7
$\begingroup$

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.

$\endgroup$
1
  • 4
    $\begingroup$ It's false. Exercise: come up with an example. Almost any pair of similar $2\times2$ matrices will do. $\endgroup$ – Gerry Myerson Dec 10 '12 at 5:12
11
$\begingroup$

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
  • $\begingroup$ +1 A beautiful explanation. $\endgroup$ – user56202 Dec 17 '20 at 3:42
4
$\begingroup$

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
  • 2
    $\begingroup$ This is false, take a look at $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}, S=\begin{bmatrix}1&2\\3&5\end{bmatrix}$. Then $$B=SAS^{-1}=\begin{bmatrix}3&-1\\9&-3\end{bmatrix}$$ $\endgroup$ – Invisible Jul 31 '20 at 7:17
  • 2
    $\begingroup$ Eigenvector of $A$ is $\begin{bmatrix}1\\3\end{bmatrix}$. $\endgroup$ – Invisible Jul 31 '20 at 7:23
  • $\begingroup$ @Invisible surely you mean eigenvector of B? I don't see how that contradicts my answer though. I'm sure you would agree the corresponding eigenvector of $A$ is $\begin{bmatrix} 1 \\ 0\end{bmatrix}$, and we have $\begin{bmatrix} 1 \\ 3\end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 5\end{bmatrix}\cdot\begin{bmatrix} 1 \\ 0\end{bmatrix}$, i.e. these two concrete eigenvectors are related by the change of basis matrix you provided. $\endgroup$ – Chris Brooks Jul 31 '20 at 15:58
  • $\begingroup$ If they are related by a (nontrivial) change of basis, @Chris, then it is not the same vectors. I think you are answering a different questions. $\endgroup$ – Arctic Char Jul 31 '20 at 16:51
  • $\begingroup$ @ArcticChar a vector is not just its coordinates in some basis. To the linear transformation from Invisible's comment (written as A in one basis and B in another) we can associate an eigenvector to the eigenvalue 0. In the first basis this vector is written (1 0) and in the second basis it's written (1 3). It's the same abstract vector though. Do you also take issue with referring to the two matrices as the same linear transformation? $\endgroup$ – Chris Brooks Jul 31 '20 at 17:07
3
$\begingroup$

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
3
$\begingroup$

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 $\Lambda \begin{bmatrix} 1 \\ 1 \end{bmatrix}= \begin{bmatrix} -1 \\ 1 \end{bmatrix}$.

$\endgroup$
2
$\begingroup$

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 $\lambda$.

$\endgroup$
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'.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.