Similar Matrices: Are Eigenvalues the Same?

[Sudharaka]

saravananbs's question from Math Help Forum,

if A and B are similar matrices then the eigenvalues are same. is the converse is true? why?
thank u

Hi saravananbs,

No. The converse is not true in general. Take the two matrices, \(A=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\mbox{ and }B=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\).

\[\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}\]

\[\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}\]

Therefore both matrices have the same eigenvalue 1. However it can be easily shown that \(A\mbox{ and }B\) are not similar matrices.

 

[Opalg]

saravananbs's question from Math Help Forum,
Hi saravananbs,

No. The converse is not true in general. Take the two matrices, \(A=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\mbox{ and }B=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\).

\[\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}\]

\[\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\end{pmatrix}=1.\begin{pmatrix}1 \\ 1\end{pmatrix}\]

Therefore both matrices have the same eigenvalue 1. However it can be easily shown that \(A\mbox{ and }B\) are not similar matrices.

That example does not really work, because $A$ and $B$ do not have the same eigenvalues: $A$ has a (repeated) eigenvalue 1, whereas the eigenvalues of $B$ are 1 and –1.

However, if $C = \begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix}$, then $A$ and $C$ have the same eigenvalues (namely 1, repeated), but are not similar.

 

[Sudharaka]

That example does not really work, because $A$ and $B$ do not have the same eigenvalues: $A$ has a (repeated) eigenvalue 1, whereas the eigenvalues of $B$ are 1 and –1.

However, if $C = \begin{pmatrix}1 & 1 \\0 & 1\end{pmatrix}$, then $A$ and $C$ have the same eigenvalues (namely 1, repeated), but are not similar.

Thanks for correcting that. Of course I now see that only the eigenvalue 1 is common to both \(A\) and \(B\).