Matrix-Vector Product

Question

Suppose I have the expression $\lVert\mathbf B \cdot\hat n\lVert=1$, where $\mathbf B$ is a matrix and $\hat n$ is a unit vector (both can have any dimensions, as long as they are compatible). What restrictions must I set on $\mathbf B$ for this to be true?

If the expression was $\lVert\mathbf B \cdot\hat n\lVert\leq1$, I have concluded that you would only need to make sure that the absolute values of the elements of the columns of $\mathbf B^T\cdot\mathbf B$ add up to one or less. Would it be correct to say that to respect $\lVert\mathbf B \cdot\hat n\lVert=1$ they would need to add up to exactly one?

Much appreciated.

Answer 1

You demand that $(Bn,Bn) = 1$ for any unit vector $n$. From this it necessarily follows that $B^TB=I$.

To see this, first show that $B$ preserves lengths for all vectors.

Here is that step: Let $v$ be any non-zero vector. Then $(Bv,Bv) = ||v||^2(Bn,Bn)$ where $n$ is $v/||v||$. Note that $n$ is a unit vector. By assumption then, $(Bn,Bn) = 1$ and we have $(Bv,Bv) = ||v||^2$. But this says $||Bv||^2 = ||v||^2$, so that $||Bv|| = ||v||$. Clearly this holds when $v=0$ as well, so we have that $||Bv|| = ||v||$ for all vectors $v$.

From there it is standard to show that $B$ is also an isometry. From there it is standard to show that $B^TB=I$.

So you are not going to find any examples besides orthogonal matrices.

Answer 2

No, that is not sufficient. For instance, let $\mathbf{B} = \begin{bmatrix}1/2 & 1/2\\1/2 & 1/2 \end{bmatrix}$ and $\hat n = \begin{bmatrix} 1/\sqrt{5} \\ 2/\sqrt{5}\end{bmatrix}$. Then $\hat n$ is a unit vector and $\mathbf B^T\mathbf B = \mathbf B$ has the sum of absolute values of the elements in its columns equal to 1. Yet $\mathbf B \cdot \hat n = \begin{bmatrix} \frac{3}{2\sqrt{5}} \\ \frac{3}{2\sqrt{5}} \end{bmatrix}$ has norm .9, not 1.

It is sufficient that $\mathbf B$ be orthogonal, i.e. that $\mathbf B^T \mathbf B = \mathbf I$. In fact, it's not hard to show that if $\mathbf B$ has the property you state (that all columns of $\mathbf B^T \mathbf B$ sum to (at most) 1), and $\lVert \mathbf B \cdot \hat n \rVert = 1$ for any unit vector $\hat n$, then $\mathbf B$ is orthogonal. ($\mathbf B \cdot \hat{e_i}$ is the $i$th column of $B$, which therefore has norm 1; the squared norm of the $i$th column is the $i$th diagonal entry of $\mathbf B^T \mathbf B$.)

It is also necessary that $\mathbf B$ be orthogonal: notice that requiring that $\lVert \mathbf B \cdot \hat n \rVert = 1$ for any unit vector $\hat n$ is equivalent to requiring $\lVert \mathbf B \cdot \hat v \rVert = \lVert \hat v \rVert$ for any vector $\hat v$, i.e., that the linear transformation determined by $\mathbf B$ is an isometry. Then see, for instance, Proving that a linear isometry on $\mathbb{R}^{n}$ is an orthogonal matrix.