Prove that $\|A\|_2 = \sqrt{\|A^* A \|_2}$

Question

This is a homework exercise, so hints are welcome.

Prove $\|A\|_2 = \sqrt{\|A^* A \|_2}$.

The $\|\cdot\|_2$ is the induced 2-norm, and $A \in M_{n,m}(\mathbb{C})$. $A^*$ is the complex conjugate. Note that the matrix $2$-norm is defined as

$$ \sup_x\{\|Ax\|_2 \,\big|\,\|x\|_2 \leq 1 \}\text{,} $$

where $\|x\|_2$ is the Euclidean norm.

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So far, I tried expanding the norms, resulting in

$$ \sup_x\{\sum_i|(Ax)_i|^2\,\big|\, \|x\|_2 \leq 1 \} = \sup_x\{\sqrt{\sum_i |(A^* A x)_i|^2} \,\big|\,\|x\|_2 \leq 1 \}\text{.} $$

Unfortunately, I don't have a clue where to go from here.

Answer 1

Hint: $ |Av|^2 = |\langle Av, Av \rangle| = |\langle v, A^*A v \rangle| \leq |v| |A^*A v| $ by Cauchy-Schwarz. What does this tell you about $ ||A|| $?

Answer 2

From Cauchy-Schwarz you get an inequality as Starfall already mentioned. To get equality you need probably to know something about diagonalization of self-adjoint matrices (like $A^* A$). The fact is that you may find orthonormal vectors $e_1,...,e_n$ and associated positive real eigenvalues $\lambda_1\geq ...\geq \lambda_n\geq 0$ so that $A^*A e_k=\lambda_k e_k$. You may then proceed to show that equality in Cauchy-Schwarz is obtained for $e_1$ (and the value is $\lambda_1$).

Answer 3

Let the SVD of $\mathrm A$ be

$$\mathrm A = \mathrm U \Sigma \mathrm V^* = \begin{bmatrix} \mathrm U_1 & \mathrm U_2\end{bmatrix} \begin{bmatrix} \hat\Sigma & \mathrm O\\ \mathrm O & \mathrm O\end{bmatrix} \begin{bmatrix} \mathrm V_1^*\\ \mathrm V_2^* \end{bmatrix}$$

where the zero matrices may be empty. Hence, $\mathrm A^* = \mathrm V \Sigma^T \mathrm U^*$ and, thus,

$$\mathrm A^* \mathrm A = \mathrm V \Sigma^T \mathrm U^* \mathrm U \Sigma \mathrm V^* = \mathrm V \Sigma^T \Sigma \mathrm V^* = \begin{bmatrix} \mathrm V_1 & \mathrm V_2\end{bmatrix} \begin{bmatrix} \hat\Sigma^2 & \mathrm O\\ \mathrm O & \mathrm O\end{bmatrix} \begin{bmatrix} \mathrm V_1^*\\ \mathrm V_2^*\end{bmatrix}$$

Can you take it from here?