91Ó°ÊÓ

Prove that if a symmetric matrix \(A\) has only one eigenvalue \(\lambda\) then \(A=\lambda I\).

Demonstrate the Cayley-Hamilton Theorem for the given matrix. The Cayley- Hamilton Theorem states that a matrix satisfies its characteristic equation. For example, the characteristic equation of \(A=\left[\begin{array}{rr}1 & -3 \\\ 2 & 5\end{array}\right]\) is \(\lambda^{2}-6 \lambda+11=0,\) and by the theorem you have \(A^{2}-6 A+11 I_{2}=O\). $$\left[\begin{array}{rr} 6 & -1 \\ 1 & 5 \end{array}\right]$$

Demonstrate the Cayley-Hamilton Theorem for the given matrix. The Cayley- Hamilton Theorem states that a matrix satisfies its characteristic equation. For example, the characteristic equation of \(A=\left[\begin{array}{rr}1 & -3 \\\ 2 & 5\end{array}\right]\) is \(\lambda^{2}-6 \lambda+11=0,\) and by the theorem you have \(A^{2}-6 A+11 I_{2}=O\). $$\left[\begin{array}{rr} 4 & 1 \\ -2 & 1 \end{array}\right]$$

Demonstrate the Cayley-Hamilton Theorem for the given matrix. The Cayley- Hamilton Theorem states that a matrix satisfies its characteristic equation. For example, the characteristic equation of \(A=\left[\begin{array}{rr}1 & -3 \\\ 2 & 5\end{array}\right]\) is \(\lambda^{2}-6 \lambda+11=0,\) and by the theorem you have \(A^{2}-6 A+11 I_{2}=O\). $$\left[\begin{array}{lll} 3 & 1 & 4 \\ 2 & 4 & 0 \\ 5 & 5 & 6 \end{array}\right]$$

Can a matrix be similar to two different diagonal matrices? Explain your answer.

For an invertible matrix \(A\), prove that \(A\) and \(A^{-1}\) have the same eigenvectors. How are the eigenvalues of \(A\) related to the eigenvalues of \(A^{-1} ?\)

Prove that nonzero nilpotent matrices are not diagonalizable. Getting Started: From Exercise 73 in Section \(7.1,\) you know that 0 is the only eigenvalue of the nilpotent matrix \(A\). Show that it is impossible for \(A\) to be diagonalizable. (i) Assume \(A\) is diagonalizable, so there exists an invertible matrix \(P\) such that \(P^{-1} A P=D,\) where \(D\) is the zero matrix. (ii) Find \(A\) in terms of \(P, P^{-1},\) and \(D\) (iii) Find a contradiction and conclude that nonzero nilpotent matrices are not diagonalizable.

Prove that the constant term of the characteristic polynomial is \(\pm|A|\).

Prove that a triangular matrix is nonsingular if and only if its eigenvalues are real and nonzero. Getting Started: Because this is an "if and only if" statement, you must prove that the statement is true in both directions. Review Theorems 3.2 and 3.7. (i) To prove the statement in one direction, assume that the triangular matrix \(A\) is nonsingular. Use your knowledge of nonsingular and triangular matrices and determinants to conclude that the entries on the main diagonal of \(A\) are nonzero. (ii) Because \(A\) is triangular, you can use Theorem 7.3 and part (i) to conclude that the eigenvalues are real and nonzero. (iii) To prove the statement in the other direction, assume that the eigenvalues of the triangular matrix \(A\) are real and nonzero. Repeat parts (i) and (ii) in reverse order to prove that \(A\) is nonsingular.

Find all values of the angle \(\theta\) for which the matrix $$A=\left[\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$$ has real eigenvalues. Interpret your answer geometrically.