MHB Solving for Invertible Matrix: What Am I Doing Wrong?

[Yankel]

Hello all again,

A is a matrix with order nXn, such that:

\[A^{3}-2A^{2}+I=0\]

I need to choose the correct answer:

1) A is not invertible
2) It is not possible to say if A is invertible
3)
\[(A^{-1})^{2}=2I-A\]
4)
\[A^{-1}=2I-A\]

I can't find the solution here. I tried my own, and got:

\[A^{3}-2A^{2}=-I\]

\[2A^{2}-A^{3}=I\]

\[A(2A-A^{2})=I\]

and therefore:

\[A^{-1}=2A-A^{2}\]

what am I doing wrong here?

 

[I like Serena]

Hello all again,

A is a matrix with order nXn, such that:

\[A^{3}-2A^{2}+I=0\]

I need to choose the correct answer:

1) A is not invertible
2) It is not possible to say if A is invertible
3)
\[(A^{-1})^{2}=2I-A\]
4)
\[A^{-1}=2I-A\]

I can't find the solution here. I tried my own, and got:

\[A^{3}-2A^{2}=-I\]

\[2A^{2}-A^{3}=I\]

\[A(2A-A^{2})=I\]

and therefore:

\[A^{-1}=2A-A^{2}\]

what am I doing wrong here?

Hi again Yankel! :)

Let's start with invertibility.

If $A$ is not invertible, there must be some $v\ne 0$ such that $Av=0$.
What is $(A^{3}-2A^{2}+I)v$?

Assuming that $A$ is invertible, then you've found that:
$$A^{-1}=2A-A^{2} = A(2I-A)$$
Suppose we multiply on the left with $A^{-1}$?

 

[Yankel]

Oh, I see, you multiply on the left and get that A^-1 squared is exactly what I was looking for.

I did not understand the condition for A not being invertible.

 

[I like Serena]

Indeed.

One of the equivalent definitions of a matrix $A$ being invertible, is (see wiki):

The equation $Ax = 0$ has only the trivial solution $x = 0$.​

Let's suppose that $A$ is not invertible.
Then there must be some $v\ne 0$ such that $Av = 0$.
Therefore:
$$(A^{3}-2A^{2}+I)v = A^3v - 2A^2v + Iv = A^2(Av) - 2A(Av) + v= A^20 - 2A 0 + v = v \ne 0$$
This is a contradiction since it's given that $A^{3}-2A^{2}+I = 0$.
Therefore $A$ is invertible.