# The shape of an orthonormal matrix

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 $$u_i \cdot X = 0$$.