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

1. Apr 2, 2017

Mr Davis 97

1. The problem statement, all variables and given/known data
Prove that if rank(A) = 0, then A = 0.

2. Relevant equations

3. 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.

2. Apr 2, 2017

Staff: Mentor

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

3. Apr 2, 2017

Mr Davis 97

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

4. Apr 2, 2017

Staff: Mentor

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.

5. Apr 2, 2017

Mr Davis 97

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.

6. Apr 2, 2017

Staff: Mentor

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

7. Apr 2, 2017

vela

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

8. Apr 3, 2017

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
$$\mbox{dim}(U) = \mbox{dim}(\mbox{ker}A ) + \mbox{dim}(\mbox{ran}A)$$
What is the only linear operator whose nullity is $\mbox{dim}(U)$? By nullity I mean the dimension of the kernel of $A$.

Last edited: Apr 3, 2017