Finding inverse from matrix equation

[Mr Davis 97]

Homework Statement

Suppose that a square matrix $A$ satisfies $(A - I)^2 = 0$. Find an explicit formula for $A^{-1}$ in terms of $A$

Homework Equations

The Attempt at a Solution

From manipulation we find that $A^2 - 2A + I = 0$ and then $A(2I - A) = I$. This shows that if we right-multiply $A$ by $2I - A$, we get the identity matrix. However, to show that $2I - A$ is an inverse, don't we also have to show that we can left-multiply $A$ by $2I - A$ such that $(2I - A)A = I$? I don't see how we can do that...

 

[DrClaude]

I don't think that showing that $(2I - A)A = A(2I - A)$ should be too difficult.

 

[fresh_42]

Homework Statement

Suppose that a square matrix $A$ satisfies $(A - I)^2 = 0$. Find an explicit formula for $A^{-1}$ in terms of $A$

Homework Equations

The Attempt at a Solution

From manipulation we find that $A^2 - 2A + I = 0$ and then $A(2I - A) = I$. This shows that if we right-multiply $A$ by $2I - A$, we get the identity matrix. However, to show that $2I - A$ is an inverse, don't we also have to show that we can left-multiply $A$ by $2I - A$ such that $(2I - A)A = I$?

Yes.

I don't see how we can do that...

Multiply it.

 

[Ray Vickson]

Homework Statement

Suppose that a square matrix $A$ satisfies $(A - I)^2 = 0$. Find an explicit formula for $A^{-1}$ in terms of $A$

Homework Equations

The Attempt at a Solution

From manipulation we find that $A^2 - 2A + I = 0$ and then $A(2I - A) = I$. This shows that if we right-multiply $A$ by $2I - A$, we get the identity matrix. However, to show that $2I - A$ is an inverse, don't we also have to show that we can left-multiply $A$ by $2I - A$ such that $(2I - A)A = I$? I don't see how we can do that...

There is a standard theorem which states that if a square matrix $A$ has a right (left) inverse $B$, then it also has a left (right) inverse, and that is equal to $B$ as well. You should expand your understanding by trying to prove that theorem.