# Show that A is identical

• MHB
Gold Member
MHB
Hello! (Wave)

I want to prove that if a $m \times m$ matrix $A$ has rank $m$ and satisfies the condition $A^2=A$ then it will be identical.

$A^2=A \Rightarrow A^2-A=0 \Rightarrow A(A-I)=0$.

From this we get that either $A=0$ or $A=I$.

Since $A$ has rank $m$, it follows that it has $m$ non-zero rows, and so it cannot be $0$.

Thus $A=I$. Is everything right? Or could something be improved? (Thinking)

## Answers and Replies

Homework Helper
MHB
Hello! (Wave)

I want to prove that if a $m \times m$ matrix $A$ has rank $m$ and satisfies the condition $A^2=A$ then it will be identical.

$A^2=A \Rightarrow A^2-A=0 \Rightarrow A(A-I)=0$.

From this we get that either $A=0$ or $A=I$.

Hey evinda!

I don't think we can draw that conclusion.
Consider $A=\begin{pmatrix}1&0\\1&0\end{pmatrix}$, what is $A(A-I)$? (Worried)

Gold Member
MHB
Hey evinda!

I don't think we can draw that conclusion.
Consider $A=\begin{pmatrix}1&0\\1&0\end{pmatrix}$, what is $A(A-I)$? (Worried)

It is $0$ although neither $A$ nor $A-I$ is $0$.

How else can we get to the conclusion that $A=I$ ? (Thinking)

Do we use the linear independence? (Thinking)

Homework Helper
MHB
It is $0$ although neither $A$ nor $A-I$ is $0$.

How else can we get to the conclusion that $A=I$ ?

Do we use the linear independence?

How about multiplying by $A^{-1}$?
$A$ is invertible isn't it? (Thinking)

Gold Member
MHB
How about multiplying by $A^{-1}$?
$A$ is invertible isn't it? (Thinking)

So is it as follows? (Thinking)

Since the rank of the $m\times m$ matrix $A$ is $m$ we have that at the echelon form the matrix has no zero-rows. This implies that at the diagonal all elements are different from zero and this implies that the determinant is different from zero, since the determinant is equal to the product of the diagonal elements. Since the determinant is not equal to zero, the matrix is invertible. So $A^{-1}$ exists.

Thus $A^2-A=0 \Rightarrow A(A-I)=0 \Rightarrow A^{-1}A(A-I)=0 \Rightarrow A-I=0 \Rightarrow A=I$.

Is it complete now? Could we improve something? (Thinking)

Homework Helper
MHB
So is it as follows?

Since the rank of the $m\times m$ matrix $A$ is $m$ we have that at the echelon form the matrix has no zero-rows. This implies that at the diagonal all elements are different from zero and this implies that the determinant is different from zero, since the determinant is equal to the product of the diagonal elements. Since the determinant is not equal to zero, the matrix is invertible. So $A^{-1}$ exists.

Thus $A^2-A=0 \Rightarrow A(A-I)=0 \Rightarrow A^{-1}A(A-I)=0 \Rightarrow A-I=0 \Rightarrow A=I$.

Is it complete now? Could we improve something?

All good. (Nod)

It may suffice to simply state that an $m\times m$ matrix of rank $m$ is invertible though.
It's a property of matrices:
If A is a square matrix (i.e., m = n), then A is invertible if and only if A has rank n (that is, A has full rank).
(Nerd)

Gold Member
MHB
All good. (Nod)

It may suffice to simply state that an $m\times m$ matrix of rank $m$ is invertible though.
It's a property of matrices:
If A is a square matrix (i.e., m = n), then A is invertible if and only if A has rank n (that is, A has full rank).
(Nerd)

I see... Thanks a lot! (Party)