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

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 $$\mathbf{A}$$.

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?

• In general, the two matrices will have different sizes. One must be a bit careful about what "same eigenvalues" actually means. – Rodrigo de Azevedo Apr 11 '19 at 6:15

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.
• So do matrices $\mathbf{A A}^H$ and $\mathbf{A}^H \mathbf{A}$ have the same eigenvalues? – Wei-Cheng Liu Apr 10 '19 at 5:45
• Wait a minute. The document hkumath.hku.hk/course/temp/Matrix.pdf says "$A^H A$ and $A A^H$ have the same NON-ZERO eigenvalues." So do you sure that $A^H A$ and $A A^H$ have the same eigenvalues? – Wei-Cheng Liu Apr 11 '19 at 5:32