The rank of the matrix $A$ Let $A$ be a $5\times 4$ matrix with real entries such that $Ax=0$ iff $x=0$ where $x$ is a $4\times 1$ vector and $0$ is a null vector. What is the rank of $A$? I can't understand how to do it, please someone help.
 A: The hypothesis suggests that the system  $Ax=0$ has a unique solution. And we know that such system has a unique solution if-f the rank of the coefficient matrix i.e. A is equal to the number of variables which in this case is 4.
You can read something about this in the wikipedia entry:
http://en.wikipedia.org/wiki/Rank_(linear_algebra)
A: Hints: The information you're given about $A$ relates to solutions of the equation $Ax = 0$; you should be able to determine the null space of $A$ from this information. Do you know of a theorem that relates the rank of a matrix and the dimension of its null space?
A: The dimension of the kernel of this transformation is $0$, since the kernel contains only the $0$ vector. Hence the rank of this matrix is $4-0=4$.
A: Since $A$ is a $5 \times 4$ matrix over $\mathbb R$, we can say that $A:\mathbb R^4 -> \mathbb R^5$. dim($\mathbb R^4$)=$4$ and dim($\mathbb R^5$)=5.
From the theorem,we know that  
dim $V$ ($\mathbb R^4$)= Rank($A$) + Nullity($A$).
since $0$ is the only element in $Ker A$, $A$ is one-one (Nullity =$0$),
hence dim(\mathbb R^4) = Rank(A).
$\implies Rank(A)=4$. 
