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If $A$ and $B$ are similar matrices then every eigenvector of $A$ is an eigenvector of $B$.

Is the above statement is true? I know that similar matrices have same eigenvalue but I'm not sure about the eigenvectors.

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It's false. Exercise: come up with an example. Almost any pair of similar $2\times2$ matrices will do. –  Gerry Myerson Dec 10 '12 at 5:12

4 Answers 4

False. For an easy reasoning, suppose $A$ is diagonalizable and $B$ is the diagonalization of $A$. $A$ and $B$ are similar by the definition of diagonalizable: there exists an invertible matrix $P$ such that $P^{-1}AP=B$, and $B$ is a diagonal matrix.

Then the standard basis vectors are the eigenvectors of $B$, but clearly these need not be the eigenvectors of $A$. (In fact, if your claim was true, then the every eigenvector of a diagonalizable matrix would be a standard basis vector, which is false.)

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If $A$ and $B$ are similar matrices, then they represent the same linear transformation $T$, albeit written in different bases. So really the two matrices have the same eigenvectors, they just look different because you're expressing them in terms of a different basis.

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You can, and often should, think of similar matrices $A,B$ as being matrices of a same linear transformation $f:V\to V$ in different bases of $V$. Then if $f$ has eigenvalues $\lambda$, the corresponding eigenvectors are (abstract) vectors of $V$, and expressing these in the bases used repectively for $A$ and for $B$ gives (concrete) eigenvectors for the matrices $A$ and $B$ respectively. It is clear that $A$ and $B$ have the same eigenvalues as $f$, which are independent of any basis, but what you are essentially asking is wheter the concrete coordinates of these vectors are the same in the bases used for $A$ and for $B$, and the answer is of course "no".

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Take $A=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$, $V=\frac{1}{\sqrt{2}} \begin{bmatrix} -1 & 1 \\ 1 & 1\end{bmatrix}$, and let $\Lambda = V^{-1} A V = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$.

Then $A \begin{bmatrix} 1 \\ 1 \end{bmatrix}= \begin{bmatrix} 1 \\ 1 \end{bmatrix}$, but $\Lambda \begin{bmatrix} 1 \\ 1 \end{bmatrix}= \begin{bmatrix} -1 \\ 1 \end{bmatrix}$.

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