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

Properties

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Properties of Eigenvalues and Eigenvectors

Properties

  • The eigenvalues of a triangular matrix (upper, lower, or diagonal) are the entries on its main diagonal E.g. [abc0de00f]\left[ \begin{array}{rrr} \colorTwo{a} & b & c\\ 0 & \colorTwo{d} & e\\ 0 & 0 & \colorTwo{f}\\ \end{array} \right] has eigenvalues λ1=a,λ2=d,λ3=f\lambda_1 = \colorTwo{a},\quad \lambda_2 = \colorTwo{d},\quad \lambda_3 = \colorTwo{f}
  • AA is invertible if and only if 0 is not an eigenvalue of AA
  • If x⃗\vec x is an eigenvector of AA with eigenvalue λ\lambda, then Amx⃗=λmx⃗A^m\vec x = \lambda^m\vec x
Checking Eigenvalues
For any square matrix An×nA_{n \times n} with eigenvalues λ1,λ2,…,λn\lambda_1, \lambda_2,\dots,\lambda_n:
  • det(A)=λ1×λ2×…×λn\text{det}(A)=\lambda_1\times\lambda_2 \times \ldots \times \lambda_n
  • tr(A)= a11+a22+⋯+ann=λ1+λ2 +…+λn\text{tr}(A)=\ a_{11}+a_{22}+\cdots+a_{nn}=\lambda_1+\lambda_{2\ }+\ldots+\lambda_n
Wize Concept
tr(A)\text{tr}(A) is the trace of a matrix An×nA_{n \times n}, which is defined as the sum of the entries on the main diagonal.

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Similar Matrices

Given An×nA_{n \times n} and any invertible matrix PP, then we say AA is similar to P−1APP^{-1}AP.
Letting B=P−1APB=P^{-1}AP, then BB is similar to AA which we denote B∼AB \sim A.
Similar matrices have the same eigenvalues!
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Example: Similar Matrices

Suppose A=[3241]A=\begin{bmatrix}3&2\\4&1\end{bmatrix} . Consider the matrix P=12[1−111]P=\dfrac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\1&1\end{bmatrix}, which has inverse P−1=PTP^{-1}=P^T.
Find the eigenvalues of AA using the invertible matrix PP.

We want to use the formula for similar matrices: B=P−1APB=P^{-1}AP.
First, find P−1P^{-1}:
P−1=PT=12[11−11]P^{-1} = P^T =\dfrac{1}{\sqrt{2}}\begin{bmatrix}1&1\\-1&1\end{bmatrix}
Now we can find a similar matrix using the formula:
P−1AP=12[11−11][3241]12[1−111]=12[731−1][1−111]=12[10−40−2]=[5−20−1]=B\begin{aligned} P^{-1}AP &= \dfrac{1}{\sqrt{2}}\begin{bmatrix}1&1\\-1&1\end{bmatrix} \begin{bmatrix}3&2\\4&1\end{bmatrix} \dfrac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\\[1.5em] &=\dfrac{1}{2}\begin{bmatrix}7&3\\1&-1\end{bmatrix}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\\[1.5em] &=\frac{1}{2}\begin{bmatrix}10&-4\\0&-2\end{bmatrix}\\[1.5em] &=\begin{bmatrix}5&-2\\0&-1\end{bmatrix}=B \end{aligned}
Since BB is upper triangular, its eigenvalues are on the main diagonal.
Therefore, the eigenvalues of BB are λ1=5\lambda_1 = 5 and λ2=−1\lambda_2 = -1. Since A∼BA \sim B, these are also the eigenvalues of AA.
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Example: Eigenvectors and Linear Transformations

Let L:R3→R3L:\mathbb{R}^3\to\mathbb{R}^3 be the linear transformation defined by:
L(x,y,z)=(−2x−y−3zx+zx+y+2z)L(x,y,z) = \left( \begin{array}{c} -2x-y-3z\\ x+z\\ x+y+2z\\ \end{array} \right)

Part A)

If it is known that x⃗=[1−10]\vec x= \begin{bmatrix} 1\\ -1\\ 0\\ \end{bmatrix} is an eigenvector of the transformation LL, find the vector (L∘L∘L∘L∘L)(x⃗)(L \circ L \circ L \circ L \circ L)(\vec x).
Since LL is linear, it can be represented by a matrix AA such that L(v⃗)=Av⃗L(\vec v) = A\vec v.
Let's start by finding the matrix AA that induces this transformation:
L(xyz)=(−2x−y−3zx+zx+y+2z)=(−2x−1y−3z1x+0y+1z1x+1y+2z)=[−2−1−3101112][xyz]L\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}-2x-y-3z\\x+z\\x+y+2z\end{array}\right) =\left(\begin{array}{r}-2x-1y-3z\\ 1x+0y+1z\\1x+1y+2z\end{array}\right) =\left[\begin{array}{rrr}-2&-1&-3\\1&0&1\\1&1&2\end{array}\right]\left[\begin{array}{c}x\\y\\z\end{array}\right]
  ⟹  A=[−2−1−3101112]\implies A=\left[\begin{array}{rrr}-2&-1&-3\\1&0&1\\1&1&2\end{array}\right]
Since x⃗\vec x is an eigenvector of LL, it must also be an eigenvector of AA. Let's find its eigenvalue:
L(x⃗)=Ax⃗=[−2−1−3101112][1−10]=[−110]=−1x⃗L(\vec x) = A\vec x = \left[\begin{array}{rrr}-2&-1&-3\\1&0&1\\1&1&2\end{array}\right] \left[\begin{array}{r}1\\-1\\0\end{array}\right] = \left[\begin{array}{r}-1\\1\\0\end{array}\right] = \textcolor{green}{\mathbf{-1}}\vec x
So x⃗\vec x is an eigenvector with associated eigenvalue λ=−1\lambda=\textcolor{green}{\mathbf{-1}}.
Then notice that the vector (L∘L∘L∘L∘L)(x⃗)(L \circ L \circ L \circ L \circ L)(\vec x) is equivalent to A5x⃗A^5\vec x, since applying LL five times is equivalent to multiplying by AA five times.
L(L(L(L(L(x⃗)))))=A5x⃗L(L(L(L(L(\vec x))))) = A^5 \vec x
and by a property of eigenvalues: A5x⃗=λ5x⃗A^5 \vec x = \lambda^5\vec x
Therefore:
(L∘L∘L∘L∘L)(x⃗)=A5x⃗=λ5x⃗=(−1)5x⃗=−x⃗=−[1−10]= [−110] (L \circ L \circ L \circ L \circ L)(\vec x) =A^5\vec x = \lambda^5\vec x = (\textcolor{green}{\bm{-1}})^5\vec x = -\vec x = -\begin{bmatrix} 1\\ -1\\ 0\\ \end{bmatrix} = \boxed{\ \begin{bmatrix} -1\\ 1\\ 0\\ \end{bmatrix}\ }
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Part B)

Given that two eigenvalues of LL are λ1=1\lambda_1=1 and λ2=−1\lambda_2 = -1, determine whether LL is invertible.
LL is invertible if AA is invertible, and AA is invertible if and only if 0 is not one of its eigenvalues.
The question tells us A=[−2−1−3101112]A=\left[\begin{array}{rrr}-2&-1&-3\\1&0&1\\1&1&2\end{array}\right] has eigenvalues λ1=1\lambda_1=1 and λ2=−1\lambda_2=-1 (since AA and LL must have the same eigenvalues).
From the properties of eigenvalues, the sum of the eigenvalues must equal the trace of matrix AA. In other words:
λ1+λ2+λ3=a11+a22+a331+(−1)+λ3=(−2)+0+2λ3=0\begin{aligned} \lambda_1+\lambda_2+\lambda_3 &=a_{11}+a_{22}+a_{33}\\[0.5em] 1+(-1)+\lambda_3 &=(-2)+0+2\\[0.5em] \lambda_3&=0 \end{aligned}
The third eigenvalue of AA is 0, so AA is not invertible, therefore LL is not invertible.

Practice: Similar Matrices

Consider the matrix A=[−1421]A= \left[ \begin{array}{rrr} -1 & 4\\ 2 & 1 \end{array} \right].
Use the invertible matrix P=[1101]P= \left[ \begin{array}{rrr} 1 & 1\\ 0 & 1 \end{array} \right] to find a matrix similar to AA.
Extra Practice
133 - FML 3 - 18.1W e.g. 19
If   ⁣A‾\bcb{\ul{A}} be a 3×3\bcb{3 \times 3} matrix where trace(  ⁣A‾)=−1\bcb{\text{trace}(\ul{A}) = -1} and det(  ⁣A‾)=−1\bcb{\det{\ul{A}} = -1}, then the possible eigenvalues of   ⁣A‾\bcb{\ul{A}} are:
133 - FML 3 - 18.1W e.g. 18
If   ⁣A‾\bcb{\ul{A}} is a 4×4\bcb{4 \times 4} matrix with eigenvalues λ1=λ2=λ3=1\bcb{\lambda_1 = \lambda_2 = \lambda_3 = 1} and λ4=2\bcb{\lambda_4 = 2}, which of the following statements are true?
133 - FML 3 - 18.1W e.g. 70.5
Suppose 5\bcb{5} and −2\bcb{-2} are eigenvalues of the matrix   ⁣A‾\bcb{\ul{A}}, which of the following statements must be true?
det(  ⁣A‾−5  ⁣I‾)=det(  ⁣A‾+2  ⁣I‾)\bcb{\det{\ul{A} - 5\ul{I}} = \det{\ul{A} + 2 \ul{I}}}.
133 - FML 3 - 18.1W e.g. 70.4
Suppose 5\bcb{5} and −2\bcb{-2} are eigenvalues of the matrix   ⁣A‾\bcb{\ul{A}}, which of the following statements must be true?
trace(  ⁣A‾−5  ⁣I‾)=trace(  ⁣A‾+2  ⁣I‾)\bcb{\text{trace}(\ul{A} - 5\ul{I}) = \text{trace}(\ul{A} + 2\ul{I})}.
133 - FML 3 - 18.1W e.g. 70.3
Suppose 5\bcb{5} and −2\bcb{-2} are eigenvalues of the matrix   ⁣A‾\bcb{\ul{A}}, which of the following statements must be true?
det(  ⁣A‾−5  ⁣I‾)=0\bcb{\det{\ul{A} - 5\ul{I}} = 0}, and   ⁣A‾v⃗=−2v⃗\bcb{\ul{A} \ol{v} = -2 \ol{v}} has only the trivial solution.
133 - FML 3 - 18.1W e.g. 70.2
Suppose 5\bcb{5} and −2\bcb{-2} are eigenvalues of the matrix   ⁣A‾\bcb{\ul{A}}, which of the following statements must be true?
A−5I\bcb{A-5I} is invertible and det(  ⁣A‾+2  ⁣I‾)=0\bcb{\det{\ul{A} + 2\ul{I} } = 0}.
133 - FML 3 - 18.1W e.g. 70.1
Suppose 5\bcb{5} and −2\bcb{-2} are eigenvalues of the matrix   ⁣A‾\bcb{\ul{A}}, which of the following statements must be true?
A−5I\bcb{A - 5I} is not invertible and there is a nontrivial solution to   ⁣A‾v⃗=−2v⃗\bcb{\ul{A} \ol{v} = -2 \ol{v}}.
Properties
Practice: Similar Matrices
Consider the matrix A=[−1421]A= \left[ \begin{array}{rrr} -1 & 4\\ 2 & 1 \end{array} \right].
Practice Question: Eigenvalues of Triangular Matrices
Prove that any element on the main diagonal of a triangular matrix is an eigenvalue.