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Theorem: Suppose $U,V$ are orthogonal subspaces spans $X$, let $A:U\to U,B:V\to V$. then the operator $C=\begin{pmatrix}A&0\\0&B\end{pmatrix}:X\to X$

and $\|C\|=\max\{\|A\|,\|B\|\}$

using above theorem I am told to show the following

$A$ be any linear operator on finite dimensional vector spaces then $\|A\|=\max\{|s_1,\dots,\|s_k\|\}$ where $s_i$'s are singular values of $A$

from svd of $A$ we get $A=U\Sigma V^T \Rightarrow \|A\|=\|U\Sigma V^T\|\le\|U\|\|\Sigma\|\|V^T\|=\|\Sigma\|$ since unitary matrices has rank $1$...please help how to proceed

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In a finite dimensional space $A$ has an adjoint $A^*$ and $||A^*|| = ||A||$

Let $B = AA^*$ then $||B|| = ||A A^*|| = ||A||^2$

$B$ is self adjoint and non-negative therefore has a spectrum of real non-negative eigenvalues {$\lambda_i$}, corresponding orthogonal eigenspaces {$E_i$}, and projections {$P_i$} onto those eigenspaces. Furthermore $B = \Sigma \lambda_i P_i$.

Then by your given initial result, $||B|| = max(|| \lambda_i P_i||) = max(|\lambda_i|.|| P_i||) $. But for each projection, $||P_i|| = 1$.

So, $||B|| = max(|\lambda_i| )$ and $||A|| = ||B||^{1/2} = (max(|\lambda_i|) ^{1/2} = max(|\lambda_i|^{1/2}) $.

Since the eigenvalues of $B$ are non-negative then $||A|| = max(\lambda_i^{1/2})$ which by definition is the maximum of the singular values of $A$

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