1. ## Similar Matrices II

Find the eigenvalues of the following pair of matrices, and use them to decide wich of the pairs are similar:

(ii) $A= \begin{pmatrix} 1 & 0 & 0 \\ 4 & -3 & -2 \\ -4 & 1 & 0 \end{pmatrix}$ and $B= \begin{pmatrix} 1 & 0 & 0 \\ -2 & 2 & 0 \\ 4 & 10 & -1 \end{pmatrix}$

Now $A-\lambda I_{3}= \begin{pmatrix} 1-\lambda & 0 & 0 \\ 4 & -3-\lambda & -2 \\ -4 & 1 & -\lambda \end{pmatrix}
$
and $B-\lambda I_{3}= \begin{pmatrix} 1-\lambda & 0 & 0 \\ -2 & 2-\lambda & 0 \\ 4 & 10 & -1-\lambda \end{pmatrix}$ and expanding by the first row we have

$det(A-\lambda I_{3})=(1-\lambda) \begin{vmatrix} -3-\lambda & -2 \\ 1 & -\lambda \end{vmatrix}$ = $(1-\lambda)(\lambda^2+3\lambda+2)=-\lambda^3-2\lambda^2+\lambda+2$

and

$det(B-\lambda I_{3})=(1-\lambda) \begin{vmatrix} 2-\lambda & 0 \\ 10 & -1-\lambda \end{vmatrix}$ = $(1-\lambda)(\lambda^2-\lambda-2)=-\lambda^3+2\lambda^2+\lambda-2$

Am right saying that since these characteristic polynomials are not the same, the matrices are not similar?

2. Originally Posted by nmatthies1
Find the eigenvalues of the following pair of matrices, and use them to decide wich of the pairs are similar:

(ii) $A= \begin{pmatrix} 1 & 0 & 0 \\ 4 & -3 & -2 \\ -4 & 1 & 0 \end{pmatrix}$ and $B= \begin{pmatrix} 1 & 0 & 0 \\ -2 & 2 & 0 \\ 4 & 10 & -1 \end{pmatrix}$

Now $A-\lambda I_{3}= \begin{pmatrix} 1-\lambda & 0 & 0 \\ 4 & -3-\lambda & -2 \\ -4 & 1 & -\lambda \end{pmatrix}
$
and $B-\lambda I_{3}= \begin{pmatrix} 1-\lambda & 0 & 0 \\ -2 & 2-\lambda & 0 \\ 4 & 10 & -1-\lambda \end{pmatrix}$ and expanding by the first row we have

$det(A-\lambda I_{3})=(1-\lambda) \begin{vmatrix} -3-\lambda & -2 \\ 1 & -\lambda \end{vmatrix}$ = $(1-\lambda)(\lambda^2+3\lambda+2)=-\lambda^3-2\lambda^2+\lambda+2$

and

$det(B-\lambda I_{3})=(1-\lambda) \begin{vmatrix} 2-\lambda & 0 \\ 10 & -1-\lambda \end{vmatrix}$ = $(1-\lambda)(\lambda^2-\lambda-2)=-\lambda^3+2\lambda^2+\lambda-2$

Am right saying that since these characteristic polynomials are not the same, the matrices are not similar?
More precisely, they have different eigenvalues. Since B is a triangular matrix, its eigenvalues are 1, 2, and -1. But 2 is not an eigenvalue of A.