If Two Matrices Have the Same Eigenvalues with Linearly Independent Eigenvectors, then They Are Equal

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 424

Let $A$ and $B$ be $n\times n$ matrices.
Suppose that $A$ and $B$ have the same eigenvalues $\lambda_1, \dots, \lambda_n$ with the same corresponding eigenvectors $\mathbf{x}_1, \dots, \mathbf{x}_n$.
Prove that if the eigenvectors $\mathbf{x}_1, \dots, \mathbf{x}_n$ are linearly independent, then $A=B$.

 
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Proof.

Since $A$ and $B$ have $n$ linearly independent eigenvectors $\mathbf{x}_1, \dots, \mathbf{x}_n$, they are diagonalizable.
Specifically, if we put $S=[\mathbf{x}_1, \dots, \mathbf{x}_n]$.

Then $S$ is invertible (as column vectors of $S$ are linearly independent) and we have
\[S^{-1}AS=D \text{ and } S^{-1}BS=D,\] where $D$ is the diagonal matrix whose diagonal entries are eigenvalues:
\[D=\begin{bmatrix}
\lambda_1 & 0 & \cdots & 0 \\
0 & \lambda_2 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \lambda_n
\end{bmatrix}.\] It follows that we have
\[S^{-1}AS=D=S^{-1}BS,\] and hence $A=B$. This completes the proof.


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