The Hadamard gate is described by this matrix
\begin{equation}
H=\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}
\end{equation}
Conjugate transpose of $H$ is again $H$. Hence we have to check if $HH$ is $I$.

Multiplication goes as follows ($h_{ij}$ denotes elements of resulting matrix):

- $h_{11} = 1\cdot1 + 1\cdot1 = 2$
- $h_{12} = 1\cdot 1 + 1\cdot (-1) = 0$
- $h_{21} = 1 \cdot1 + (-1) \cdot 1= 0$
- $h_{21} = 1 \cdot1 + (-1)\cdot (-1) = 2$

So

\begin{equation}
HH=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}
\begin{pmatrix}
2 & 0 \\
0 & 2
\end{pmatrix}
=
\frac{1}{2}
\begin{pmatrix}
2 & 0 \\
0 & 2
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
=I.
\end{equation}

1Have you tried taking the conjugate transpose of the H matrix (which is equal to itself) and multiplying it by H to see what the result looks like? I'm not sure what exactly the question is... – Mariia Mykhailova – 2020-01-03T08:36:11.463

sorry i am trying to learn. when i look at the matrix, i see a probability on the left (0.707), which can treated be a scalar and multiplied across each value in the matrix on the right. The first column would be .707, .707, and the second would be .707, -.707. So in my head i am picturing something other than the I matrix. I appreciate your help :) – neutrino – 2020-01-03T08:42:45.973

2@VP9 but what you're describing in your previous comment is just the matrix, not the matrix multiplied by itself. – DaftWullie – 2020-01-03T10:17:44.807