Proof verification for Identity matrices

• Rather than saying that moving the identiy to the LHS, it is due to we add $-I$ to both sides.

• We have $A^2-I=(A-I)(A+I)$, we just have to expand the right hand side to verify that.

• In matrices, $AB=0$ doesn't imply that $A=0$ or $B=0$. For example $$\begin{bmatrix} 2 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix} 0 & 0 \\ 0 & -2\end{bmatrix}= \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}$$

• In particular,

$$\left(\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}+\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\right)\left(\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}-\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\right)= \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}$$

that is we cant' conclude that $(A+I)(A-I)=0$ implies $A+I=0$ or $A-I=0$ as well.

Consider any diagonal matrix with diagonal elements $\pm 1$. Show that$A^{2}=I_n$. you get $2^{n}$ matrices whose square is $I_n$.

Furthermore if you want a concrete example of a matrix whose square is the identity but not itself a simple matrix consider for example this one:

$$\begin{bmatrix}\frac{1}{2} & \frac{3}{4} \\ 1 & -\frac{1}{2}\end{bmatrix}$$

These matrices are called involuntory