Question

Question: In Exercises 21 and 22, A is an\(n \times n\)matrix. Mark each statement True or False. Justify each answer.

1. If\(A{\bf{x}} = \lambda {\bf{x}}\)for some vector x, then\(\lambda \)is an eignvalue of A.
2. A matrix A is not invertible if and only if 0 is an eigenvalue of A.
3. A number of c is an eigenvalue of A if and only if the equation\(\left( {A - cI} \right){\bf{x}} = {\bf{0}}\)has a nontrivial solution.
4. Finding an eigenvector of A may be difficult, but checking whether a given vector is in face an eigenvector is easy.
5. To find the eigenvalues of A, reduce A to echelon form.

Short answer

a. The given statement is False.

b. The given statement is True.

c. The given statement is True

d. The given statement is True.

e. The given statement is False.

Step-by-step solution

Find an answer for part (a)

The equation \(A{\bf{x}} = \lambda {\bf{x}}\) essentially has a nontrivial solution, then only \(\lambda \) will be the eigenvalue of A.

Thus, statement (a) is false.

Find an answer for part (b)

If A is invertible, then 0 is an eigenvalue of A.

Thus, statement (b) is true.

Find an answer for part (c)

If a scalar c is the eigenvalue of A, then;

\(\begin{array}{c}A{\bf{x}} = c{\bf{x}}\\\left( {A - c} \right){\bf{x}} = 0\end{array}\)

Find an answer for part (d)

To find an eigenvector of a matrix, first, it needs to determine the eigenvalue and then the eigenvector.

On the other hand, to check the eigenvector, we need to solve the equation \(A{\bf{x}} = \lambda {\bf{x}}\), where x is an eigenvector.

Thus, statement (d) is true.

Find the answer for part (e)

The characteristic equation of the matrix A is:

\(\det \left( {A - \lambda I} \right) = 0\)

The eigenvalues cannot be created by reducing in echelon form. The echelon form is used to determine the eigenvectors.

Thus, the statement (e) is false.