7
$\begingroup$

Let $ \boldsymbol{A}=(a_{ij})_{n\times n} $ be a complex square matrix with eigenvalues: $\lambda_1 ,\lambda_2, \dots , \lambda_n $. Prove that $$ \sum_{r=1}^{n} |\lambda_r|^2 \le \sum_{i,j=1}^{n} |a_{ij}|^2\,.$$ Furthermore, show that the equality holds if and only if $$\boldsymbol{A^H A=AA^H}\,.$$
Here, $\boldsymbol{A^H}$ is the conjugate transpose of $\boldsymbol{A}$.

We can write $$\sum_{i,j=1}^n\,|a_{ij}|^2=\operatorname{Tr}(\boldsymbol{AA^H})=\|\boldsymbol{A}\|_F^2,$$ where $\|\_\|_F$ is the Frobenius norm. So this inequality gives a lower bound on $\|\boldsymbol{A}\|_F$, namely $$\|\boldsymbol{A}\|_F\geq \sqrt{\sum_{r=1}^n\,|\lambda_r|^2}\,.$$

Edit. Another proof of the inequality is found here. However, there is no discussion about the equality case.

$\endgroup$
10
  • $\begingroup$ Does $\mathbf{A}^{H}$ represent the complex conjugate of $\mathbf{A}$? $\endgroup$ Dec 28, 2013 at 23:49
  • $\begingroup$ No, it is the conjugate transpose of $\boldsymbol{A}$. Sorry I forgot to mention that. $\endgroup$ Dec 28, 2013 at 23:54
  • $\begingroup$ and your attempt? $\endgroup$
    – Lost1
    Dec 29, 2013 at 0:10
  • $\begingroup$ I am trying to figure out the prove. I need help. $\endgroup$ Dec 29, 2013 at 0:14
  • $\begingroup$ I am afraid that's not good enough, sorry. $\endgroup$
    – Igor Rivin
    Dec 29, 2013 at 0:16

1 Answer 1

7
$\begingroup$

If $\mathbf{A}$ is a general square complex matrix (i.e. $\mathbf{A}\in\mathbb{C}^{n\times n}$), we have by Schur decomposition:

$$\mathbf{A}=\mathbf{Q}\mathbf{U}\mathbf{Q}^{\dagger}$$

Where $\mathbf{Q}$ is a unitary matrix (i.e. $\mathbf{Q}^{-1}=\mathbf{Q}^{\dagger}$, where $\mathbf{Q}^{\dagger}$ represents the conjugate transpose of $\mathbf{Q}$) and $\mathbf{U}$ is an upper triangular matrix. Multiplying each side by its Hermitian transpose, we get:

$$\mathbf{A}\mathbf{A}^{\dagger}=(\mathbf{Q}\mathbf{U}\mathbf{Q}^{\dagger})(\mathbf{Q}\mathbf{U}\mathbf{Q}^{\dagger})^{\dagger}=\mathbf{Q}\mathbf{U}\mathbf{Q}^{\dagger}\mathbf{Q}\mathbf{U}^{\dagger}\mathbf{Q}^{\dagger}=\mathbf{Q}\mathbf{U}\mathbf{U}^{\dagger}\mathbf{Q}^{\dagger}$$

Taking the trace of both sides and using the property that the Frobenius norm of $\mathbf{A}$ is $\operatorname{Tr}(\mathbf{A}\mathbf{A}^{\dagger})$, we have:

$$\sum_{1\leq i,j \leq n}|A_{ij}|^{2}=\operatorname{Tr}(\mathbf{A}\mathbf{A}^{\dagger})=\operatorname{Tr}(\mathbf{Q}\mathbf{U}\mathbf{U}^{\dagger}\mathbf{Q}^{\dagger})=\operatorname{Tr}(\mathbf{U}\mathbf{U}^{\dagger})=\operatorname{Tr}((\mathbf{\Lambda}+\mathbf{N})(\mathbf{\Lambda}+\mathbf{N})^{\dagger})$$

We get the final equality from the property of an upper triangular matrix that it can be written as the sum of a strictly upper triangular matrix $\mathbf{N}$ and a diagonal matrix, which in this case is the matrix with eigenvalues along its diagonal $\mathbf{\Lambda}=\operatorname{diag}(\lambda_{1},\dots,\lambda_{n})$. We also use the property that the conjugate transpose is distributive over matrix addition, i.e. $(\mathbf{\Lambda}+\mathbf{N})^{\dagger}=\mathbf{\Lambda}^{\dagger}+\mathbf{N}^{\dagger}$. Therefore we have:

$$\sum_{1\leq i,j \leq n}|A_{ij}|^{2}=\operatorname{Tr}((\mathbf{\Lambda}+\mathbf{N})(\mathbf{\Lambda}^{\dagger}+\mathbf{N}^{\dagger}))=\operatorname{Tr}(\mathbf{\Lambda}\mathbf{\Lambda}^{\dagger}+\mathcal{O}(\mathbf{\Lambda}))$$

And therefore because $\operatorname{Tr}(\mathbf{\Lambda}\mathbf{\Lambda}^{\dagger}) \leq \operatorname{Tr}(\mathbf{\Lambda}\mathbf{\Lambda}^{\dagger} + \mathcal{O}(\mathbf{\Lambda}))$, we have:

$$\sum_{i=1}^{n}|\lambda_{i}|^{2}\leq \sum_{1\leq i,j \leq n}|A_{ij}|^{2}$$

As required.

Note that if a matrix $\mathbf{A}$ is normal, i.e. it satisfies $\mathbf{A}^{\dagger}\mathbf{A}=\mathbf{A}\mathbf{A}^{\dagger}$ then $\mathbf{A}$ is diagonalizable, i.e. $\exists \mathbf{P}\in\mathbb{C}^{n\times n}$ such that $\mathbf{A}=\mathbf{P}\mathbf{\Lambda}\mathbf{P}^{\dagger}$ and $\mathbf{\Lambda}=\operatorname{diag}(\lambda_{1},\dots,\lambda_{n})$ and $\mathbf{P}$ is unitary.

Multiplying by the complex conjugate of both sides:

$$\mathbf{A}\mathbf{A}^{\dagger}=(\mathbf{P}\mathbf{\Lambda}\mathbf{P}^{\dagger})(\mathbf{P}\mathbf{\Lambda}\mathbf{P}^{\dagger})^{\dagger}=\mathbf{P}\mathbf{\Lambda}\mathbf{P}^{\dagger}\mathbf{P}\mathbf{\Lambda}^{\dagger}\mathbf{P}^{\dagger}=\mathbf{P}\mathbf{\Lambda}\mathbf{\Lambda}^{\dagger}\mathbf{P}^{\dagger}$$

Again using the property that the Frobenius norm is $\operatorname{Tr}(\mathbf{A}\mathbf{A}^{\dagger})$, we take the trace of both sides:

$$\sum_{1 \leq i,j \leq n}|A_{ij}|^{2}=\operatorname{Tr}(\mathbf{A}\mathbf{A}^{\dagger})=\operatorname{Tr}(\mathbf{P}\mathbf{\Lambda}\mathbf{\Lambda}^{\dagger}\mathbf{P}^{\dagger})=\operatorname{Tr}(\mathbf{\Lambda}\mathbf{\Lambda}^{\dagger})=\sum_{i=1}^{n}\lambda_{i}\lambda_{i}^{*}=\sum_{i=1}^{n}|\lambda_{i}|^{2}$$

So we have equality iff $\mathbf{A}$ is normal.

$\endgroup$
2
  • 2
    $\begingroup$ You did not say what $\mathcal O(\mathbf \Lambda)$ is. I think it should be $\text{Tr}(\mathbf N \mathbf N^\dagger)$. Then you can show equality holds in the case when $\mathbf A$ is normal without the second part of the proof. Also, you did not show the only if. Otherwise I would upvote your answer. $\endgroup$ Dec 30, 2013 at 15:10
  • 2
    $\begingroup$ Also, you did not show why $\text{Tr}(\mathcal O(\mathbf \Lambda) \ge 0$. $\endgroup$ Dec 30, 2013 at 15:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .