Sum of squared dot products of a unit vector with columns of an orthogonal matrix

Question

Say we have an orthogonal matrix $\ U=[u_1 \ u_2 \ \cdots \ u_n]$, where $u_i, \ i={1,\ldots,n}$, is the ith columnn of $U$. If I took the dot product of a unit vector $x\in \mathbb{R}^n$ with each of the columns of $ \ U$, squared each of them, and summed it all up, would the result of $\sum_{i=1}^n(x^Tu_i)^2?$ be 1? In other words, is the vector with components $x^Tu_i, \ i={1,\ldots,n}$, a unit vector?

Answer 1

For an orthogonal matrix, $U$, we have $$ \color{#C00}{UU^T}=I $$ The sum of squares of the dot product of a unit vector, $v$, with the rows of $U$ is $$ \begin{align} \|vU\|^2 &=vU(vU)^T\\ &=v\color{#C00}{UU^T}v^T\\ &=vv^T\\ &=\|v\|^2 \end{align} $$

Answer 2

An orthogonal matrix represent a change of basis between orthonormal basis. Then if $A\in\Bbb R^{n\times n}$ is orthogonal and $x\in\Bbb R^n$ then what you are describing is just $(Ax|Ax)$.

And because orthogonal matrices preserves inner products then

$$(Ax|Ax)=(x|x)=\|x\|^2$$

If $\|x\|^2=1$ it imply that $x$ is an unit vector, and viceversa.