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Show that A is identical

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,711
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)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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)