Question

Let the sets of eigenvalues and eigenvectors of a matrix $B$ be $\{\lambda_k \mid 1 \le k \le n\}$ and $\{\mathbf{v}_k \mid 1 \le k \le n\}$, respectively. For any invertible matrix $P$, the sets of eigenvalues and eigenvectors of the matrix $A$, where $B=P^{-1}AP$, respectively, are

• A. $\{\lambda_k \det(A) \mid 1 \le k \le n\}$ and $\{P\mathbf{v}_k \mid 1 \le k \le n\}$
• B. $\{\lambda_k \mid 1 \le k \le n\}$ and $\{\mathbf{v}_k \mid 1 \le k \le n\}$
• C. $\{\lambda_k \mid 1 \le k \le n\}$ and $\{P\mathbf{v}_k \mid 1 \le k \le n\}$
• D. $\{\lambda_k \mid 1 \le k \le n\}$ and $\{P^{-1}\mathbf{v}_k \mid 1 \le k \le n\}$

Hint:

Similarity ($A=PBP^{-1}$ or $B=P^{-1}AP$) preserves eigenvalues. Eigenvectors transform by the change of basis: $\mathbf{v}_B \mapsto P\mathbf{v}_B$ for $A$.

Answer 1

The Correct Option is C

Step 1: Use similarity.
Given $B=P^{-1}AP$ (so $A$ is similar to $B$). If $B\mathbf{v}_k=\lambda_k \mathbf{v}_k$, then \[ A(P\mathbf{v}_k)=PBP^{-1}(P\mathbf{v}_k)=PB\mathbf{v}_k=P(\lambda_k \mathbf{v}_k)=\lambda_k (P\mathbf{v}_k). \] Hence $P\mathbf{v}_k$ is an eigenvector of $A$ corresponding to the same eigenvalue $\lambda_k$. 
Step 2: Read off sets.
Therefore, the eigenvalues of $A$ are $\{\lambda_k\}$ (unchanged under similarity), and the corresponding eigenvectors are $\{P\mathbf{v}_k\}$. \[ \boxed{\text{Option (C)}} \]