Prove or disprove that if $I+3A-A^2 = 0$ then $A$ is invertible

Question

I have the following statement:

Prove or disprove that if $I_n+3A-A^2 = 0$ then $A$ is invertible and the inverse is $A-3I_n$

My attempt was:

Suppose $A$ is a matrix of $m \times k$ order. Since $I_n = A^2 -3A$, from matrix equality $A$ must have $n\times n$ order, so $A$ is a square matrix of order $n$.

On the other hand, $I_n = A^2 -3A \iff I_n = A(A-3I_n)$, so only i need to prove that $I=(A-3I_n)A$. But I could not do it.

Any hint is appreciated.

What makes you think you haven't proved $A(A - 3I) = (A - 3I)A = I$? Remember, powers of $A$ commute! — Robert Lewis, Aug 30 '20 at 01:20
Note that, for $A^2$ to make sense, $A$ should be a square matrix — J. W. Tanner, Aug 30 '20 at 01:21
@Fawkes4494d3 Well, is good, but i think we can prove from $A(A-3I_n)$ to reach $(A-3I_n)A$ adding up some $(1-1)$ to factorize..? — ESCM, Aug 30 '20 at 01:25
@EduardoSebastian: not it is not but $A$ commutes with itself and with $I$. So $A(A - 3I) = A^2 - 3A = (A - 3I)A$ — Robert Lewis, Aug 30 '20 at 01:27
@RobertLewis so $A$ conmutes with any linear combination of $A$ and $I$ ? i mean with $cA + bI, c,b \in \mathbb{R}$ — ESCM, Aug 30 '20 at 01:28
Well it's a standard result that if $A, B$ are square matrices order $n$ and $AB=I$ then $BA=I$. The proof is non-trivial. — Paramanand Singh, Aug 30 '20 at 09:04

score 1 · Answer 1 · answered Aug 30 '20 at 03:16

You can use this fact also, take $B=A-3I$.

Now , as , $AB=I$ , we want to prove $BA=I$

Clearly, $A,B$ are of full rank here , otherwise $(AB)$ can't be of full rank.

So, $AB=I \implies (A^{-1}A)B=A^{-1} \implies B=A^{-1} $

Now, $B=BI \implies B=B(AB) \implies B=(BA)B \implies (I-BA)B = 0 $

Now, as $B$ is of full rank, $(I-BA)BB^{-1}=0_{n×n}B^{-1} \implies (I-BA)=0_{n×n} \implies BA=I \implies (A-3I)A=I $

Prove or disprove that if $I+3A-A^2 = 0$ then $A$ is invertible

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