True or False: Matrices with linearly independent row and column vectors are square.
Here is the answer of my textbook:
True; if the row vectors are linearly independent then $\text{nullity}(A)=0$ and $\text{rank}(A)=n=\text{the number of rows}$.
But since $\text{rank}(A)+\text{nullity}(A)=\text{the number of columns}$, $A$ must be square.
Why must a matrice with linearly independent vectors have $\text{nullity}(A)=0$?
That is where I lose track of the question.
Are zero rows considered to be linearly dependent?
Best Answer
Let's say the matrix has $m$ rows and $n$ columns. Either $m < n$, $m > n$, or $m = n$.
If $m < n$, then we have $n$ columns which lie in $\mathbb{R}^m$. Since $\mathbb{R}^m$ has dimension $m$, we can't have more than $m$ linearly independent vectors in $\mathbb{R}^m$. So the $n$ columns must be linearly dependent, a contradiction. Thus, we cannot have $m < n$.
If $m > n$, then we have $m$ rows which lie in $\mathbb{R}^n$. Since $\mathbb{R}^n$ has dimension $n$, we can't have more than $n$ linearly independent vectors in $\mathbb{R}^n$. So the $m$ rows must be linearly dependent, a contradiction. Thus, we cannot have $m > n$.
The only remaining possibility is $m = n$, which means that the matrix must be square.