Showing that the 0 matrix is the only one with rank = 0

[Mr Davis 97]

Homework Statement

Prove that if rank(A) = 0, then A = 0.

Homework Equations

The Attempt at a Solution

This seems like a very easy problem, but I just want to make sure I am not missing anything.

rank(A) = dim(Im(A)) = 0, so Im(A) = {0}. Thus, A is by definition the zero matrix.

My only concern is whether I can conclude from Im(A) = {0} that A is the zero matrix. That is, that there is no other matrix that satisfies this property.

 

[fresh_42]

Homework Statement

Prove that if rank(A) = 0, then A = 0.

Homework Equations

The Attempt at a Solution

This seems like a very easy problem, but I just want to make sure I am not missing anything.

rank(A) = dim(Im(A)) = 0, so Im(A) = {0}. Thus, A is by definition the zero matrix.

My only concern is whether I can conclude from Im(A) = 0 that A is the zero matrix. That is, that there is no other matrix that satisfies this property.

What does $\operatorname{Im}A = 0$ mean?

 

[Mr Davis 97]

What does $\operatorname{Im}A = 0$ mean?

I meant to write something like $\text{Im}(A)= \{\vec{0} \}$

 

[fresh_42]

Yes, of course, but what does it mean? What is the definition of $\operatorname{Im} A=0$? Or $= \{\vec{0}\}$, that isn't the point. Then compare it to what $A=0$ means.

 

[Mr Davis 97]

Yes, of course, but what does it mean? What is the definition of $\operatorname{Im} A=0$? Or $= \{\vec{0}\}$, that isn't the point. Then compare it to what $A=0$ means.

Well, that means that the matrix $A$ maps everything to $\vec{0}$. So of course the matrix matrix satisfies this. I just don't really know how I would show that the zero matrix is the only matrix that satisfies this.

 

[fresh_42]

Well, that means that the matrix $A$ maps everything to $\vec{0}$. So of course the matrix matrix satisfies this. I just don't really know how I would show that the zero matrix is the only matrix that satisfies this.

$\operatorname{Im}A =0 \,\Longleftrightarrow\,A(v)=0 \,\forall\, v\in V \,\Longleftrightarrow\,A=0$
It's simply the definitions, i.e. it works in both directions.

 

[vela]

Well, that means that the matrix $A$ maps everything to $\vec{0}$. So of course the matrix matrix satisfies this. I just don't really know how I would show that the zero matrix is the only matrix that satisfies this.

You could find the columns of A by multiplying by the appropriate vector, e.g., (1,0,0,...,0), (0,1,0,...,0).

 

[nuuskur]

A corollary of Rank-nullity says that the rank of a matrix is the same as the order of the highest order nonzero minor in that matrix. Therefore, if a matrix has rank zero, it means that its every entry must be the zero element.

Alternatively, you could invoke the Rank-Nullity theorem. You could view the matrix $A$ as a linear operator (there's another theorem that allows you to do this) $A : U\to V$.
The rank of $A$ is then defined to be the dimension of the subspace $A(U) =:\mbox{ran}A = \mbox{im}A$. R-N gives you
<br /> \mbox{dim}(U) = \mbox{dim}(\mbox{ker}A ) + \mbox{dim}(\mbox{ran}A)<br />
What is the only linear operator whose nullity is $\mbox{dim}(U)$? By nullity I mean the dimension of the kernel of $A$.