Vrftjkry

So I have the following question:

Analyze the following 'Claim' (which may or may not be true) and the corresponding 'Proof', by writing 'TRUE' or 'FALSE' (together with the reason) for each step. [Note: $I_n$ is the $n times n$ identity matrix.]

Claim: Let $A$ be any $n times n$ matrix satisfying $A^2=I_n$. Then either $A=I_n$ or $A=-I_n$.

'Proof'.

Step 1: $A$ satisfies $A^2-I_n = 0$ (True or False)

True.

My reasoning: Clearly, this is true. $A^2=I_n$ is not always true, but because it is true, I should have no problem moving the Identity matrix the the LHS.

Step 2: So $(A+I_n)(A-I_n)=0$ (True or false)

True.

My reasoning: Because $I_n$ is the identity matrix, there should be no issues with factoring just like normal algebra.

Step 3: $A+I_n=0$ or $A-I_n=0$

I'm not sure about this part. I'm very tempted to say this is fine but I am not sure how I can justify this, if I even can.

Therefore $A=-I_n$ or $A=I_n$. (End of 'Proof'.)

Is what I am doing right so far or am I messing up somewhere?

• 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 $$begin{bmatrix} 2 & 0 \ 0 & 0end{bmatrix}begin{bmatrix} 0 & 0 \ 0 & -2end{bmatrix}= begin{bmatrix} 0 & 0 \ 0 & 0end{bmatrix}$$




• In particular,

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

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