Question-70
Let \(\displaystyle A,B,P\) be square matrices such that \(\displaystyle A=PBP^{-1}\) for some invertible matrix \(\displaystyle P\). Which of the following statements are true?
Option-1: \(\displaystyle A\) and \(\displaystyle B\) have the same eigenvalues
To see this, consider the characteristic polynomial of \(\displaystyle A\):
\[ \begin{equation*} \begin{aligned} |A-\lambda I| & =|PBP^{-1} -\lambda I|\\ & =|PBP^{-1} -\lambda PIP^{-1} |\\ & =|P( B-\lambda I) P^{-1} |\\ & =|P|\cdot |B-\lambda I|\cdot |P^{-1} |\\ & =|B-\lambda I| \end{aligned} \end{equation*} \]
We see that \(\displaystyle A\) and \(\displaystyle B\) have the same characteristic polynomials and hence the same eigenvalues.
Option-2: \(\displaystyle A\) and \(\displaystyle B\) needn’t have the same eigenvectors
Consider:
\[ \begin{equation*} A=\begin{bmatrix} 1 & 2\\ 2 & 1 \end{bmatrix} ,\ P=\frac{1}{\sqrt{2}}\begin{bmatrix} -1 & 1\\ 1 & 1 \end{bmatrix} ,\ B=\begin{bmatrix} -1 & 0\\ 0 & 3 \end{bmatrix} \end{equation*} \]
Verify that \(\displaystyle A=PBP^{-1}\). In fact, here \(\displaystyle P^{-1} =P^{T}\). The columns of \(\displaystyle P\) are the eigenvectors of \(\displaystyle A\), while the vectors of the standard basis are the eigenvectors of \(\displaystyle B\).
Option-3: \(\displaystyle v\) is an eigenvector of \(\displaystyle B\)
Let \(\displaystyle ( \lambda ,v)\) be an eigenpair of \(\displaystyle B\). We know that \(\displaystyle A=PBP^{-1}\) which implies that \(\displaystyle AP=PB\). Now:
\[ \begin{equation*} \begin{aligned} A( Pv) & =PBv\\ & =P( Bv)\\ & =P( \lambda v)\\ & =\lambda ( Pv) \end{aligned} \end{equation*} \]
Option-4: \(\displaystyle Pv\) is an eigenvector of \(\displaystyle A\)
Let \(\displaystyle ( \lambda ,Pv)\) be an eigenpair of \(\displaystyle A\). Then:
\[ \begin{equation*} \begin{aligned} A( Pv) & =\lambda ( Pv)\\ \left( P^{-1} AP\right) v & =\lambda v\\ Bv & =\lambda v \end{aligned} \end{equation*} \]
We see that \(\displaystyle ( \lambda ,v)\) is an eigenpair of \(\displaystyle B\).