Angle and distance between subspaces.

Question

Let $\text{Sin}(\Theta)$ be the diagonal matrix containing the $Sine$'s of principal angles between two subspaces $\mathcal{A}$ and $\mathcal{B}$, and let $\Pi_A$ and $\Pi_B$ be their respective projectors. How do I show that $\|\text{Sin}(\Theta)\| = \frac{1}{\sqrt{2}}\cdot\|\Pi_A - \Pi_B\|$ (where $\|\cdot\|$ is the spectral norm)?

Edit: the $\frac{1}{\sqrt{2}}$ factor should not be there as per the answer below.

Answer 1

I seem to obtain a slightly different result. In particular, I think that the factor of $\frac 1{\sqrt{2}}$ should not be there.

I will focus on the case that both subspaces are $\ell$-dimensional subspaces of $\Bbb C^n$ with $2 \ell \leq n$. Bhatia's Matrix Analysis proves this result (rather, sets up an exercise in which one may prove this result) by using the following consequence of the existence of the CS-decomposition.

Theorem VII.1.8: Let $X,Y$ be $n \times \ell$ matrices with orthonormal columns. Then there exist $\ell \times \ell$ unitary matrices $U,V$ and an $n \times n$ unitary matrix $Q$ with the following properties.

If $2\ell \leq n,$ then

$$ QX U = \pmatrix{I_\ell \\ 0\\0}, \quad QY V = \pmatrix{C \\ S\\ 0} $$ where $C,S$ are diagonal matrices with diagonal entries $0 \leq c_1 \leq \cdots \leq c_\ell \leq 1$ and $1 \geq s_1 \geq \cdots \geq s_\ell \geq 0$ (respectively) and $C^2 + S^2 = I$.

There is also a version of this theorem for the case that $2\ell > n$ which I am ignoring for the purposes of my answer. Now, suppose that $X$ and $Y$ have columns that form orthonormal bases for $\mathcal A, \mathcal B$ respectively. Let $Q,U,V$ be as in the above theorem. Let $\mathcal A',\mathcal B'$ denote the images of $\mathcal A,\mathcal B$ respectively under $Q$. That is, $\mathcal A' = \{Qx : x \in \mathcal A\}$, and $\mathcal B'$ is defined similarly.

Note that the columns of $QXU$ form an orthonormal basis for $\mathcal A'$, and the columns of $QYV$ form an orthonormal basis for $\mathcal B'$. Moreover, the principal angles between $\mathcal A'$ and $\mathcal B'$ must be equal to those between $\mathcal A$ and $\mathcal B$. With that, we can see that $S = \sin(\Theta)$, as you have defined it.

Now, we note that $\Pi_{\mathcal A} = XX^*$, so that \begin{align} \left\|\Pi_{\mathcal A} - \Pi_{\mathcal B} \right\| & = \left\|XX^* - YY^* \right\| \\ & = \left\|Q(XX^* - YY^*)Q^* \right\| \\ & = \left\|(QXU)(QXU)^* - (QYV)(QYV)^* \right\| \\ & = \left\|\pmatrix{I_\ell & 0 & 0\\0 & 0 & 0\\0 & 0 & 0} - \pmatrix{C^2 & CS& 0\\CS & S^2 & 0\\0 & 0 & 0} \right\| \\ & = \left\|\pmatrix{I - C^2 & -CS & 0\\ -CS & -S^2 & 0\\0 & 0 & 0}\right\| = \left\|\pmatrix{S^2 & -CS & 0\\ -CS & -S^2 & 0\\0 & 0 & 0}\right\|. \end{align} So, we now need to compute the spectral norm of the matrix $$ \pmatrix{S^2 & -CS\\ -CS & -S^2}. $$ We see that there is a permutation-similarity between this matrix and the block-diagonal matrix $$ \pmatrix{M_1 \\ & \ddots \\ && M_\ell}, \quad M_k = s_k\pmatrix{s_k & -c_k\\ -c_k & -s_k}, \quad k = 1,\dots,\ell. $$ We see that $M_k$ is $s_k$ multiplied by an orthogonal matrix, which means that we have $\|M_k\| = s_k$. Thus, we conclude that $\|\Pi_{\mathcal A} - \Pi_{\mathcal B}\|$ is equal to $\max_{k = 1,\dots,\ell} s_k$, which coincides with $\|S\| = \|\sin(\Theta)\|,$ which was what we wanted.