Chapter 5, Problem 1SE

Mark each statement as True or False. Justify each answer.

1. a. If A is invertible and 1 is an eigenvalue for A, then 1 is also an eigenvalue of A⁻¹.
2. b. If A is row equivalent to the identity matrix I, then A is diagonalizable.
3. c. If A contains a row or column of zeros, then 0 is an eigenvalue of A.
4. d. Each eigenvalue of A is also an eigenvalue of A².
5. e. Each eigenvector of A is also an eigenvector of A².
6. f. Each eigenvector of an invertible matrix A is also an eigenvector of A⁻¹.
7. g. Eigenvalues must be nonzero scalars.
8. h. Eigenvectors must be nonzero vectors.
9. i. Two eigenvectors corresponding to the same eigenvalue are always linearly dependent.
10. j. Similar matrices always have exactly the same eigenvalues.
11. k. Similar matrices always have exactly the same eigen vectors.
12. l. The sum of two eigenvectors of a matrix A is also an eigenvector of A.
13. m. The eigenvalues of an upper triangular matrix A are exactly the nonzero entries on the diagonal of A.
14. n. The matrices A and A^T have the same eigenvalues, counting multiplicities.
15. ○.      If a 5 × 5 matrix A has fewer than 5 distinct eigenvalues, then A is not diagonalizable.
16. p. There exists a 2 × 2 matrix that has no eigenvectors in ℝ².
17. q. If A is diagonalizable, then the columns of A are linearly independent.
18. r. A nonzero vector cannot correspond to two different eigenvalues of A.
19. s. A (square) matrix A is invertible if and only if there is a coordinate system in which the transformation x i ↦ Ax is represented by a diagonal matrix.
20. t. If each vector e_j in the standard basis for ℝⁿ is an eigenvector of A, then A is a diagonal matrix.
21. u. If A is similar to a diagonalizable matrix B, then A is also diagonalizable.
22. v. If A and B are invertible n × n matrices, then AB is similar to BA.
23. w. An n × n matrix with n linearly independent eigenvectors is invertible.
24. x. If A is an n × n diagonalizable matrix, then each vector in ℝⁿ can be written as a linear combination of eigenvectors of A.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A is invertible and 1 is an eigenvalue for A, then 1 is also an eigenvalue of $A^{-1}$.

Explanation:

Consider the equation as follows:

$Ax=1\cdot x$ (1)

Left-multiply with $A^{-1}$ on both sides in Equation (1).

$\begin{array}{l}{A}^{-1}Ax=A^{-1}\cdot 1\cdot x\\ x=A^{-1}x\end{array}$

Rewrite the equation as follows:

$A^{-1}x=1\cdot x$

In this equation, x cannot be equal to zero. That is, $x\ne 0$.

Therefore, 1 is an eigenvalue of $A^{-1}$.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

If A is a row equivalent to the identity matrix I, then A is diagonalizable.

Explanation:

If A is a row that is equivalent to the identity matrix, then A is invertible.

Refer to Example 4 in section 5.3:

The given matrix is invertible. However, matrix A is not diagonalizable.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A contains a row or column of zeros, then 0 is an eigenvalue of A.

Explanation:

Theorem: The Invertible Matrix Theorem

Let Abe an $n\times n$ matrix. Then,A is invertible if and only if:

s. The number 0 is not an eigenvalue of A.

t. The determinant of A is not zero.

As given in the statement, if A contains a row or column of zeros, then A is not a row equivalent to the identity matrix.

Therefore, matrix A is not invertible.

Refer to the invertible matrix theorem, o is an eigenvalue of matrix A.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Each eigenvalue of A is also an eigenvalue of $A^2$.

Explanation:

Consider a diagonal matrix whose eigenvalues are 1 and 2.

$D=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$

The value of $D^2$ is as follows:

$D^2=\left[\begin{array}{cc}1& 0\\ 0& 4\end{array}\right]$

Then, the diagonal entries in the matrix are squaresof the eigenvalues of matrix A.

Therefore, the eigenvalues of $A^2$ are the squares of the eigenvalues of A.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Each eigenvector of A is also an eigenvector of $A^2$.

Explanation:

Consider a nonzero vector x.

The condition to be satisfied is $Ax=\lambda x$

Left-multiply with A on both sides.

$\begin{array}{c}{A}^{2}x=A\left(Ax\right)\\ =A\left(\lambda x\right)\\ =\lambda Ax\\ ={\lambda }^{2}x\end{array}$

The derived equation shows that x is also an eigenvector of $A^2$.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Each eigenvector of an invertible matrix A is also an eigenvector of $A^{-1}$.

Explanation:

Consider a nonzero vector x.

The condition to be satisfied is $Ax=\lambda x$

Left-multiply with $A^{-1}$ on both sides.

$\begin{array}{c}A{A}^{-1}x=A^{-1}\left(\lambda x\right)\\ =\lambda A^{-1}x\end{array}$

Matrix A is invertible and the eigenvalue of the matrix is nonzero.

${\lambda }^{-1}x=A^{-1}x$

The derived equation shows that x is also an eigenvector of $A^{-1}$.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Eigenvalues must be nonzero scalars.

Explanation:

For each singular square matrix, 0 will be the eigenvalue.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Eigenvectors must be nonzero vectors.

Explanation:

Definition:

An eigenvector of an $n\times n$ matrix A is a nonzero vector x, that is, $Ax=\lambda x$ for some scalar $\lambda$. Scalar $\lambda$ is called an eigenvalue of A if there is a nontrivial solution of $Ax=\lambda x$; such an x is called an eigenvector corresponding to ${\lambda }^{-1}$.

Refer to the definition; an eigenvector must be nonzero.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Two eigenvectors corresponding to the same eigenvalue are always linearly independent.

Explanation:

Refer to example 4 in section 5.1.

The eigenvalue for the given matrix is 2.

However, the calculated eigenvector for the eigenvalue is not linearly independent.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Similar matrices always have exactly the same eigenvalues.

Explanation:

Theorem 4 (Section 5.2):

If $n\times n$ matrices A and B are similar, then they have the same characteristic polynomial, and hence, the same eigenvalues (with the same multiplicities).

Refer to theorem 4; the given statement is correct.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

Similar matrices always have exactly the same eigenvectors.

Explanation:

Refer to example 3 in section 5.3.

Consider $3\times 3$ matrix as $A=\left[\begin{array}{ccc}1& 3& 3\\ -3& -5& -3\\ 3& 3& 1\end{array}\right]$.

The eigenvalues of the matrix are $\lambda =1,-2$.

The diagonalizable matrix is $D=\left[\begin{array}{ccc}1& 0& 0\\ 0& -2& 0\\ 0& 0& -2\end{array}\right]$.

Consider the matrix A is similar to the matrix D.

Suppose the eigenvalues of matrix D to be in column $I_3$ of matrix A. However, in the given problem, it is not the case.

Therefore, the given statement is incorrect.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

The sum of two eigenvectors of matrix A is also an eigenvector of A.

Explanation:

Consider a $2\times 2$ matrix as $A=\left[\begin{array}{cc}4& 0\\ 0& 6\end{array}\right]$

The value of ${\text{e}}_1$ is given as follows:

${\text{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right]$.

The value of ${\text{e}}_2$ is given as follows:

${\text{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$.

Here, ${\text{e}}_1$ and ${\text{e}}_2$ are eigenvectors of matrix A. However, the sum of ${\text{e}}_1$ and ${\text{e}}_2$ are not the eigenvector of matrix A.

Therefore, the given statement is incorrect.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

The eigenvalues of upper triangular matrix A are the exact nonzero entries on the diagonal of A.

Explanation:

Theorem 1: (Section 5.1):

The eigenvalues of a triangular matrix are the entries on its main diagonal.

In the given problem, the upper triangular matrix A has exactly nonzero entries. In reality, it is not necessary for that to be the case. Zero entries can also be available.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

Matrices A and $A^T$ have the same eigenvalues, counting multiplicities.

Explanation:

Apply the concept of determinant transpose property.

Show the determinant calculation of matrix A as follows:

$\det \left(A^T-\lambda I\right)=\det {\left(A-\lambda I\right)}^T=\det \left(A-\lambda I\right)$

Therefore, matrices A and $A^T$ have a polynomial with the same characteristics.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

If a $5\times 5$ matrix A has fewer than 5 distinct eigenvalues, then A is not diagonalizable.

Explanation:

Refer to practice problem 3 in section 5.3.

Consider an identity matrix A that is $5\times 5$. Even for lesser eigenvalues, the given matrix A is diagonalizable.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

A $2\times 2$ matrix that has no eigenvectors in ${ℝ}^2$ exists.

Explanation:

Consider matrix A.

Matrix A rotates the vectors through $\pi /2$ radians about the origin.

The value of Ax is not a multiple of x when x is a nonzero value.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

If A is diagonalizable, then the columns of A are linearly independent.

Explanation:

Consider a diagonal matrix A with 0 as diagonals.

Then, the columns of the diagonal matrix A are not linearly independent.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

A nonzero vector cannot correspond to two different eigenvalues of A.

Explanation:

Consider the following example:

$Ax={\lambda }_{1}x$ and $Ax={\lambda }_{2}x$.

Equate both the values of $Ax$.

$\begin{array}{l}{\lambda }_{1}x={\lambda }_{2}x\\ \left({\lambda }_1-{\lambda }_2\right)x=0\end{array}$

Suppose x cannot be equal to zero, that is, $x\ne 0$.

${\lambda }_1={\lambda }_2$

Therefore, for a nonzero vector, two different eigenvalues are the same.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

A (square) matrix A is invertible if and only if there is a coordinate system in which the transformation $x\mapsto Ax$ is represented by a diagonal matrix.

Explanation:

Theorem 8 (Section 5.4):

Diagonal Matrix Representation:

Suppose $A=PD{P}^{-1}$, where D is a diagonal $n\times n$ matrix. If B is the basis for ${ℝ}^n$(formed from the columns of P), then D is the B-matrix for the transformation $x\mapsto Ax$.

Consider a singular matrix A that is diagonalizable.

Refer to the theorem; the transformation $x\mapsto Ax$ is represented by a diagonal matrix relative to a coordinate system, which is determined by the eigenvectors of matrix A.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If each vector ${\text{e}}_j$ in the standard basis for ${ℝ}^n$ is an eigenvector of A, then A is a diagonal matrix.

Explanation:

Refer to the definition of matrix multiplication.

Show the multiplication of matrix A with I as follows:

$\begin{array}{c}A=AI=A\left[\begin{array}{cccc}{\text{e}}_1& {\text{e}}_2& \cdots & {\text{e}}_n\end{array}\right]\\ =\left[\begin{array}{cccc}A{\text{e}}_1& A{\text{e}}_2& \cdots & A{\text{e}}_n\end{array}\right]\end{array}$

If the values are considered as $A{\text{e}}_j=d_j{\text{e}}_j$ for $j=1,\dots ,n$, and so on.

Then, matrix A becomes a diagonal matrix with the diagonal entries of $d_1,\dots ,d_n$.

Therefore, the given statement is correct.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A is similar to diagonalizable matrix B, then A is also diagonalizable.

Explanation:

Consider matrix B.

For a diagonalizable matrix:

$B=PD{P}^{-1}$

Here, D is a diagonalizable matrix.

Consider matrix A. Write the equation as follows:

$A=QB{Q}^{-1}$ (2)

Substitute $PD{P}^{-1}$ for B in Equation (2).

$\begin{array}{c}A=Q\left(PD{P}^{-1}\right)Q^{-1}\\ =\left(QP\right)D{\left(PQ\right)}^{-1}\end{array}$

Therefore, matrix A is diagonalizable.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A and B are invertible $n\times n$ matrices, then AB is similar to BA.

Explanation:

If the matrix B is invertible, the value of AB is similar to $B\left(AB\right)B^{-1}$.

This calculated value is equal to BA.

The given statement is true.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is false.

Explanation of Solution

Given statement:

An $n\times n$ matrix with n linearly independent eigenvectors is invertible.

Explanation:

Theorem 5 (Section 5.3):

The diagonalization theorem:

An $n\times n$ matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

In fact, $A=PD{P}^{-1}$, with D is a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors on P.

Refer to the theorem; the $n\times n$ matrix A is diagonalizable if and only if A has n linearly independent eigenvectors, but it is not compulsory that it should be invertible. Any of the eigenvalues can be zero.

The given statement is false.

To determine

To mark: Each statement as true of false.

To justify: The answer.

Answer to Problem 1SE

The given statement is true.

Explanation of Solution

Given statement:

If A is an $n\times n$ diagonalizable matrix, then each vector in ${ℝ}^n$ can be written as a linear combination of the eigenvectors of A.

Explanation:

Theorem 5 (Section 5.3):

The diagonalization theorem:

An $n\times n$ matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

In fact, $A=PD{P}^{-1}$, with D as a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors on P.

A is diagonalizable if and only if A has n linearly independent eigenvectors ${\text{v}}_1,\dots ,{\text{v}}_n$ in ${ℝ}^n$.

Refer to basis theorem:

$\left\{{\text{v}}_1,\dots ,{\text{v}}_n\right\}$ spans ${ℝ}^n$.

That is, each vector in ${ℝ}^n$ can be written as a linear combination of ${\text{v}}_1,\dots ,{\text{v}}_n$

The given statement is true.