Do matrices $\mathbf{A A}^H$ and $\mathbf{A}^H \mathbf{A}$ have the same eigenvalues?

Question

Let $\mathbf{A}$ be any complex matrix. Do matrices $\mathbf{A A}^H$ and $\mathbf{A}^H \mathbf{A}$ have the same eigenvalues?

Note: The matrix $\mathbf{A}^H$ is the conjugate transpose of the matrix $\mathbf{A}$.

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Here is my trial:

Let $\lambda_1$ be an eigenvalue of the matrix $\mathbf{AA}^H$. Let the vector $\mathbf{x}_1$ be the corresponding eigenvector. Then $$ \mathbf{A A}^H \mathbf{x}_1 = \lambda_1 \mathbf{x}_1. \tag{1} $$ Multiplying (1) by matrix $\mathbf{A}^H$ from the left, we have $$ \mathbf{A}^H (\mathbf{A A}^H \mathbf{x}_1) = \mathbf{A}^H (\lambda_1 \mathbf{x}_1). \tag{2} $$ Equation (2) can be rewritten as $$ (\mathbf{A}^H \mathbf{A}) (\mathbf{A}^H \mathbf{x}_1) = \lambda_1 (\mathbf{A}^H \mathbf{x}_1). \tag{3} $$ Therefore, $\lambda_1$ is also an eigenvalue of the matrix $\mathbf{A}^H \mathbf{A}$. The corresponding eigenvector is $\mathbf{A}^H \mathbf{x}_1$.

Similarly, let $\lambda_2$ be an eigenvalue of the matrix $\mathbf{A}^H \mathbf{A}$. Let the vector $\mathbf{x}_2$ be the corresponding eigenvector. Then $$ \mathbf{A}^H \mathbf{A} \mathbf{x}_2 = \lambda_2 \mathbf{x}_2. \tag{4} $$ Multiplying (4) by the matrix $\mathbf{A}$ from the left, we have $$ \mathbf{A} (\mathbf{A}^H \mathbf{A} \mathbf{x}_2) = \mathbf{A} (\lambda_2 \mathbf{x}_2). \tag{5} $$ Equation (5) can be written as $$ (\mathbf{A} \mathbf{A}^H) (\mathbf{A} \mathbf{x}_2) = \lambda_2 (\mathbf{A} \mathbf{x}_2). \tag{6} $$ Therefore, $\lambda_2$ is also an eigenvalue of the matrix $\mathbf{A} \mathbf{A}^H$. The corresponding eigenvector is $\mathbf{A} \mathbf{x}_2$.

Thus, we can conclude that matrices $\mathbf{A A}^H$ and $\mathbf{A}^H \mathbf{A}$ have the same eigenvalues.

Am I right?

Answer 1

Actually the eigenvalues of such form is explained by singular value decomposition. The square of a singular value of $A$ is just the eigenvalue of $A^HA$ or $AA^H$, which are all positive semi-definite forms.