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What will be the operator norm of the matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix},$ where $a,b,c,d \in \Bbb C\ $?

According to the definition of the operator norm it turns out that $$\|A\|^2 = \sup \left \{ \left (|a|^2 + |c|^2 \right ) |z|^2 + \left (|b|^2 + |d|^2 \right ) |w|^2 + 2\ \mathfrak R\ (a \overline b z \overline w) + 2\ \mathfrak R\ (c \overline d z \overline w) \ \bigg |\ |z|^2 + |w|^2 = 1,\ z,w \in \Bbb C \right \}.$$

Is there any way to simplify the above expression? Any help will be highly appreciated.

Thanks for your time.

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2 Answers 2

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Let $A$ be $n\times n$ matrix, $\|{A}\| = \sup\left\{ \|Ax\| : \|x\| = 1\right\}$. For any $x$ with $\| x\| = 1$ we have $\|Ax\|^2 = (Ax, Ax) = (A^* A x, x)$, where $A^*$ is conjugate transpose matrix for A. $A^* A$ is self-adjoint, so $A^* A = U D U^*$ for some unitary matrix $U$ and $D = \mathrm{diag}(\lambda_1, ..., \lambda_n)$, where $\lambda_k$ are non-negative eigenvalues. Hence, $(A^* A x, x) = (U D U^* x, x) = (D U^* x, U^*x)$. Let $y = U^* x$. As $U$ is unitary, $\| U^* x\| = \|x\|$, so $\|Ax\|^2 = (Dy, y)$ and $\|{A}\| = \sup\left\{ \sqrt{(Dy, y)} : \|y\| = 1\right\}$. $$ (Dy, y) = \lambda_1|y_1|^2 + \lambda_2|y_2|^2 + ... + \lambda_n|y_n|^2 \leq \underset{1\leq k \leq n}{\max} \lambda_k \cdot \sum_{j=1}^n |y_j|^2 = \underset{1\leq k \leq n}{\max} \lambda_k \cdot ||y||^2 = \underset{1\leq k \leq n}{\max} \lambda_k$$ Moreover, for $y = (0, ..., 0, 1, 0, ..., 0)$ with $1$ at $k$-th place ($k$ is index of largest eigenvalue) there is an equality.

So, $\|A\| = \sqrt{\underset{1\leq k \leq n}{\max} \lambda_k}$, where $\lambda_k$ are eigenvalues of $A^* A$.

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    $\begingroup$ I think in the end you would like to say "eigenvalues of $A^*A$", right? $\endgroup$ Feb 21, 2021 at 21:48
  • $\begingroup$ Also the part with the equality is not exactly true. You want $k$ to be the index corresponding to the maximum. Anyway, nice answer. $\endgroup$ Feb 21, 2021 at 21:52
  • $\begingroup$ Yes, my bad, eigenvalues of $A^*A$ and index of maximum, thank you :) $\endgroup$ Feb 21, 2021 at 22:13
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Yes, there is. Please, keep in mind that I do not know the definition of operator norm, but if you did that part of job correctly, then there is a simplification.

Firstly, focus on the last half of your expression, the one with real parts. That sum is equal to $\Re(C_3\cdot z\overline{w})$, where $C_3=2(a\overline{b}+c\overline{d})$. Rotating $z$ or $w$ around the zero would affect only this summand, so you are free to multiply $z$ and/or $w$ with appropriate constant $e^{i\varphi}$ to achieve $\Re(C_3\cdot z\overline{w})=|C_3|\cdot|zw|$.

Now, your expression becomes $$\sup\{C_1\cdot|z|^2+C_2\cdot|w|^2+|C_3|\cdot|zw|\mid|z|^2+|w|^2=1,z,w\in\mathbb{C}\}.$$ By agreeing that $C_3$ replaces the $|C_3|$ from now on, your problem reduces to problem of finding the maximum of $$C_1\cdot a^2+C_2\cdot b^2+C_3\cdot ab,\text{ where }a^2+b^2=1,$$where all numbers involved are real and positive. Now you can apply methods like Lagrangian multipliers to find the maximum and then substitute the expressions behind the constants to get what you want. But it will still be an ugly expression.

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