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Let $T$ be a linear operator on a finite-dimensional inner product space $V$. Suppose that $T$ is a projection such that $\|T(x)\| \le \|x\|$ for $x \in V$. Prove that $T$ is an orthogonal projection.

I can't understand well. The definition of orthogonal operator is $\|T(x)\| = \|x\|$. But why that $\|T(x)\| \le \|x\|$ for $x \in V$ means orthogonal projection?

Michael Albanese
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nonam
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2 Answers2

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Note that $\|Tx\|=\|x\|$ is the definition of an isometry. Over a finite-dimensional real inner product space, this is equivalent to the matrix of $T$ in an orthonormal basis being "orthogonal", i.e. $A^TA=AA^T=I_n$.

Orthogonal projection means $T^2=T$ and $T^*=T$, i.e. self-adjoint idempotent. The first thing you should remark is that this is equivalent to $T$ being idempotent with $\ker T \perp \mbox{im} T$. Indeed, if $T$ is idempotent, then $T^*$ is idempotent with $\ker T^*=(\mbox{im} T)^\perp$ and $\mbox{im} T^*=(\ker T)^\perp$.

Claim: if $T$ is idempotent and $\|T\|\leq 1$ (i.e. $\|Tx\|\leq \|x\|$ for all $x\in V$), then $T$ is self-adjoint (i.e. $T$ is an orthogonal projection).

Remark: the converse is true since then the orthogonal direct sum $V=\ker T\oplus \mbox{im} T$ yields $\|x+y\|^2=\|x\|^2+\|y\|^2\geq \|y\|^2=\|0+Ty\|^2=\|T(x+y)\|^2$ for all $x\in \ker T$ and all $y\in \mbox{im} T$. Note that all this holds on a general inner product space. No need to assume finite dimension.

Proof: we need to prove that $(x,y)=0$ for every $x\in \ker T$ and every $y\in\mbox{im} T$. Let us take two such vectors, which are characterized by $Tx=0$ and $Ty=y$. Then for every $t\in\mathbb{R}$ $$ t^2\|y\|^2=\|ty\|^2=\|T(x+ty)\|^2\leq \|x+ty\|^2=\|x\|^2+2t\,\mbox{Re}(x,y)+t^2\|y\|^2 $$ whence $$ \mbox{Re}(x,y)\geq -\frac{\|x\|^2}{2t}\;\forall t>0\qquad\mbox{and}\qquad \mbox{Re}(x,y)\leq -\frac{\|x\|^2}{2t}\;\forall t<0 $$ which implies $\mbox{Re}(x,y)=0$ by letting $t$ tend to $\pm \infty$. In the real case, we are done since $\mbox{Re}(x,y)=(x,y)=0$. In the complex case, take $e^{i\theta}$ such that $|(x,y)|=e^{i\theta}\mbox{Re}(x,y)=\mbox{Re}(x,e^{i\theta}y)$ and apply the above to $x,e^{i\theta}y$ to conclude that $|(x,y)|=0$ whence $(x,y)=0$. QED.

Julien
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    Is it possible to prove it by converse direction? I mean, if $T$ is not orthogonal, then there exist some vector $u \in N(t)^\perp$ such that $T(u) \neq u$. Then $T(u)=u+(T(u)-u)$ and since $T$ is a projection, $T(u)-u \in N(T)$. By pythagoras' theorem, $\parallel T(u)\parallel^2=\parallel u\parallel^2+\parallel T(u)-u \parallel^2$ so $\parallel u \parallel < \parallel T(u)\parallel$. – nonam Jun 09 '13 at 13:25
  • +Therefore, if $\parallel T(x) \parallel \le \parallel x\parallel$ then $T$ is an orthogonal projection. Is this the right way to prove this problem? – nonam Jun 09 '13 at 13:31
  • @nonam As far as I know, the standard argument is the one I gave above. What you say proves that $(\ker T )^\perp \subseteq \mbox{im} T$. What we want is the other direction. But if you apply this to $T^*$, you get what you want. So that works too. – Julien Jun 09 '13 at 13:38
  • In the solution book, the reverse version of your proof is shown. It also says that if it is not orthogonal, $ \parallel tx+y \parallel ^2= \parallel tx \parallel ^2+2\Re<x,y>+ \parallel y \parallel ^2= \parallel tx \parallel ^2- \parallel y \parallel ^2$ where $x \in W$, $y \in W'$ and $t=2 \parallel y \parallel ^2/2\Re<x,y>$. Therefore $\parallel tx+y \parallel ^2 < \parallel T(tx+y) \parallel ^2= \parallel tx \parallel ^2$ and $T$ must be an orthogonal. I can't understand why $ \parallel tx \parallel ^2 - \parallel y \parallel ^2$ in the result. – nonam Jun 09 '13 at 13:50
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    @nonam Note that it is usually better to give direct proofs when they are available. Probably they meant $t=-|y|^2/Re(x,y)$ so that $|tx+y|^2=t^2|x|^2-|y|^2$. – Julien Jun 09 '13 at 13:57
  • Yeah, you're right. Something is wrong with this solution... Thanks a lot! – nonam Jun 09 '13 at 14:03
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Recall that $T$ is and orthogonal proyection iff $T^2 = T = T^*$, as T is already a projection, we have that $T^2 = T$, also for the adjoint ($(T^*)^2 = T^*$), then we show that $T = T^*$:

\begin{align} \|(T^* - TT^*)(x)\|^2 &= <T^*(x) - TT^*(x), T^*(x) - TT^*(x)> \\ &= <T^*(x), T^*(x) - TT^*(x)> - <TT^*(x), T^*(x) - TT^*(x)> \\ &= <x, TT^*(x) - TT^*(x)> - <TT^*(x), T^*(x)> + <TT^*(x), TT^*(x)> &(T^2 = T)\\ &= - <T^*(x), T^*(x)> + <TT^*(x), TT^*(x)> &((T^*)^2 = T^*)\\ &\leq -<T^*(x), T^*(x)> + <T^*(x), T^*(x)> &(\|T(x)\| \leq \|x\|)\\ &= 0 \end{align}

as $x$ is arbitrary we conclude that $T^* = TT^*$, and also $T = T^*T$, now:

\begin{align} \|(T-T^*)(x)\|^2 &= <T(x)-T^*(x), T(x)-T^*(x)> \\ &= <T(x), T(x)-T^*(x)> - <T^*(x), T(x)-T^*(x)> \\ &= <x, T^*T(x) - T^*(x)> - <x, T(x) - TT^*(x)> &((T^*)^2 = T^*, \;T^2 = T)\\ &= <x, T(x) - T^*(x)> - <x, T(x) - T^*(x)> &(T^* = TT^*, \; T = T^*T) \\ &= 0 \end{align}

as $x$ is arbitrary, the claim follows.

Daniel Martinez
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