Vtyjkyuk

This may be a trivial question yet I was unable to find an answer:

$$left | A right | _2=sqrtlambda_textmax(A^^*A)=sigma_textmax(A)$$

where the spectral norm $left | A right | _2$ of a complex matrix $A$ is defined as $$textmax leftx$$

How does one prove the first and the second equality?

Put $B=A^*A$ which is a Hermitian matrix.

As a linear transformation of Euclidean vector space $E$ is Hermite iff there exists an orthonormal basis of E consisting of all the eigenvectors of $B$

Let $lambda_1,...,lambda_n$ be the eigenvalues of $B$ and $left e_1,...e_n right $ be an orthonormal basis of $E$

Let $x=a_1e_1+...+a_ne_n$

we have $left | x right |=left langle sum_i=1^na_ie_i,sum_i=1^na_ie_i right rangle^1/2 =sqrtsum_i=1^na_i^2$,

$Bx=Bleft ( sum_i=1^na_ie_i right )=sum_i=1^na_iB(e_i)=sum_i=1^nlambda_ia_ie_i$

Denote $lambda_j_0$ to be the largest eigenvalue of $B$.

Therefore,

$left | Ax right |=left langle Ax,Ax right rangle=left langle x,A^*Ax right rangle=left langle x,Bx right rangle=left langle sum_i=1^na_ie_i,sum_i=1^nlambda_ia_ie_i right rangle=sqrtsum_i=1^na_ioverlinelambda_ia_i leq underset1leq jleq nmaxsqrtlambda_j right times (left | x right |)$

So, if $left | A right |$ = $textmax left : $ then $left | A right |leq underset1leq jleq nmaxsqrtlambda_j right $ (1)

Consider: $x_0=e_j_0$ $Rightarrow left | x right |=1$ so that $left | A right | geq left langle x,Bx right rangle=left langle e_j_0,B(e_j_0) right rangle=left langle e_j_0,lambda_j_0 e_j_0 right rangle=sqrtleft $ (2)

Combining (1) and (2) gives us $left | A right |= underset1leq jleq nmaxsqrt lambda_j right $ where $lambda_j$ is the eigenvalue of $B=A^*A$

Conclusion: $$left | A right | _2=sqrtlambda_textmax(A^^*A)=sigma_textmax(A)$$

First of all,
$$beginalign*sup_||Ax||_2 & = sup_||USigma V^Tx||_2 = sup_||Sigma V^Tx||_2endalign*$$
since $U$ is unitary, that is, $||Ux_0||_2^2 = x_0^TU^TUx_0 = x_0^Tx_0 = ||x_0||_2^2$, for some vector $x_0$.

Then let $y = V^Tx$. By the same argument above, $||y||_2 = ||V^Tx||_2 = ||x||_2 = 1$ since $V$ is unitary.
$$sup_||Sigma V^Tx||_2 = sup_y||Sigma y||_2$$
Since $Sigma = mboxdiag(sigma_1, cdots, sigma_n)$, where $sigma_1$ is the largest singular value. The max for the above, $sigma_1$, is attained when $y = (1,cdots,0)^T$. You can find the max by, for example, solving the above using a Lagrange Multiplier.