# Orthogonal Block Matrix

Given $$A \in \mathbb{R^{\text{nxn}}}$$, $$B \in \mathbb{R^{n\text{x}m}}$$ and $$C \in \mathbb{R^{m\text{x}m}}$$ such that $$M = \begin{bmatrix} A & B \\ 0 & C \end{bmatrix} \in \mathbb{R^{(n+m)\text{x}(n+m)}}$$ a block Matrix.

Prove that: $$M$$ is orthogonal $$\iff$$ $$A$$ and $$C$$ are orthogonal and $$B = 0$$

My solution idea:

$$\Longleftarrow$$

Since $$A$$ is orthogonal, its columns form a linearly independent orthonormal set of vectors. It follows that the first $$n$$ columns of $$M$$ are orthonormal, because a vector $$v_{i}$$ of the form $$(a_{1,i},a_{2,i},...a_{n,i},0,..,0)$$ for $$i = 1,..n$$ still has Norm equal to $$1$$ and $$\langle\ v_{i},v_{j} \rangle = 0$$ for $$j = 1,..,n$$ given $$i \neq j$$

Same logic applies to columns vectors of $$M$$ from $$n+1$$ to $$m$$ since $$C$$ is orthogonal. So, taken together $$\{v_{1},...,v_{n},v_{n+1},...,v_{m}\}$$ they build an orthonormal Basis for $$\mathbb{R^{(n+m)}}$$ and it follows that $$M$$ is orthogonal.

$$\Longrightarrow$$

Reversing the argument:

Since $$M$$ is an orthogonal Matrix, its columns vectors build an orthonormal Basis for $$\mathbb{R^{(n+m)}}$$

Given that $$a_{i,j} = 0$$ for $$i = m,..,n+m$$ and $$j =1,..,n$$ It follows that the columns vectors of $$A$$ build an orthonormal Basis for $$\mathbb{R^{n}}$$ and $$A$$ is orthogonal.

Now to prove $$B = 0$$ confuses me. Because the columns vectors of $$M$$ are orthonormal, it has to be that the inner product of each column vector of $$A$$ with the columns vectors of $$B$$ is zero. Is this enough to follow that $$B= 0$$?

If $$B = 0$$ then it's easy to follow that $$C$$ is orthogonal.

The $$n+i$$-th column of $$M$$ must be orthogonal to all the first $$n$$ columns of $$M$$. If you write the dot product, it's clear that the dot product of the $$n+i$$-th column of $$M$$ and the $$k$$-th column of $$M$$ (for $$k\leq n$$) is equal to the dot product of the $$i$$-th column of $$B$$ and the $$k$$-th column of $$A$$.
Therefore, the above demand is equivalent to demanding that the $$i$$-th column of $$B$$ must be orthogonal to all the $$n$$ columns of $$A$$.
Since the columns of $$A$$ span $$\mathbb R^n$$ (orthogonal nonzero vectors are always linearly independent), the only vector orthogonal to all the columns of $$A$$ is the zero vector.
• Yes, the last part is what I was missing. Since the columns of $A$ span $R^{n}$, it has to be that every column of $B$ is the zero vector. Thanks a lot :) – yarafoudah Sep 25 '18 at 13:06