Let $A$ be a real symmetric matrix and $D$ a real diagonal matrix with non-zero entries along the diagonal.

Is it true that the eigenvectors of $D^{-1}AD$ are the same as those of $A$?

If not, can anything be said relating the eigenvectors/eigenvalues of these two matrices?

  • $\begingroup$ This conjugation amounts to a change of basis in which each of the standard basis vectors is scaled by some, possibly different, amount. $\endgroup$ – amd May 11 '17 at 23:09

Let $A$ be an arbitrary matrix and $B=P^{-1}AP$, where $P$ is an arbitrary invertible matrix.

If $Av=\lambda v$, then $Bw=\lambda w$ for $w=P^{-1}v$.

In this sense, similar matrices have the same eigenvalues but not necessarily the same eigenvectors.

In particular, the eigenvectors are different when $P$ is a diagonal matrix that is not a scalar multiple of the identity matrix.

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  • $\begingroup$ The eigenvalues are the same, but the eigenvectors most certainly aren't $\endgroup$ – eyeballfrog May 11 '17 at 22:41
  • $\begingroup$ See also math.stackexchange.com/a/87705/589. $\endgroup$ – lhf May 11 '17 at 22:51
  • $\begingroup$ More specific to OP's question, pretty much any non-diagonal $2 \times 2$ real symmetric matrix and any nonscalar diagonal $D$ should do the trick. $\endgroup$ – Dustan Levenstein May 11 '17 at 22:55
  • $\begingroup$ An further question is under what conditions is the OP's statement true. Clearly having $A$ be diagonal itself is sufficient. Is it also necessary? $\endgroup$ – eyeballfrog May 11 '17 at 23:01

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