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

Can anyone help me proving the following:

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

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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 $\|P\| \leq 1.$

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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 $$\|Ax\|^2=x^tA^tAx.$$

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