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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$\begingroup$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".
$\endgroup$ 1 $\begingroup$There exists an invertible (change of basis) matrix $P$ such that $A=P^{−1}BP$. Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$. Then, $v\neq0$ and $Av=\lambda v$.
So, $(P^{−1}BP)v=P^{−1}B(Pv)=\lambda v$.
Hence, $B(Pv)=P(\lambda v)=\lambda Pv$.
Since $P$ is invertible, it is one-to-one, hence it cannot take a nonzero vector $v$ to $0$ (it already takes $0$ to $0$). Thus, $Pv\neq 0$.
Therefore, $Pv$ is an eigenvector of $B$ with eigenvalue $\lambda$.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 15 $\begingroup$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 $\Lambda \begin{bmatrix} 1 \\ 1 \end{bmatrix}= \begin{bmatrix} -1 \\ 1 \end{bmatrix}$.
$\endgroup$ $\begingroup$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.)
$\endgroup$ 0 $\begingroup$No. If that was the case, if a matrix was diagonalazible in one basis, it would be diagonalizable in every basis.
In order to be diagonalizable in basis B, for M an$ n \times n $ we need a full basis of eigenvectors, i.e., n ( linearly -independent) eigenvectors $B_1, B_2,.., B_n$. If these were eigenvectors in every other basis B', then they would be a full basis in B' and so M would be diagonalizable in B' for every other basis B'.
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