Why does such an orthogonal vector exist?

Question

Suppose $W$ and $V$ are $k$-dimensional subspaces of $\mathbb{R}^m$ for some $m \geq k$. Let $V$ have orthonormal basis $\mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_k}$. Then is it possible to choose an orthonormal basis $\mathbf{w_1}, \mathbf{w_2}, \ldots, \mathbf{w_k}$ for $W$ such that $\mathbf{w_k}$ is orthogonal to $\mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v}_{k-1}$?

Answer 1

Let $V_0$ be the span of $\mathbf{v_1},\dots,\mathbf{v}_{k-1}$ and let $P:\mathbb{R}^m\to \mathbb{R}^m$ be the orthogonal projection onto $V_0$. We can restrict $P$ to a linear map $W\to V_0$. Since $\dim W=k$ and $\dim V_0=k-1$, the kernel of this linear map must be nontrivial. Thus there is a nonzero vector $\mathbf{w}_k\in W$ (which you can scale to be a unit vector) whose projection onto $V_0$ is zero, which just means it is orthogonal to every element of $V_0$. Extending $\mathbf{w}_k$ to an orthonormal basis of $W$, you have the desired basis.

(In fact, by iterating this construction, you can get an orthonormal basis such that $\mathbf{w}_i$ is orthogonal to $\mathbf{v}_1,\dots,\mathbf{v}_{i-1}$ for all $i$, not just for $i=k$. For instance, after choosing $\mathbf{w}_k$, you can let $W'$ be the orthogonal complement of $\mathbf{w}_k$ in $W$ and consider the orthogonal projection map from $W'$ to the span of $\mathbf{v_1},\dots,\mathbf{v}_{k-2}$. Again, for dimension reasons, this projection must have nontrivial kernel, and you can let $\mathbf{w}_{k-1}$ be any unit vector in its kernel. You can then choose $\mathbf{w}_{k-2}$ similarly, and so on.)