Ytukyg

Let $U$ be an orthonormal matrix. $U$ has $n$ rows and $n$ columns. We know that the first row is given by the following vector
$$bigg ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} bigg).$$
Let's consider a column vector $X$.
$$X = (mu, ... mu)^{T}.$$
I am to prove that
$$UX = big( sqrt{n} mu, 0, 0, ldots, 0)^{T} tag{1}.$$
Both $U$ and $X$ are such matrices that their dimension fits and they can be multiplied in the way shown above.

How can I prove $(1)$?

It should be clear that the first entry in $UX$ is given by

$( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} ) cdot(mu, ... mu)^{T}= sqrt{n} mu$.

Now suppose that the $j -th$ row of $U$ has the form $(a_1,a_2,...,a_n)$. Since $U$ is orthonormal we have

$0=(a_1,a_2,...,a_n) cdot ( frac{1}{sqrt{n}}, ldots frac{1}{sqrt{n}} )^T =frac{1}{sqrt{n}}(a_1+a_2+...+a_n)$, thus $a_1+a_2+...+a_n=0$.

The $j -th$ entry in $UX$ is therefore

$$= mu(a_1+a_2+...+a_n)=0.$$

Let me give an alternative to @Fred's answer, or at least a more geometric version, based on three observations:

1. The rows $u_1, u_2, ldots u_n$ of $U$ are orthonormal vectors, i.e., each has length $1$, and they are all perpendicular, i.e., $u_i cdot u_j = 0$ if $i ne j$, and $u_i cdot u_j = 1$ if $i = j$.




2. The product $UX$ consists of the dot product $u_i cdot X$ of each row with the vector $X$.




3. The vector $X$ is a scalar multiple $c u_1$ of the first row, with the scalar being $c = sqrt{n} mu$.

So what's the dot product of the first row with $X$? It's just $c u_1 cdot u_1 = c $.

What about the dot product $u_i cdot X$ for any $i ne 1$? Well, because $X$ is parallel to $u_1$, but $u_1$ is perpendicular to all the other $u_i$, we must have $u_i cdot X = 0$.