# Eigenvectors of Similar Matrices (TIFR 2013 problem 7)

Question:

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 $$B$$?