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Wize University Linear Algebra Textbook > Eigenvalues and Eigenvectors (Spectral Theory)

Eigenvalues and Eigenvectors

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Eigenvalues and Eigenvectors

Let An×nA_{n \times n} be a square matrix. We say that x⃗∈Rn\vec x \in \reals^n is an eigenvector of AA with associated eigenvalue λ∈R\lambda \in \reals if:
Ax⃗=λx⃗\boxed{\quad A\vec x = \lambda\vec x \quad}
Wize Concept
Idea: the eigenvectors of AA are the vectors that only get stretched by a factor of λ\lambda when multiplied by AA.

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Characteristic Polynomial

The characteristic polynomial of matrix An×nA_{n \times n} is a degree nn polynomial defined to be:
pA(λ)=det(A−λI)\boxed{\quad p_A(\lambda)=\det{A-\lambda I} \quad}
The roots of the characteristic polynomial are the eigenvalues of AA. These are the values of λ\lambda such that pA(λ)=0p_A(\lambda)=0.
An eigenvalue's algebraic multiplicity is the number of times it appears as a root (the exponent).
Example
The characteristic polynomial pA(λ)=(λ+1)(λ−3)2p_A(\lambda)=(\lambda+1)(\lambda-3)^2 has two roots:
  • λ1=−1\lambda_1=-1, with algebraic multiplicity 1 since the factor (λ+1)(\lambda +1) has an exponent of 1
  • λ2=3\lambda_2=3, with algebraic multiplicity 2 since the factor (λ−3)2(\lambda-3)^{\bm{\colorOne{2}}} has an exponent of 2
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How to Find Eigenvalues

Given a square matrix AA, set the characteristic polynomial equal to 00 and solve for λ\lambda:
det(A−λI)=0or, equivalently:det(λI−A)=0\begin{array}{c} \boxed{\quad \det{A-\lambda I}=0 \quad}\\[0.8em] \text{or, equivalently:}\\[0.8em] \boxed{\quad \det{\lambda I - A}=0 \quad}\\[0.8em] \end{array}
Wize Concept
If det(A−λI)=0\det{A-\lambda I}=0, we can always find non-trivial solution vectors x⃗\vec x to the equation Ax⃗=λx⃗A\vec x = \lambda\vec x.

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How to Find Eigenvectors

For each λ\lambda, we find the associated eigenvectors x⃗\vec x by finding all solutions to the linear system:
(A−λI) x⃗=0⃗or, equivalently:(λI−A) x⃗=0⃗\begin{array}{c} \boxed{\quad (A-\lambda I)\,\vec x=\vec 0 \quad}\\[0.8em] \text{or, equivalently:}\\[0.8em] \boxed{\quad (\lambda I - A)\,\vec x=\vec 0 \quad}\\[0.8em] \end{array}
Wize Concept
This system has infinitely many solutions because λ\lambda was chosen so that det(A−λI)=0\text{det}(A-\lambda I)=0.
The number of basic solutions to the system is the geometric multiplicity of λ\lambda.
We may use Gauss-Jordan elimination to solve the system of linear equations, but we can often solve by inspection!

Shortcut for 2×2\colorThree{2 \times 2} Matrices
Given a 2×22 \times 2 matrix AA and an eigenvalue λ\lambda, we can find eigenvalues easily by inspection:
A−λI=[ab??]  ⟹    x⃗=[−ba]    (or any scalar multiple)A-\lambda I = \begin{bmatrix} a & b\\ ? & ?\\ \end{bmatrix} \quad\implies \quad \boxed{\ \ \vec x = \begin{bmatrix} -b\\ a\\ \end{bmatrix}\ \ } \ \ \text{(or any scalar multiple)}
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Example: Eigenvalues and Eigenvectors

Find the eigenvalues and eigenvectors of A=[−332−4]A=\left[\begin{array}{rr} -3&3\\2&-4 \end{array}\right].

Find the Eigenvalues

We solve the equation:
det(A−λI)=0∣−3−λ32−4−λ∣=0(−3−λ)(−4−λ)−(3)(2)=0λ2+7λ+12−6=0λ2+7λ+6=0(λ+6)(λ+1)=0\begin{array}{rcl} \det{A-\lambda I}&=&0\\[1em] \left|\begin{array}{cc}-3-\lambda & 3\\2&-4-\lambda\end{array}\right|&=&0\\[1em] (-3-\lambda)(-4-\lambda)-(3)(2)&=&0\\[1em] \lambda^2+7\lambda+12-6&=&0\\[1em] \lambda^2+7\lambda+6&=&0\\[1em] (\lambda+6)(\lambda+1)&=&0 \end{array}
Then the eigenvalues are λ1=−6\boxed{ \lambda_1 = -6 } and λ2=−1\boxed{ \lambda_2=-1 }. Note that each eigenvalue has algebraic multiplicity 1.

Find the Eigenvectors

Case 1: λ1=−6\colorThree{\lambda_1 = -6}
We solve the linear system: (A−(−6)I)x⃗=0⃗(A-(\colorThree{-6})I)\vec x=\vec 0.
The augmented matrix of this system is:
[−3−(−6)302−4−(−6)0]=[330220]  →RREF  [110000]\left[\begin{array}{cc|r} -3-(-6)& 3 &0\\ 2&-4-(-6)& 0 \end{array}\right] = \left[\begin{array}{rr|r} 3& 3 &0\\ 2&2& 0 \end{array}\right] \ \ \xrightarrow{RREF}\ \ \left[\begin{array}{rr|r} 1& 1 &0\\ 0&0& 0 \end{array}\right]
From here we find that the solution is any vector of the form: [−tt]=t[−11]\left[\begin{array}{r}-t\\t\end{array}\right]=t\left[\begin{array}{r}-1\\1\end{array}\right]
So an eigenvector associated with λ1=−6\lambda_1 = -6 is: x⃗1=[−11]\boxed{ \vec x_1=\left[\begin{array}{r}-1\\1\end{array}\right] }
Case 2: λ2=−1\colorThree{\lambda_2 = -1}
We solve the linear system: (A−(−1)I)x⃗=0⃗(A-(\colorThree{-1})I)\vec x = \vec 0
The augmented matrix of this system is:
[−3−(−1)302−4−(−1)0]=[−2302−30]  →RREF  [1−320000]\left[\begin{array}{cc|r} -3-(-1)& 3 &0\\ 2&-4-(-1)& 0 \end{array}\right] = \left[\begin{array}{rr|r} -2& 3 &0\\ 2&-3& 0 \end{array}\right] \ \ \xrightarrow{RREF}\ \ \left[\begin{array}{rr|r} 1& -\frac{3}{2} &0\\ 0&0& 0 \end{array}\right]
From here we find that the solution is any vector of the form: [32tt]=t[321]\left[\begin{array}{c}\frac{3}{2}t\\t\end{array}\right]=t\left[\begin{array}{r}\frac{3}{2}\\1\end{array}\right]
We prefer not to have fractions, so choosing t=2t=2 gives us the basic eigenvector: x⃗2=[32]\boxed{ \vec x_2=\left[\begin{array}{r}3\\2\end{array}\right] }
Let's check our work:
Ax⃗1=[−332−4][−11]=[6−6]=−6⋅x⃗1    ✓A\vec x_1=\left[\begin{array}{rr} -3&3\\2&-4 \end{array}\right]\left[\begin{array}{r}-1\\1\end{array}\right] = \left[\begin{array}{r}6\\-6\end{array}\right]=-6\cdot\vec x_1 \;\;\textcolor{green}{\checkmark}
Ax⃗2=[−332−4][32]=[−3−2]=−1⋅x⃗2    ✓A\vec x_2=\left[\begin{array}{rr} -3&3\\2&-4 \end{array}\right]\left[\begin{array}{r}3\\2\end{array}\right] = \left[\begin{array}{r}-3\\-2\end{array}\right]=-1\cdot\vec x_2\;\;\textcolor{green}{\checkmark}

Practice: Eigenvalues and Eigenvectors

Find all eigenvalues and eigenvectors of the matrix A=[010010−111]A=\left[\begin{array}{rrr} 0&1&0\\ 0&1&0\\ -1&1&1 \end{array}\right].
State the eigenvalues of AA in ascending order, along with their algebraic multiplicity.

Practice: Eigenvalues and Eigenvectors

Determine which of the following vectors are eigenvectors of A=[133−3−5−3331]A=\left[\begin{array}{rrr} 1&3&3\\-3&-5&-3\\3&3&1 \end{array}\right].
[Select all that apply]
Extra Practice
Eigenvalues and Eigenvectors
Practice: Eigenvalues and Eigenvectors
Find all eigenvalues and eigenvectors of the matrix A=[010010−111]A=\left[\begin{array}{rrr} 0&1&0\\ 0&1&0\\ -1&1&1 \end{array}\right].
Eigenvalues and Eigenvectors
Practice: Eigenvalues and Eigenvectors
Determine which of the following vectors are eigenvectors of A=[133−3−5−3331]A=\left[\begin{array}{rrr} 1&3&3\\-3&-5&-3\\3&3&1 \end{array}\right].
[Select all that apply]
133 - FML 2 - 18.1W e.g. 6
Find the characteristic polynomial CA‾(x)C_{\A}(x) for the matrix   ⁣A‾=[31−122−1220]\ul{A} = % \begin{bmatrix} 3 & 1 & -1 \\ 2 & 2 & -1 \\ 2 & 2 & 0 \\ \end{bmatrix}.
Practice Problems
The matrix
A‾=[31−122−1220]\A=\begin{bmatrix} 3&1&-1\\ 2&2&-1\\ 2&2&0 \end{bmatrix}
has characteristic polynomial:
Practice Problems
The matrix
  ⁣A‾=[5−6−6−1423−6−4]\ul{A} = \begin{bmatrix} 5 & -6 & - 6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix}
has characteristic polynomial
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg6}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{6}$_
ξ⃗1=[31]\vXi_1 = \colvec{3}{1} and ξ⃗2=[12]\vXi_2 = \colvec{1}{2} are eigenvectors for the matrix
A‾=[53−212] \A = \sm{5}{3}{-2}{12}
where λ1=6\lambda_1 = 6 and λ2=11\lambda_2 = 11.
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg5}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{5}$_
Find the eigenvectors and corresponding eigenvalues of the matrix
A‾=[53−212] \A = \sm{5}{3}{-2}{12}
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg9}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{9}$_
Show that if x⃗\vx is an eigenvector of   ⁣M ⁣‾  \M with eigenvalue kk, then it is also an eigenvector of the matrix B ⁣‾ =10  ⁣M ⁣‾  \B = 10\M.
Use your result to find the eigenvalue of x⃗\vx with respect to the matrix B ⁣‾ \B, i.e. B ⁣‾  x⃗=λ x⃗\B \, \vx = \lambda \, \vx where λ=\lambda =
10k
19.4F_Final_Builder_Ch_12.6_Extra_Eigen_Things_$\tkcth{eg8}$_$\key{Final}$_Builder_$\tkcth{12.6.}\tkcf{8}$_
Find a basis for the eigenspace of the matrix A‾=[00−2121103]\A = \begin{bmatrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{bmatrix} corresponding to the eigenvalue λ2=2\lambda_2=2 (i.e. find the vector(s) that span the space A‾x⃗=2x⃗\A \vx = 2\vx).
19.4F_Final_Builder_Ch_12.6_Extra_Eigen_Things_$\tkcth{eg7}$_$\key{Final}$_Builder_$\tkcth{12.6.}\tkcf{7}$_
Find a basis for the eigenspace of the matrix
A‾=[00−2121103]\A = \begin{bmatrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{bmatrix}
corresponding to the eigenvalue λ1=1\lambda_1=1 (i.e. find the vector(s) that span the space A‾x⃗=2x⃗\A \vx = 2\vx).
19.4F_Final_Builder_Ch_12.6_Extra_Eigen_Things_$\tkcth{eg6}$_$\key{Final}$_Builder_$\tkcth{12.6.}\tkcf{6}$_
Find the eigenvalues of the matrix A‾=[00−2121103]\A = \begin{bmatrix} 0 & 0 & -2 \\ 1 & 2 & 1 \\ 1 & 0 & 3 \end{bmatrix}.
19.4F_Final_Builder_Ch_12.6_Extra_Eigen_Things_$\tkcth{eg4}$_$\key{Final}$_Builder_$\tkcth{12.6.}\tkcf{4}$_
Find the eigenvalues and corresponding eigenvectors of the matrix A‾=[−1320].\A = \sm{-1}{3}{2}{0}.
19.4F_Final_Builder_Ch_12.6_Extra_Eigen_Things_$\tkcth{eg3}$_$\key{Final}$_Builder_$\tkcth{12.6.}\tkcf{3}$_
Find the eigenvector of the matrix A‾=[308−1]\A = \sm{3}{0}{8}{-1} corresponding to the eigenvalue λ2=3\lambda_2 = 3.
19.4F_Final_Builder_Ch_12.6_Extra_Eigen_Things_$\tkcth{eg1}$_$\key{Final}$_Builder_$\tkcth{12.6.}\tkcf{1}$_
Find the eigenvalues of the matrix A‾=[308−1].\A = \sm{3}{0}{8}{-1}.
Practice: Eigenvalues and Eigenvectors
Given A=[6−31211111]A=\begin{bmatrix}6&-3&1\\2&1&1\\1&1&1\end{bmatrix} and the eigenvector [875]T\begin{bmatrix}8&7&5\end{bmatrix}^T, find the eigenvalue to which it is associated.
$\tkct{Mock F1}$ $\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W - e.g. 27
What are the eigenvectors of the standard matrix representation of the transformation T\bcb{T} that projects a vector v⃗\bcb{\vec{v}} onto the line x+y=0\bcb{x + y = 0}?
NB it does not matter which vector / value is called "1" or "2", so long as the vector and value you choose, match. If there is no way to create a match, select the appropriate option.
Vector Space and Subspace