Let $\hat{Q_i}$ be the householder reflector for the vector $A_{i:k,k}$, which is just $i$ through $k$ elements of column $i$. Now let:
$${Q_i} = \begin{pmatrix} I_{i-1} & 0\\ 0 & \hat{Q_i}\\ \end{pmatrix}$$
where $I_{i-1}$ is the (i-1) identity matrix. Prove that ${Q_i}$ is symmetric and orthogonal.
OK, so clearly ${Q_i}$ is symmetric. To show its orthogonal, I need to show that ${Q_i}{Q_i}^T$ is the identity. So since ${Q_i}$ is symmetric, then I just get the following:
$${Q_i}{Q_i} = \begin{pmatrix} I_{i-1}I_{i-1} & 0\\ 0 & \hat{Q_i}\hat{Q_i}\\ \end{pmatrix}$$
which reduces to: $${Q_i}{Q_i} = \begin{pmatrix} I_{i-1} & 0\\ 0 & \hat{Q_i}\hat{Q_i}\\ \end{pmatrix}$$
However, I'm stuck here. There might be some property of Householder reflectors I'm completely forgetting. And honestly, I have no idea if we're actually allowed to multiply block matrices in the way I did. Can anybody help? I'd appreciate it.
Householder reflectors are orthogonal. \begin{align} &\left(I-2\frac{vv^T}{v^Tv}\right)\left(I-2\frac{vv^T}{v^Tv}\right)\\ &=I-4\frac{vv^T}{v^Tv}+4\frac{v(v^Tv)v^T}{(v^Tv)^2}\\ &=I-4\frac{vv^T}{v^Tv}+4\frac{vv^T}{(v^Tv)}\\ &=I \end{align}