Rank of Matrix, and row of zero's Suppose $A$ is a $m \times n$ matrix with $Rank(A)=k <m$ How would you prove that the $A$ contains a row of zero's
it makes sense logically if m is the number of columns and the rank is the number of leading 1's in A then if the number of leading ones is less then m then there should be a row of zero's to make up for this discrepancy, but im having trouble showing this algebraically 
 A: Suppose $\widetilde{A}$ is row-reduced echelon form of matrix $A$.
First entry (from left) of each NON-ZERO row in $\widetilde{A}$, is $1$, known as pivotal entry. 
To prove your statement, it is sufficient to show that each NON-ZERO row can have only a single pivotal entry.
Suppose, we have proven this. Let us look at its implications.
We know that: rank $A=$total count of pivotal entries.
$\implies$ rank $A=$ count of NON-ZERO rows in $\widetilde{A}$ ($\because$ each NON-ZERO row can have a single pivotal entry)
So, rank $A<m\implies$ count of NON-ZERO rows is less than $m$.
$\implies$ count of ALL-ZERO rows $\ge 1$
($\because$ rows of $\widetilde{A}$ are of two types: either NON-ZERO or ALL-ZERO).
Hence your conclusion in the post follows!  
Now the required proof of the proposition: each NON-ZERO row can have only a single pivotal entry
Proof: Suppose NOT. Assume that there are two pivotal entries (in two distict columns $i$ and $j$) of a row, say $k_i$ and $k_j$.
But, by definition, if $i<j$, then entry $k_i$ must be zero, because $k_j$ has to be the first non-zero entry (so entries before it must be zero).
We know by trichotomy, that either $i<j$ or $j<i$.
Thus, either of the two entries $k_i$ and $k_j$ is NOT pivotal entry.
But this contradicts our assumption that they both are pivotal.
Hence, the conclusion follows.
