# Similar Matrices II

• Mar 10th 2009, 07:29 PM
nmatthies1
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?
• Mar 11th 2009, 04:30 AM
HallsofIvy
Quote:

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.