If A is a matrix satisfying $A^3 + 4A - 2I = 0$, explain why A is invertible...

Question

If A is a matrix satisfying $A^3 + 4A - 2I = 0$, explain why A is invertible.

-I understand that I can easily find a matrix that fits this condition and prove that its determinant is not zero, but how would I prove this for all A matrices as opposed to a specific example. Any help is appreciated.

Answer 1

You can see that $$\det(A^3+4A)=\det(A)\times\det(A^2+4I)=\det(2I)\neq 0,$$ which implies $\det(A)\neq0$ and so $A$ is invertible.

Answer 2

One has: $$A(A^2+4A)=2I.$$ Therefore: $$A\left(\frac{1}{2}A^2+2A\right)=I.$$ Hence $A$ is right invertible and thus invertible.

N.B. You have to assume that $2$ is invertible in the ring from which $A$ takes its entries.

Answer 3

Observe that your relation is equivalent to $$A\left(\frac{1}{2}A^2+2A\right)=I.$$

Answer 4

Rewrite the relation as $$\frac12 A(A^2+4I)=I,$$ and you get $$A^{-1}=\frac12(A^2+4I).$$