Defining the matrix norm as $||A||=\max_{u^\dagger u=1}|u^\dagger Au|$ then show that when $A=\begin{bmatrix}1&0\\1&1\end{bmatrix}$ we have $||A||=3/2$

My Attempt

If $u$ is a real vector, say $u=\begin{bmatrix}a\\b\end{bmatrix}$

\begin{align} u^\dagger Au&=\begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}a&b\end{bmatrix}\begin{bmatrix}a\\a+b\end{bmatrix}\\ &=a^2+ab+b^2=1+ab=1+a\sqrt{1-a^2}=1+\sqrt{a^2(1-a^2)}\\ |u^\dagger Au|&=|1+\sqrt{a^2(1-a^2)}|=1+\sqrt{a^2(1-a^2)} \end{align} where we used the fact that $u^\dagger u=a^2+b^2=1$. Now, $|u^\dagger Au|$ is maximum when $f(a)=a^2(1-a^2)=a^2-a^4$ is maximum.

$f'(a)=2a-4a^3=2a(1-2a^2)=0\implies a=0$ or $a=\pm 1/\sqrt{2}$

$\therefore\max_{u^\dagger u=1}|u^\dagger Au|=1+\sqrt{1/2(1-1/2)}=1+\sqrt{1/2.1/2}=1+\sqrt{1/4}=1+1/2=3/2$

What if $u$ is complex, how do I evaluate $||A||$?

When $u=\begin{bmatrix}a\\b\end{bmatrix}$ is possibly complex then $|u^\dagger Au|=|1+ab^*|$ \begin{align} |u^\dagger Au|^2&=|1+ab^*|^2=|1+(x_1+ix_2)(y_1-iy_2)|^2\\ &=|1+((x_1y_1+x_2y_2)+i(x_2y_1-x_1y_2))|^2\\ &=(1+x_1y_1+x_2y_2)^2+(x_2y_1-x_1y_2)^2\\ \end{align}

Thanks @Abezhiko for the hint,

\begin{align} u^\dagger Au&=1+\frac{1}{2}e^{i\theta}\sin(2t)=1+\frac{1}{2}\cos\theta\sin(2t)+i\frac{1}{2}\sin\theta\sin(2t)\\ f(\theta,t)&=|u^\dagger Au|^2=1+\cos\theta\sin(2t)+\frac{1}{4}\cos^2\theta\sin^2(2t)+\frac{1}{4}\sin^2\theta\sin^2(2t)\\ &=1+\cos\theta\sin(2t)+\frac{1}{4}\sin^2(2t)=1+\cos\theta\sin(2t)+\frac{2}{8}\sin^2(2t)\\ &=1+\cos\theta\sin(2t)+\frac{1}{8}(1-\cos(4t))=\frac{9}{8}+\cos\theta\sin(2t)-\frac{1}{8}\cos(4t)\\ f(\theta,t)&=\frac{9}{8}+\cos\theta\sin(2t)-\frac{1}{8}\cos(4t)\\ \frac{\partial f}{\partial\theta}&=-\sin\theta\sin(2t)=0\implies \sin\theta=0\text{ or }\sin(2t)=0\\ \frac{\partial f}{\partial t}&=2\cos(2t)\cos\theta+\frac{1}{2}\sin(4t)=\cos(2t)\Big(2\cos\theta+\sin(2t)\Big)=0\\ \sin(2t)&=0\implies\cos(2t)=1:\cos\theta=0\implies \boxed{t=0,\theta=\pi/2}\\ \sin\theta&=0\implies\cos\theta=1: \cos(2t)\Big(2+\sin(2t)\Big)=0\implies\cos(2t)=0\\ &\implies \boxed{t=\pi/4,\theta=0}\\ &t=\pi/4,\theta=0:|u_1^\dagger Au_1|=\sqrt{\frac{9}{8}+\cos\theta\sin(2t)-\frac{1}{8}\cos(4t)}=\sqrt{9/8+1+1/8}=\sqrt{18/8}=\sqrt{9/4}=3/2\\ &t=0,\theta=\pi/2:|u_2^\dagger Au_2|=\sqrt{\frac{9}{8}+\cos\theta\sin(2t)-\frac{1}{8}\cos(4t)}=\sqrt{9/8+0-1/8}=\sqrt{8/8}=1\\ \end{align}

$\therefore$, $\max_{u^\dagger u=1}|u^\dagger Au|=\max(1,3/2)=3/2$

  • $\begingroup$ The quantity $1+ab^*$ cannot be maximized as such, because it is complex $-$ a norm is alway positive. In the first place, you have to define a generalization of this matricial norm to $\mathrm{Mat}_2(\mathbb{C})$. When done, you can proceed the same way as you did for the real case. $\endgroup$
    – Abezhiko
    Mar 18 at 12:37
  • 1
    $\begingroup$ @Abezhiko Sorry there was a typo in the post. The matrix norm is rather $||A||=\max_{u^\dagger u=1}|u^\dagger Au|$ $\endgroup$
    – Sooraj S
    Mar 18 at 12:42

2 Answers 2


You can proceed the same way; however, it is perhaps easier to reparametrize the $u$-vectors beforehand. Indeed, if $u = (a,b)$, then the condition $u^\dagger u = |a|^2 + |b|^2 = 1$ implies $u = (e^{i\theta_1} \cos t, e^{i\theta_2} \sin t)$. Without loss of generality, we can always choose a frame of reference such that $\theta_1 = 0$ and $\theta_2 = \theta$ (alternatively, you can work with a variable $\phi = \theta_2 - \theta_1$). In consequence, we find $u^\dagger Au = 1 + ab^* = 1 + e^{i\theta} \sin t \cos t = 1 + \frac{1}{2}e^{i\theta}\sin(2t)$ and $$ f(t,\theta) := |u^\dagger Au| = \sqrt{1 + \sin(2t)\cos(\theta) + \frac{1}{4}\sin^2(2t)}\,,$$ which can be maximised as any bivariate function (i.e. by setting its gradient to zero, etc.).


Hint: By triangle inequality and AM-GM, we have $$|1 + ab^\ast| \le 1 + |ab^\ast| = 1 + |a|\cdot |b| \le 1 + \frac{|a|^2 + |b|^2}{2} = \frac32.$$


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