0
$\begingroup$

I was trying to prove this following theorem , Let

$$ W=\begin{bmatrix} \underset{\scriptscriptstyle n\times k} {W_1} && \underset{\scriptscriptstyle n\times (n-k)} {W_2} \end{bmatrix} $$ $$ Z= \begin{bmatrix} \underset{\scriptscriptstyle n\times k}{Z_1} && \underset{\scriptscriptstyle n\times (n-k)}{Z_2} \end{bmatrix} $$ be two orthogonal matrices. If $span \{W_1\}=S_1$ and $span\{Z_1\}=S_2$, then prove that $$ d(S_1,S_2)=\lVert{W_1^{T}Z_2}\rVert=\lVert{Z_1^{T}W_2}\rVert$$

This theorem is from Matrix computation by Gene H. Golub.

In the proof of that, we get enter image description here

Later it is proven that, $$\lVert{Q_{21}}\rVert_2= \lVert{Q_{12}}\rVert_2$$

But from this how do we directly get our desired result?

Is it something related to singular value of the block matrices? I'm not following this part.

Kindly guide through this.

$\endgroup$

1 Answer 1

2
$\begingroup$

It is true that it could be made more explicit. Perhaps it would be more clear if the expression for $d(S_1,S_2)$ contained one important equality, namely $$ \left\|\begin{bmatrix}0&W_1^TZ_2\\-W_2^TZ_1&0\end{bmatrix}\right\|_2=\max\{\|W_1^TZ_2\|_2,\|W_2^TZ_1\|_2\}. $$ The goal of the last part of the proof is then to show that the norms of both blocks are equal and hence no $\max$ is needed.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.