Prove that Ax=Ix has only the trivial solution

[drawar]

Homework Statement

Let A be a square matrix of order n such that $2{A^2} + A = 4I$. Prove that the only $x \in ℝ^n$ that satisfies $Ax = Ix$ is x=0.

Homework Equations

$Ax = 0$ has only the trivial solution iff A is invertible.

The Attempt at a Solution

The problem would be pretty trivial if the given equation was Ax=0, but how am I going to tackle it when the RHS is Ix? TIA!

 

[tiny-tim]

hi drawar!

hint: if Ax = x, what is A²x ?

 

[drawar]

hi drawar!

hint: if Ax = x, what is A²x ?

Thank you for your reply :)
It seems your hint works out quite nicely, please check my working:
Observe that $Ax = Ix = x \Rightarrow {A^2}x = Ax = x$
Hence $2{A^2} + A = 4I \Leftrightarrow 2{A^2}x + Ax = 4Ix \Leftrightarrow 2x + x = 4x \Leftrightarrow x = 0$
This completes the proof.

 

[tiny-tim]

fine!

 

[drawar]

fine!

Yeah, thanks!