Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can anyone help me proving the following:

Let $A\in\mathbb R^{m\times n}$ with $\textrm{rank}(A)=n$. Show that $\|A(A^TA)^{-1}A^T\|_2=1$.

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

Let $P = A(A^TA)^{-1}A^T $.

We have $PA = A$, so $\|P\|\|A\| \geq \|PA\| = \|A\|$ and hence $\|P\| \geq 1$ as $A \neq 0$.

Next, $P = P^TP$, we have for any $x$, $x^TPx = \|Px\|^2$, but by Cauchy-Schwarz $x^TPx \leq \|x\| \|Px\|$, so $\|Px\|^2 \leq \|x\| \|Px\|$, so $\|Px\| \leq \|x\|$ for all $x$ (if $Px=0$ then this is obvious), so $\|P\| \leq 1.$

share|improve this answer
add comment

Let $$ B:=A(A^tA)^{-1}A^t. $$

Then $B$ is symmetric and $$ B^2=B. $$

So the spectrum of $B$ is contained in $\{0,1\}$.

Since $B$ is nonzero (for otherwise one can easily show $A=0$) and is diagonalizable, it follows that $1$ belongs to the spectrum.

So $$\rho(B)=\rho(B^tB)=\|A(A^tA)^{-1}A^t\|_2^2=1.$$

Note: you might want to justify why $A^tA$ is invertible. First note that by the rank-nullity theorem, $A$ is injective. Then check that $\mbox{Ker} A^tA=\mbox{Ker}A$, essentially because $$ \|Ax\|^2=x^tA^tAx. $$

share|improve this answer
    
@Dion I hope this helps (and is correct). Let me know if you need details. –  1015 Feb 14 '13 at 21:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.