Pauli matrices and the complex number matrix representation The three spin Pauli matrices are:
$
\sigma_1 = \begin{pmatrix}
 0 & 1 \\
 1 & 0 
\end{pmatrix}, 
\sigma_2 = \begin{pmatrix}
 0 & -i \\
 i & 0 
\end{pmatrix},
\sigma_3 = \begin{pmatrix}
 1 & 0 \\
 0 & -1 
\end{pmatrix}
$
According to problem 2.2.3 (Mathematical methods for physicists, Arfken Weber & Harris, Seventh Edition) Complex numbers, a + ib, with a and b real, may be represented by (or are isomorphic with) 2 × 2 matrices as follows: 
$
a + ib = \begin{pmatrix}
 a & b \\
 -b & a 
\end{pmatrix}
$
This implies that: 
$
i = \begin{pmatrix}
 0 & 1 \\
 -1 & 0 
\end{pmatrix}, 
-i = \begin{pmatrix}
 0 & -1 \\
 1 & 0 
\end{pmatrix}
$
And therefore according to the above proposition. 
$
\sigma_2 = \begin{pmatrix}
 0 & -i \\
 i & 0 
\end{pmatrix}  = 
i \begin{pmatrix}
 0 & -1 \\
 1 & 0 
\end{pmatrix} = i(-i) = 1 = 
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
$
By exchanging the rows in $\sigma_1$, we can also write it as the identity matrix. Therefore according to this we get $\sigma_1 = \sigma_2$ which to me is quite surprising. They have the property that $\sigma_1^2 = \sigma_2^2 = I$.
One can also prove that $\sigma_1 \sigma_2 = i \sigma_3$. Since $\sigma_1 = \sigma_2 = I$, $\sigma_1 \sigma_2 = I$. The RHS becomes:
$
i\sigma_3 = \begin{pmatrix}
 0 & 1 \\
 -1 & 0 
\end{pmatrix}
\begin{pmatrix}
 1 & 0 \\
 0 & -1 
\end{pmatrix} = 
\begin{pmatrix}
 0 & -1\\
 -1 & 0 
\end{pmatrix}  = 
\begin{pmatrix}
 1 & 0\\
 0 & 1 
\end{pmatrix}
$
The problem arises when I try to prove the identity $\sigma_i \sigma_j + \sigma_j \sigma_i = 2\delta_{ij}I_2 $ in Problem 2.2.11(c). This clearly does not hold since $\sigma_2 \sigma_3 + \sigma_3 \sigma_2$ $= I \sigma_3 + \sigma_3 I \neq 0$
Where am I making a mistake here ? 

The mapping $a +ib \Rightarrow \begin{pmatrix}
 a & b \\
 -b & a 
\end{pmatrix}$ where $ a,b \in \mathbb{R}$ does not map any complex number to a Pauli matrix. The Pauli matrices are not of the right form. So saying that $1$ maps to $\sigma_2$ (or even worse, that it is equal to it) is meaningless.
