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

Question: TIFR 2013 problem 7


If \(A\) and \(B\) are similar matrices then every eigenvector of \(A\) is an eigenvector of \(B\).

Hint: What does similar matrices mean?


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 \(B\)?

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