TIFR 2013 problem 7 | Eigenvectors of Similar Matrices

Try this problem from TIFR 2013 problem 7 based on Eigenvectors of Similar Matrices.

Question: TIFR 2013 problem 7

True/False?

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

Hint: What does similar matrices mean?

Discussion:

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 $Av=\lambda v$.

Then $P^{-1}BPv=P^{-1}B(Pv)=\lambda v$

Then $B(Pv)=P(\lambda v)=\lambda Pv$

Since $P$ is invertible, it is one-to-one, hence it cannot take a nonzero vector $v$ to $0$(it already takes $0$ to $0$). Therefore, $Pv\neq0$.

Therefore, $Pv$ is an eigenvector of B with eigenvalue $\lambda$.

By now it seems impossible that eigenvectors of $A$ and $B$ would be same. We make this complete by finding an example.

Consider $A=\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$

(1,0) is an eigenvector of A with eigenvalue $\lambda$. (Since applying (1,0) to a matrix gives the first column and first column is (1,0) itself. )

Let us have the change of basis as rotation by $\pi/2$ in the counterclockwise direction. Then (1,0) is no longer an eigenvector. What is $P$ in this case? What is $B$?