# Show that A is identical

#### evinda

##### Well-known member
MHB Site Helper
Hello!!! 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? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello!!! 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 $A(A-I)$? #### evinda

##### Well-known member
MHB Site Helper
Hey evinda !!

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

##### MHB Seeker
Staff member
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? #### evinda

##### Well-known member
MHB Site Helper
How about multiplying by $A^{-1}$?
$A$ is invertible isn't it? 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? #### Klaas van Aarsen

##### MHB Seeker
Staff member
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. 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). #### evinda

##### Well-known member
MHB Site Helper
All good. 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). I see... Thanks a lot!!! 