• No products in the cart.

Profile Photo

Eigenvectors of Similar Matrices (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 \(B\)?

No comments, be the first one to comment !

    Leave a Reply

    Your email address will not be published. Required fields are marked *



    GOOGLECreate an Account