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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 $(1)$?

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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

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

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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 $u_i \cdot X = 0$.

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