3
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ In general, the two matrices will have different sizes. One must be a bit careful about what "same eigenvalues" actually means. $\endgroup$ – Rodrigo de Azevedo Apr 11 '19 at 6:15
2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ So do matrices $\mathbf{A A}^H$ and $\mathbf{A}^H \mathbf{A}$ have the same eigenvalues? $\endgroup$ – Wei-Cheng Liu Apr 10 '19 at 5:45
  • 1
    $\begingroup$ Yes, but different eigenvectors. $\endgroup$ – Wenkuei P'ei Apr 10 '19 at 9:55
  • $\begingroup$ OK, I understand. Thank you. $\endgroup$ – Wei-Cheng Liu Apr 11 '19 at 3:09
  • $\begingroup$ 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? $\endgroup$ – Wei-Cheng Liu Apr 11 '19 at 5:32
  • 1
    $\begingroup$ Yeah, the zero eigenvalues are all the same as being zero. The larger matrix might have extra zeros. There are cases that the small matrix has no zero eigenvalue but the larger one has. That's the best theory concerning your question. $\endgroup$ – Wenkuei P'ei Apr 11 '19 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.