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Proof of Theorem 2.5.3

Reference : Golub G.H., Van Loan C.F.- Matrix Computations book.

I fail to understand the last line of the proof as to why,

$$||A-B||^2_2 \geq ||(A-B)z||^2_2 $$ I fail to understand how $z$ is so special.

1 Answers 1

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By definition, $$\|A-B\|_2=\max\{\|(A-B)z\|_2:\|z\|\leq1\}. $$ Thus, for any $z\in \mathbb R^n$ with $\|z\|_2\leq 1$, we have $$ \|(A-B)z\|_2\leq\|A-B\|_2, $$ and squaring both sides of the inequality gives the resulting inequality.

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    @Awygan Thanks for the explanation, but $z$ is the 2-norm vector in the intersection. I don't think $z \in R^n$ but it only belongs in the intersection subspace. – 2017-01-20
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    @learner I don't understand. If $z$ is an element of a subspace of $\mathbb R^n$ then $z\in\mathbb R^n$. – 2017-01-20
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    @ Awygan I believe the intersection of subspace, i.e. $z \in$ span($x_1, x_2, ... x_k$) $\cap$ span($v_1, v_2, .. $) which is a subset of $R^n$ and not the entire space $R^n$. – 2017-01-20
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    @learner the intersection of subspaces is again a subspace, so it contains a unit vector. – 2017-01-20
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    @ Awygan I believe the intersection of subspace, i.e. $z \in$ span($x_1, x_2, ... x_k$) $\cap$ span($v_1, v_2, .. $) which is a subset of $R^n$ and not the entire space $R^n$. Sorry for re-commenting. I had to edit the last comment. Have I understood $z$ wrong ? – 2017-01-20
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    @learner The set $\text{span}(x_1,\ldots,x_{n-k})\cap\text{span}(v_1,\ldots,v_{k+1})$ is a **nontrivial subspace** of $\mathbb{R}^n$, which means it contains a $2$-norm unit vector. – 2017-01-20