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5.1 Eigenvalues and Eigenvectors: Class Notes

Consider the equation $\left[\begin{array}{ccc}6& 3& -8\\ 0& -2& 0\\ 1& 0& -3\end{array}\right]\left[\begin{array}{c}8\\ 0\\ 1\end{array}\right]=5\left[\begin{array}{c}8\\ 0\\ 1\end{array}\right]$.

Definition

If $A$ is an $n\times n$ matrix, then a nonzero vector $𝐱$ in $R^n$ is called an eigenvector of $A$ (or of the matrix operator $T_A$) if $A𝐱$ is a scalar multiple of $𝐱$; that is, $A𝐱=\lambda 𝐱$ for some scalar $\lambda$. The scalar $\lambda$ is called an eigenvalue of $A$ (or of $T_A$), and $𝐱$ is said to be an eigenvector corresponding to $\lambda$.

In the equation above, $\lambda =5$ is an eigenvalue of the matrix, and $\overrightarrow{x}=\left[\begin{array}{c}8\\ 0\\ 1\end{array}\right]$ is its associated eigenvector.

Note that if $A\overrightarrow{x}=\lambda \overrightarrow{x}$, then $\lambda \overrightarrow{x}-A\overrightarrow{x}=\overrightarrow{0}$. Then $(\lambda I-A)\overrightarrow{x}=\overrightarrow{0}$. Since $\overrightarrow{x}\ne \overrightarrow{0}$ (by the definition, an eigenvector is nonzero), we seek nontrivial solutions.

Theorem 5.1.1

If $A$ is an $n\times n$ matrix, then $\lambda$ is an eigenvalue of $A$ if and only if it satisfies the equation $det(\lambda I-A)=0$.

Definition

If $A$ is an $n\times n$ matrix, then the characteristic equation of $A$ is $det(\lambda I-A)=0$.

Tip: The characteristic equation is sometimes written $det(A-\lambda I)=0$.

Definition

The characteristic polynomial of an $n\times n$ matrix $A$ results from expanding the characteristic equation of $A$ and has the form $p(\lambda )={\lambda }^n+c_1{\lambda }^{n-1}+\dots +c_n$.

Definition

The solution space of the system $(\lambda I-A)𝐱=\mathrm{𝟎}$ is called the eigenspace of $A$ corresponding to $\lambda$.

#6

In each part, find the characteristic equation, the eigenvalues, and bases for the eigenspaces of the matrix.

a. $\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]$    b. $\left[\begin{array}{cc}2& -3\\ 0& 2\end{array}\right]$    c. $\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$    d. $\left[\begin{array}{cc}1& 2\\ -2& -1\end{array}\right]$

Solution

a. Form $\lambda I-A$ and set its determinant equal to zero.

$$\lambda I-A=\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]-\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}\lambda -2& -1\\ -1& \lambda -2\end{array}\right]$$

$$\left|\begin{array}{cc}\lambda -2& -1\\ -1& \lambda -2\end{array}\right|=0\Rightarrow (\lambda -2{)}^2-1=0\Rightarrow (\lambda -2{)}^2=1$$

The characteristic equation is $(\lambda -2{)}^2-1=0$. Then

$$\lambda -2=\pm 1\Rightarrow \lambda =1,3\text{(eigenvalues)}$$

Substitute ${\lambda }_1=1$ into $(\lambda I-A)\overrightarrow{x}=\overrightarrow{0}$ to find $\overrightarrow{x}$:

$$\left[\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right]\overrightarrow{x}=\overrightarrow{0}\Rightarrow x_1=-x_2\Rightarrow {\overrightarrow{x}}_1=\left[\begin{array}{c}1\\ -1\end{array}\right]t$$

Basis for the eigenspace: $\left\{\left[\begin{array}{c}1\\ -1\end{array}\right]\right\}$.

For ${\lambda }_2=3$:

$$\left[\begin{array}{cc}1& -1\\ -1& 1\end{array}\right]\overrightarrow{x}=\overrightarrow{0}\Rightarrow {\overrightarrow{x}}_2=\left[\begin{array}{c}1\\ 1\end{array}\right]t\Rightarrow \text{basis: }\left\{\left[\begin{array}{c}1\\ 1\end{array}\right]\right\}$$

Check: $\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}1\\ -1\end{array}\right]=\left[\begin{array}{c}1\\ -1\end{array}\right]$, and $\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}3\\ 3\end{array}\right]=3\left[\begin{array}{c}1\\ 1\end{array}\right]$.

b. $A=\left[\begin{array}{cc}2& -3\\ 0& 2\end{array}\right]$

$$\left|\begin{array}{cc}\lambda -2& 3\\ 0& \lambda -2\end{array}\right|=0\Rightarrow (\lambda -2{)}^2=0$$

So $\lambda =2$, a repeated root (algebraic multiplicity 2).

For $\lambda =2$:

$$\left[\begin{array}{cc}0& 3\\ 0& 0\end{array}\right]\to \left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\Rightarrow x_2=0,\overrightarrow{x}=\left[\begin{array}{c}1\\ 0\end{array}\right]t$$

Basis: $\left\{\left[\begin{array}{c}1\\ 0\end{array}\right]\right\}$. Only one basis vector, so the geometric multiplicity is 1.

c. $\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$

$$\left|\begin{array}{cc}\lambda -2& 0\\ 0& \lambda -2\end{array}\right|=0\Rightarrow (\lambda -2{)}^2=0$$

So $\lambda =2$ (algebraic multiplicity 2). For $\lambda =2$, $\lambda I-A=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$, so both $x_1$ and $x_2$ are free: $x_1=t$, $x_2=s$.

$$\overrightarrow{x}=\left[\begin{array}{c}1\\ 0\end{array}\right]t+\left[\begin{array}{c}0\\ 1\end{array}\right]s$$

Basis: $\{\left[\begin{array}{c}1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\end{array}\right]\}$ (geometric multiplicity 2).

d. $\left[\begin{array}{cc}1& 2\\ -2& -1\end{array}\right]$

$$\left|\begin{array}{cc}\lambda -1& -2\\ 2& \lambda +1\end{array}\right|=0\Rightarrow {\lambda }^2-1+4=0\Rightarrow {\lambda }^2=-3$$

Theorem 5.1.3

If $A$ is an $n\times n$ matrix, the following statements are equivalent.

• a) $\lambda$ is an eigenvalue of $A$.
• b) $\lambda$ is a solution of the characteristic equation $det(\lambda I-A)=0$.
• c) The system of equations $(\lambda I-A)𝐱=\mathrm{𝟎}$ has nontrivial solutions.
• d) There is a nonzero vector $𝐱$ such that $A𝐱=\lambda 𝐱$.

#9

Find the characteristic equation, the eigenvalues, and bases for the eigenspaces of the matrix.

$$\left[\begin{array}{ccc}6& 3& -8\\ 0& -2& 0\\ 1& 0& -3\end{array}\right]$$

Solution

Set $det(\lambda I-A)=0$ and expand along the second row, which has only one nonzero entry.

$$\left|\begin{array}{ccc}\lambda -6& -3& 8\\ 0& \lambda +2& 0\\ -1& 0& \lambda +3\end{array}\right|=0$$

$$\begin{array}{cc}\hfill {\displaystyle (\lambda +2)\left|\begin{array}{cc}\lambda -6& 8\\ -1& \lambda +3\end{array}\right|}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +2)[(\lambda -6)(\lambda +3)+8]}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +2)({\lambda }^2-3\lambda -10)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +2{)}^2(\lambda -5)}& {\displaystyle =0}\hfill \end{array}$$

So $\lambda =-2$ (algebraic multiplicity 2) and $\lambda =5$.

For ${\lambda }_1=-2$:

$$\left[\begin{array}{ccc}-8& -3& 8\\ 0& 0& 0\\ -1& 0& 1\end{array}\right]\to \left[\begin{array}{ccc}-8& -3& 8\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]\to \left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$$

So $x_2=0$ and $x_1=x_3$, which gives ${\overrightarrow{x}}_1=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]t$. The basis is $\left\{\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]\right\}$.

For ${\lambda }_2=5$:

$$\left[\begin{array}{ccc}-1& -3& 8\\ 0& 7& 0\\ -1& 0& 8\end{array}\right]\to \left[\begin{array}{ccc}-1& -3& 8\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]\to \left[\begin{array}{ccc}1& 0& -8\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$$

So $x_2=0$ and $x_1=8x_3$, which gives ${\overrightarrow{x}}_2=\left[\begin{array}{c}8\\ 0\\ 1\end{array}\right]t$. The basis is $\left\{\left[\begin{array}{c}8\\ 0\\ 1\end{array}\right]\right\}$.

Tip: This is the eigenvector from the equation at the start of the section: $A\overrightarrow{x}=\lambda \overrightarrow{x}$ with $\lambda =5$ and $\overrightarrow{x}\ne \overrightarrow{0}$.

#19

In each part, find the eigenvalues and the corresponding eigenspaces of the stated matrix operator on $R^2$. Use geometric reasoning to find the answers. No computations are needed.

a. Reflection about the line $y=x$.
b. Orthogonal projection onto the $x$-axis.
c. Rotation about the origin through a positive angle of ${90}^{\circ }$.
d. Contraction with factor $k$ $(0\le k<1)$
e. Shear in the $x$-direction by a factor $k$ $(k\ne 0)$

Solution

In each part we look for nonzero vectors with $A\overrightarrow{x}=\lambda \overrightarrow{x}$: vectors that the operator sends to a scalar multiple of themselves.

a. If $A\overrightarrow{x}=\lambda \overrightarrow{x}$, then $\overrightarrow{x}=\left[\begin{array}{c}a\\ a\end{array}\right]$ (on the line $y=x$) or $\overrightarrow{x}=\left[\begin{array}{c}a\\ -a\end{array}\right]$ (perpendicular to it).

The line $y=x$ through the origin. A vector along the line (drawn in purple) is unchanged by the reflection. A vector perpendicular to the line, along $y=-x$ (drawn in blue), is reflected to its opposite.

If $\overrightarrow{x}=\left[\begin{array}{c}a\\ a\end{array}\right]$, the reflection leaves it unchanged: $\lambda =1$, with eigenspace $span\left\{\left[\begin{array}{c}1\\ 1\end{array}\right]\right\}$.

If $\overrightarrow{x}=\left[\begin{array}{c}a\\ -a\end{array}\right]$, the reflection reverses it: $\lambda =-1$, with eigenspace $span\left\{\left[\begin{array}{c}1\\ -1\end{array}\right]\right\}$.

b. $A\overrightarrow{x}=\lambda \overrightarrow{x}$ happens when $\overrightarrow{x}$ is on the $x$-axis, so it has the form $\left[\begin{array}{c}a\\ 0\end{array}\right]$, or on the $y$-axis.

A vector on the $x$-axis (blue) projects to itself. A vertical vector (green) projects to the zero vector.

On the $x$-axis: $\lambda =1$, eigenspace $span\left\{\left[\begin{array}{c}1\\ 0\end{array}\right]\right\}$.

On the $y$-axis: $\lambda =0$, eigenspace $span\left\{\left[\begin{array}{c}0\\ 1\end{array}\right]\right\}$.

c. A rotation through ${90}^{\circ }$ turns every nonzero vector to a perpendicular direction.

A vector (purple) and its image after a 90° rotation. The image is perpendicular to the original, so it is never a scalar multiple of it.

So $A\overrightarrow{x}=\lambda \overrightarrow{x}$ doesn't happen for $\overrightarrow{x}\ne \overrightarrow{0}$: the equation has no nonzero solution unless the angle of rotation is a multiple of $\pi$.

d. $A\overrightarrow{x}=k\overrightarrow{x}$ with $0\le k<1$. This applies to all vectors in $R^2$.

A vector (blue) and its image under the contraction (orange), which points in the same direction and is $k$ times as long.

So $\lambda =k$, with basis $\{(1,0),(0,1)\}$. The eigenspace is all of $R^2$.

e. A shear in the $x$-direction moves each vector horizontally by $k$ times its $y$-value: $\overrightarrow{x}=⟨x,y⟩$ goes to $⟨x+ky,y⟩$ (see Section 8.6).

The unit square (left) and its image under the shear (right). The green vector $(1,0)$ along the $x$-axis does not move. The corner $(1,1)$ moves to $(1+k,1)$, and $(0,1)$ moves to $(k,1)$ (the blue vector tilts).

If $\overrightarrow{x}=(a,0)$, then $A\overrightarrow{x}=\overrightarrow{x}$, so $\lambda =1$ with basis $(1,0)$. Shearing any other vector (one with $y\ne 0$) changes its direction, so the result is not a scalar multiple of the vector.

Theorem 5.1.2

If $A$ is an $n\times n$ triangular matrix (upper triangular, lower triangular, or diagonal), then the eigenvalues of $A$ are the entries on the main diagonal of $A$.

#14

Find the characteristic equation of the matrix by inspection.

$$\left[\begin{array}{cccc}9& -8& 6& 3\\ 0& -1& 0& 0\\ 0& 0& 3& 0\\ 0& 0& 0& 7\end{array}\right]$$

Solution

The matrix is upper triangular, so $\lambda I-A$ is too, and its determinant is the product of its diagonal entries.

#15

Find the eigenvalues and a basis for each eigenspace of the linear operator defined by the stated formula.

$T(x,y)=(x+4y,2x+3y)$

Solution

$$[T]=\left[\begin{array}{cc}1& 4\\ 2& 3\end{array}\right]\Rightarrow \left|\begin{array}{cc}\lambda -1& -4\\ -2& \lambda -3\end{array}\right|=0$$

$${\lambda }^2-4\lambda +3-8=0\Rightarrow {\lambda }^2-4\lambda -5=0\Rightarrow (\lambda -5)(\lambda +1)=0$$

So $\lambda =-1,5$.

$${\lambda }_1=-1:\left[\begin{array}{cc}-2& -4\\ -2& -4\end{array}\right]\Rightarrow {\overrightarrow{x}}_1=\left[\begin{array}{c}2\\ -1\end{array}\right]\Rightarrow \text{basis: }\left\{\left[\begin{array}{c}2\\ -1\end{array}\right]\right\}$$

$${\lambda }_2=5:\left[\begin{array}{cc}4& -4\\ -2& 2\end{array}\right]\Rightarrow {\overrightarrow{x}}_2=\left[\begin{array}{c}1\\ 1\end{array}\right]\Rightarrow \text{basis: }\left\{\left[\begin{array}{c}1\\ 1\end{array}\right]\right\}$$

Theorem 5.1.4

A square matrix $A$ is invertible if and only if $\lambda =0$ is not an eigenvalue of $A$.

Proof

Recall that $\lambda$ is an eigenvalue of $A$ if there is a nonzero vector $\overrightarrow{x}$ such that $A\overrightarrow{x}=\lambda \overrightarrow{x}$. Then $det(\lambda I-A)=0$.

($\Rightarrow$) Suppose $A$ is invertible. Then $det(A)\ne 0$, so $det(-A)=(-1{)}^{n}det(A)\ne 0$. But if $\lambda =0$ were an eigenvalue, then $det(0I-A)=det(-A)=0$. So $\lambda =0$ is not an eigenvalue of $A$.

($\Leftarrow$) If $\lambda =0$ is not an eigenvalue of $A$, then there is no nonzero vector $\overrightarrow{x}$ satisfying $A\overrightarrow{x}=\overrightarrow{0}$ (that is, $A\overrightarrow{x}=0\overrightarrow{x}$). So the system has only the trivial solution, and therefore $A$ is invertible.

Theorem 5.1.5 Equivalent Statements (extends Theorem 4.9.8)

If $A$ is an $n\times n$ matrix, then the following statements are equivalent.

• a) $A$ is invertible.
• b) $A𝐱=\mathrm{𝟎}$ has only the trivial solution.
• c) The reduced row echelon form of $A$ is $I_n$.
• d) $A$ is expressible as a product of elementary matrices.
• e) $A𝐱=𝐛$ is consistent for every $n\times 1$ matrix $𝐛$.
• f) $A𝐱=𝐛$ has exactly one solution for every $n\times 1$ matrix $𝐛$.
• g) $det(A)\ne 0$.
• h) The column vectors of $A$ are distinct and linearly independent.
• i) The row vectors of $A$ are distinct and linearly independent.
• j) The column vectors of $A$ span $R^n$.
• k) The row vectors of $A$ span $R^n$.
• l) The column vectors of $A$ form a basis for $R^n$.
• m) The row vectors of $A$ form a basis for $R^n$.
• n) $A$ has rank $n$.
• o) $A$ has nullity 0.
• p) The orthogonal complement of the null space of $A$ is $R^n$.
• q) The orthogonal complement of the row space of $A$ is $\{\mathrm{𝟎}\}$.
• r) $\lambda =0$ is not an eigenvalue of $A$.