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[SOLVED] Eigenvalues

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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 \(B\).