# Apply the Hadamard transform to different state vectors

I am in the process of understanding a proof. First, the following is said there:

$$H\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}=\frac{1}{\sqrt{N}}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}$$

This is clear so far, because the Hadamard transformation applied to the state $$|0...0\rangle$$ produces an equally distributed superposition. I know that.

The exciting part is actually here: $$H\begin{pmatrix}1\\1\\1\\1\end{pmatrix}=\sqrt{N}\begin{pmatrix}1\\0\\\vdots\\0\end{pmatrix}$$

I've already thought a few things about that, but I'm not quite sure why it really is. Here is my train of thought:

Say $$N=4$$ $$H\begin{pmatrix}1\\1\\1\\1\end{pmatrix}=H\begin{pmatrix}1\\0\\0\\0\end{pmatrix}+H\begin{pmatrix}0\\1\\0\\0\end{pmatrix}+H\begin{pmatrix}0\\0\\1\\0\end{pmatrix}+H\begin{pmatrix}0\\0\\0\\1\end{pmatrix}$$

$$=\frac{1}{\sqrt{N}}\begin{pmatrix}1\\1\\1\\1\end{pmatrix}+\frac{1}{\sqrt{N}}\begin{pmatrix}1\\-1\\1\\-1\end{pmatrix}+\frac{1}{\sqrt{N}}\begin{pmatrix}1\\1\\-1\\-1\end{pmatrix}+\frac{1}{\sqrt{N}}\begin{pmatrix}1\\-1\\-1\\1\end{pmatrix}$$

If you like, this results in: $$N\cdot \frac{1}{\sqrt{N}}|00\rangle=\sqrt{N}|00\rangle$$ to stay in the example $$2|00\rangle$$.

Honestly, that does not quite convince me (That's just the case for N = 4).

Why I do this is because I would like to understand the following:

$$H\begin{pmatrix}1&1&...&1\\1&1&...&1\\\vdots&\ddots&\ddots&\vdots\\1&1&...&1\end{pmatrix}H=\sqrt{N}\begin{pmatrix}1&1&...&1\\0&0&...&0\\\vdots&\ddots&\ddots&\vdots\\0&0&...&0\end{pmatrix}H=N\begin{pmatrix}1&0&...&0\\0&0&...&0\\\vdots&\ddots&\ddots&\vdots\\0&0&...&0\end{pmatrix}$$

Maybe someone of you can bring light into the darkness and help me a bit ...

The rows of a Hadamard matrix $$H$$ are orthogonal to each other, and each has Euclidean length $$\sqrt N$$.
[As a side note, that means $$HH^T=N\cdot I$$, in finite dimension it implies $$\exists H^{-1}=\frac1N\cdot H^T$$, so $$H^TH=N\cdot I$$, and that also the columns of $$H$$ are orthogonal to each other.]
In your setting, $$H$$ seems to be normed, i.e. it's $$\frac1{\sqrt N}H$$ with the above $$H$$.
Now, assuming the first row is $$(1,1,\dots,1)$$ (or its normed multiple), then it's orthogonal to every other row of $$H$$, which means that the product $$H\pmatrix{1\\1\\ \vdots\\1}$$ will have all zeroes except for the first row.