Linear Algebra - Eigenvalues and Eigenvectors

A matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

Give the set of eigenvalues of $A^{-1}$ .

$\left { \frac{1}{2}+ \frac{1}{2}i, \frac{1}{2}- \frac{1}{2}i , -\frac{1}{2}+ \frac{1}{2}i, -\frac{1}{2}- \frac{1}{2}i , \right }$

$\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$

$\begin{Bmatrix} 2+2i, 2-2i, -2+2i, -2-2i \end{Bmatrix}$

$\left { \frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i, \frac{\sqrt{2}}{2}- \frac{\sqrt{2}}{2}i , -\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i, -\frac{\sqrt{2}}{2}- \frac{\sqrt{2}}{2}i , \right }$

$\left { \sqrt{2} + i \sqrt{2}, \sqrt{2} - i \sqrt{2}, -\sqrt{2} + i \sqrt{2},- \sqrt{2} - i \sqrt{2} \right }$

Explanation

If is nonsingular, the set of eigenvalues of $A^{-1}$ is exactly the set of reciprocals of eigenvalues of . The eigenvalues of are $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ , so the eigenvalues of these numbers are the reciprocals of these. The reciprocal of is

$\frac{1}{1 + i} = \frac{1(1-i)}{(1 + i)(1-i)} = \frac{1 - i}{2} = \frac{1}{2}- \frac{1}{2}i$ .

Similarly,

$\frac{1}{1 - i} = \frac{1}{2}+ \frac{1}{2}i$

$\frac{1}{-1 + i} =- \frac{1}{2}- \frac{1}{2}i$

$\frac{1}{-1 - i} =- \frac{1}{2}+ \frac{1}{2}i$

The set of eigenvalues of $A^{-1}$ is $\left { \frac{1}{2}+ \frac{1}{2}i, \frac{1}{2}- \frac{1}{2}i , -\frac{1}{2}+ \frac{1}{2}i, -\frac{1}{2}- \frac{1}{2}i , \right }$ .

A $4 \times 4$ matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

True, false, or indeterminate: the matrix is singular.

False

True

Indeterminate

Explanation

A matrix is singular - that is, not having an inverse - if and only if one of its eigenvalues is 0. Since 0 is not an element of its eigenvalue set, is nonsingular.

is a nonsingular real matrix with four eigenvalues: $\left {x, \frac{1}{x} , -x , - \frac{1}{x} \right }$ .

True or false: $A^{-1}$ must have these same four eigenvalues.

True

False

Explanation

One property of eigenvalues is that if is nonsingular, the set of eigenvalues of $A^{-1}$ is exactly the set of reciprocals of eigenvalues of . The eigenvalues of are $\left {x, \frac{1}{x} , -x , - \frac{1}{x} \right }$ , so the eigenvalues of these numbers are the reciprocals of these - in order, $\left {\frac{1}{x} ,x , - \frac{1}{x}, -x \right }$ . This is the same set.

A $4 \times 4$ matrix has $\begin{Bmatrix} 1+i, 1-i, -1+i, -1-i \end{Bmatrix}$ as its set of eigenvalues.

Calculate $\det A$ .

Explanation

The determinant of a matrix is equal to the product of its eigenvalues, so

$\det A = ( 1+i )( 1-i) (-1+i) ( -1-i )$

$= (1^{2}+ 1^{2})\left [ (-1)^{2}+ 1^{2} \right ]$

$= 2 \cdot 2$

is a nonsingular real matrix with four eigenvalues: $\left { \lambda, -\lambda , \lambda i, -\lambda i \right }$ .

True or false: $A^{-1}$ must have these same four eigenvalues.

False

True

Explanation

One property of eigenvalues is that if is nonsingular, the set of eigenvalues of $A^{-1}$ is exactly the set of reciprocals of eigenvalues of . The eigenvalues of are $\left { \lambda, -\lambda , \lambda i, -\lambda i \right }$ , so the eigenvalues of these numbers are the reciprocals of these - in order, $\left { \frac{1}{\lambda} ,- \frac{1}{\lambda} ,-\frac{i}{\lambda} , \frac{i}{\lambda} \right }$ . This is not the same set.

is an eigenvalue of a nonsingular real matrix .

True or false: It follows that is an eigenvalue of $A ^{-1}$ .

True

False

Explanation

The eigenvalues of a matrix are the solutions of the characteristic polynomial equation

$\det (A- \lambda I) = 0$ .

is a matrix with only real entries, so the coefficients of the polynomial must be real as well; it follows that any imaginary solutions are in conjugate pairs. , an eigenvalue of , is a solution of the characteristic equation; consequently, its complex conjugate is a solution, and thus, an eigenvalue of .

One property of eigenvalues is that if is nonsingular, the set of eigenvalues of $A^{-1}$ is exactly the set of reciprocals of eigenvalues of . Since is an eigenvalue of , it follows that $\frac{1}{-i} = i$ is an eigenvalue of $A ^{-1}$ .

$\begin{align*}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}6&-13\8&-19\end{bmatrix}\end{align*}$

$\lambda_{1}=0.73;\lambda_{2}=-13.73$

$\lambda_{1}=1.73;\lambda_{2}=-12.73$

$\lambda_{1}=-3.27;\lambda_{2}=-17.73$

$\lambda_{1}=9.73;\lambda_{2}=-4.73$

Explanation

$\begin{align*}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}6&-13\8&-19\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}6 - \lambda &-13\8&- \lambda - 19\end{bmatrix} \&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{From that, we can find the eigenvalues:}\&13\lambda + \lambda ^{2} - 10\&\lambda_{1}=0.73;\lambda_{2}=-13.73\end{align*}$

Suppose we have a square matrix with real-valued entries with only positive eigenvalues. Is invertible? Why or why not?

It is invertible because none of its eigenvalues is

It is not invertible because none of its eigenvalues is negative.

It is not invertible because is not an eigenvalue

It is invertible because it has no negative eigenvalues.

Explanation

The square matrix is certainly invertible with the reason being that none of its eigenvalues is . We know that is not an eigenvalue, so the following

does not hold for any nonzero vector in $\mathbb{R}^n$ , i.e.

for all nonzero $v\in \mathbb{R}^n$ . So the only vector that maps to the zero vector is the zero vector. In a square matrix, this is equivalent to the null space being the zero vector: . This is also equivalent to the matrix being invertible.

$A= \begin{bmatrix} 2 & 3 & 6 \ 0 & 4 & 4 \ 2 & 0 & 4 \end{bmatrix}$ .

Is an eigenvalue of , and if so, what is the dimension of its eigenspace?

No.

Yes; the dimension is 1.

Yes; the dimension is 2.

Yes; the dimension is 3.

Explanation

Assume that is an eigenvalue of . Then, if $b = \begin{bmatrix} b_{1} \ b_{2} \b_{3} \end{bmatrix}$ is one of its eigenvectors, it follows that

, or, equivalently,

where are the $3 \times 3$ identity and zero matrices, respectively.

$A= \begin{bmatrix} 2 & 3 & 6 \ 0 & 4 & 4 \ 2 & 0 & 4 \end{bmatrix}$ , so

$A + 2 I = \begin{bmatrix} 2+2 & 3 & 6 \ 0 & 4+2 & 4 \ 2 & 0 & 4+2 \end{bmatrix}$

$= \begin{bmatrix} 4& 3 & 6 \ 0 & 6 & 4 \ 2 &6& 2\end{bmatrix}$

Changing to reduced row echelon form:

$R1 \leftrightarrow R3$

$\begin{bmatrix} 2 &6& 2 \ 0 & 6 & 4 \ 4& 3 & 6 \end{bmatrix}$

$\frac{1}{2} R1 \rightarrow R1$

$\begin{bmatrix} 1 &3&1 \ 0 & 6 & 4 \ 4& 3 & 6 \end{bmatrix}$

$-4R1+ R3 \rightarrow R3$

$\begin{bmatrix} 1 &3&1 \ 0 & 6 & 4 \ 0& -9 & 2 \end{bmatrix}$

$\frac{1}{6} R2 \rightarrow R2$

$\begin{bmatrix} 1 &3&1 \ 0 & 1 & \frac{2}{3} \ 0& -9 & 2 \end{bmatrix}$

$-3R2 + R1 \rightarrow R1$

$9 R2 + R3 \rightarrow R3$

$\begin{bmatrix} 1 &0 &-1 \ 0 & 1 & \frac{2}{3} \ 0& 0& 8 \end{bmatrix}$

$\frac{1}{8} R3 \rightarrow R3$

$\begin{bmatrix} 1 &0 &-1 \ 0 & 1 & \frac{2}{3} \ 0& 0& 1 \end{bmatrix}$

We do not need to go further to see that this matrix will not have a row of zeroes. This means the rank of the matrix is 3, and the nullity is 0. If this happens, the tested value, in this case , is not an eigenvalue.

$A = \begin{bmatrix} a & 4 & -2 \ 1 & 3 & 0 \ - 6 & 4 & a \end{bmatrix}$

Evaluate so that the sum of the eigenvalues of is 10.

The sum of the eigenvalues of is 10 regardless of the value of .

Explanation

The sum of the eigenvalues of a square matrix is equal to its trace, the sum of its diagonal elements. Examine these elements, which are in red below:

$A = \begin{bmatrix} {\color{Red}\textbf{a} } & 4 & -2 \ 1 & {\color{Red}\textbf{3}} & 0 \ - 6 & 4 & {\color{Red}\textbf{a}} \end{bmatrix}$

Set the trace equal to 10 and solve for :

$\textup{Tr} (A ) = 10$

Eigenvalues and Eigenvectors

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