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I need help with understanding a step in a proof for the following exercise:

Let $P\in\mathbb{C}^{m\times n}$ be a non-zero projector. Show that $\lVert P \rVert_2 =1$ iff $P$ is an orthogonal projector.

First it shows that $\lVert P\rVert_2\geq 1$ which is fine. After establishing this, it goes as follows:

Let $x\in\mathbb{C}^{m}$, and $P=U\Sigma V^*$ be the SVD of $P$, we have: $$\lVert Px\rVert_2^2= x^*P^*Px=x^*V\Sigma^*U^*U\Sigma V^*x = x^*V\Sigma^2V^*x. $$ Therefore: $$\lVert P\rVert_2=\max_{x\in\mathbb{C}^m\setminus\{0\}} \frac{\lVert Px\rVert_2}{\lVert x\rVert_2}=^1\max_{x\in\mathbb{C}^m\setminus\{0\}} \frac{\lVert \Sigma V^*x\rVert_2}{\lVert V^*x\rVert_2}=^2 \max_{x\in\mathbb{C}^m\setminus\{0\}} \frac{\lVert \Sigma x\rVert_2}{\lVert x\rVert_2}=\lVert \Sigma \rVert_2$$

The conclusion after this is easy and I understand it clearly so I won't show it. However, I have trouble understanding step $1$ and $2$. For $1$ I tried figuring out the following:

We know that multiplication by unitary matrices does not affect norms, hence $\lVert x\rVert_2 = \lVert V^*x\rVert_2$ since $V$ is unitary. Thus we can make the substitution on the denominator in step $1$. For the numerator: $$\lVert Px\rVert_2^2= x^*V\Sigma^2V^*x = \left( \Sigma V^* x\right)^*\left( \Sigma V^* x\right)=\lVert \Sigma V^* x\rVert_2^2 \Rightarrow\lVert Px\rVert_2 = \lVert \Sigma V^* x\rVert_2.$$

I'm not even sure this is the right explanation for step $1$. As for step $2$, the explanation for the denominator is the same as before but as for the numerator I'm clueless because $V^*$ is in the middle of a matrix and a vector in the product. Any help would be appreciated! Thanks.

davinci_07
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1 Answers1

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Your reasoning for $(1)$ looks good to me. For $(2)$ observe that \begin{align*} \max_{x\in\mathbb{C}^m\setminus\{0\}} \frac{\lVert \Sigma V^*x\rVert_2}{\lVert V^*x\rVert_2} &= \max_{y \in \text{Image}(V^\ast)\setminus \{0\}} \frac{\lVert \Sigma y\rVert_2}{\lVert y\rVert_2} \\ &= \max_{y \in \mathbb{C}^m\setminus\{0\}} \frac{\lVert \Sigma y\rVert_2}{\lVert y\rVert_2} \\ \end{align*}

Where the last line follows because unitary matrices are invertible, and hence $\text{Image}(V^\ast)\setminus \{0\} = \mathbb{C}^m \setminus \{0\}$. Then just relabel $y$ as $x$.

infinitylord
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