eigenvalues and eigenspaces of the shear $\begin{bmatrix}1&k\\0&1\end{bmatrix}$ with $k\ne0$, and give both multiplicities.
4Population. A bird population has juveniles $j_k$ and adults $a_k$ in year $k$. Each adult produces on average $3$ juveniles that survive to the next count, half of the juveniles survive to become adults, and half of the adults survive. So $\begin{bmatrix}j_{k+1}\\a_{k+1}\end{bmatrix}=\begin{bmatrix}0&3\\0.5&0.5\end{bmatrix}\begin{bmatrix}j_k\\a_k\end{bmatrix}$. Find the eigenvalues and eigenvectors, and interpret the positive eigenvalue and its eigenvector.
5Materials. A thin rubber sheet is deformed so that the point $\mathbf{x}$ moves to $A\mathbf{x}$ with $A=\begin{bmatrix}3&1\\1&3\end{bmatrix}$. Find the two directions in which fibres of the sheet are stretched without turning (the principal directions) and the stretch factors. By what factor is area multiplied?
๐ก Show solutionsโถ
1 Vectors on the mirror line $y=x$ are unchanged: $\lambda=1$, eigenvector $(1,1)$. Vectors perpendicular to it are reversed: $\lambda=-1$, eigenvector $(1,-1)$. Check: $\det\begin{bmatrix}-\lambda&1\\1&-\lambda\end{bmatrix}=\lambda^2-1=(\lambda-1)(\lambda+1)$ โ.
2 Vectors on the $x$-axis are unchanged: $\lambda=1$, eigenvector $(1,0)$. Vectors on the $y$-axis are sent to $\mathbf{0}$: $\lambda=0$, eigenvector $(0,1)$. Check: $p(\lambda)=(1-\lambda)(-\lambda)$ โ (diagonal matrix).
3 Triangular, so $p(\lambda)=(1-\lambda)^2$ and $\lambda=1$ with algebraic multiplicity $2$. $A-I=\begin{bmatrix}0&k\\0&0\end{bmatrix}$ with $k\ne0$ forces $x_2=0$: $E_1=\operatorname{span}\{(1,0)\}$, geometric multiplicity $1$. Every other vector is slid sideways, off its line.
4 $p(\lambda)=\lambda^2-0.5\lambda-1.5=(\lambda-1.5)(\lambda+1)$. For $\lambda=1.5$: $\begin{bmatrix}-1.5&3\\0.5&-1\end{bmatrix}$ gives $j=2a$, eigenvector $(2,1)$. For $\lambda=-1$: $\begin{bmatrix}1&3\\0.5&1.5\end{bmatrix}$ gives $j=-3a$, eigenvector $(3,-1)$. Interpretation: a population with twice as many juveniles as adults keeps exactly that mix and grows by $50\%$ per year; Section 7.3 shows that every starting population approaches this growth rate and this $2:1$ age mix in the long run.
5 $p(\lambda)=(3-\lambda)^2-1=(\lambda-2)(\lambda-4)$. $\lambda=4$: eigenvector $(1,1)$; $\lambda=2$: eigenvector $(1,-1)$. Fibres along the diagonal $y=x$ are stretched by a factor $4$, fibres along $y=-x$ by a factor $2$. Area is multiplied by $|\det A|=8=4\times2$.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. At what frequencies does a two-storey building naturally sway, and what do its two โmode shapesโ look like?
Why it matters. If ground shaking during an earthquake โ or the rhythm of a crowd or of machinery โ matches a natural frequency, the swaying grows dramatically (resonance). Structural engineers in earthquake zones such as Japan, Tรผrkiye, Nepal, Chile and New Zealand compute these frequencies as eigenvalues before a building is approved. The equations of motion use calculus, but the frequencies come from a pure eigenvalue problem.
What to do.
Model each floor as a mass $m=5000$ kg joined by columns of stiffness $k=2\times10^6$ N/m (illustrative values, not data for a real building). The natural frequencies satisfy $K\mathbf{x}=\omega^2m\,\mathbf{x}$ with $K=k\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$, so $\omega^2=\tfrac km\lambda=400\lambda$, where $\lambda$ is an eigenvalue of $S=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}$.
Find the eigenvalues of $S$ exactly, then $\omega=20\sqrt\lambda$ rad/s and the frequency $f=\omega/(2\pi)$ in hertz.
Find an eigenvector for each eigenvalue, scaled so the first-floor entry is $1$. Sketch each mode: do the floors move in the same direction or in opposite directions?
Repeat with the top-floor stiffness halved, $S'=\begin{bmatrix}1.5&-0.5\\-0.5&0.5\end{bmatrix}$. Does the lowest frequency go up or down?
Tools: Calculator; optionally a spreadsheet or Python/GeoGebra to check eigenvalues.
Hand in: A half-page note with both frequencies, sketches of both mode shapes, and one sentence on which frequency an earthquake engineer worries about most and why.
Going further. Find the natural frequency of a footbridge in your city or a famous case (the London Millennium Bridge wobble, 2000) and compare it with typical walking pace (about $2$ steps per second).
Instructor note: $\lambda=\tfrac{3\mp\sqrt5}{2}\approx0.382,\ 2.618$, giving $\omega\approx12.36$ and $32.36$ rad/s, $f\approx1.97$ Hz and $5.15$ Hz. Mode shapes $(1,\ 1.618)$ (floors in step) and $(1,\,-0.618)$ (floors opposite). Softening the top floor lowers the frequencies. Students should notice that the lowest mode is close to walking and typical earthquake frequencies.
๐งช Self-check quiz
0/5 correct
Q1. If $A\mathbf{v}=3\mathbf{v}$ for some vector $\mathbf{v}\ne\mathbf{0}$, then
Q2. The eigenvalues of $\begin{bmatrix}2&5\\0&-1\end{bmatrix}$ are
Q3. The scalar $\lambda$ is an eigenvalue of $A$ exactly when
Q4. A $3\times3$ matrix has eigenvalues $1$, $2$ and $4$. Its determinant is
Q5. An eigenvalue of a $3\times3$ matrix has algebraic multiplicity $3$. Its geometric multiplicity can be