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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?

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  • $\begingroup$ Yes, any set containing the zero vector is linearly dependent. $\endgroup$
    – Andrés E. Caicedo
    Commented Aug 30, 2014 at 2:12
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    $\begingroup$ Hint: if you have $m$ linearly independant vectors in ${\mathbb R}^n$, what can you say about $m$ and $n$? $\endgroup$
    – Taladris
    Commented Aug 30, 2014 at 2:13
  • $\begingroup$ m</=n Where does this lead me? $\endgroup$
    – yolo123
    Commented Aug 30, 2014 at 2:13
  • $\begingroup$ Precisely $m\leqslant n$. I misread the first sentence and didn't see the link before commenting and thought you were asking for a solution to the title rather than an explanation on a particular proof. Sorry. $\endgroup$
    – Taladris
    Commented Aug 30, 2014 at 2:15
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    $\begingroup$ My idea for a proof was: assume the matrix $A$ has $m$ rows and $n$ columns. Rows (that represents vectors in ${\mathbb R}^n$) are linearly independant so $m\leqslant n$. Similar reasoning for columns gives $n\leqslant m$, hence a square matrix. $\endgroup$
    – Taladris
    Commented Aug 30, 2014 at 2:17

1 Answer 1

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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.

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  • $\begingroup$ If I have zero rows in a matrix, does that mean that the rows vectors are linearly DEPENDENT? $\endgroup$
    – yolo123
    Commented Aug 30, 2014 at 2:17
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    $\begingroup$ If you have zero rows in a matrix, then you don't really have a matrix. $\endgroup$
    – JimmyK4542
    Commented Aug 30, 2014 at 2:18
  • $\begingroup$ Sorry, I meant a row WITH 0's in it. $\endgroup$
    – yolo123
    Commented Aug 30, 2014 at 2:19
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    $\begingroup$ A set of vectors $\{v_1,\ldots,v_m\}$ is said to be linearly dependent if you can find scalars $\alpha_1, \ldots, \alpha_m$ which are not all zero such that $\alpha_1v_1+\cdots+\alpha_nv_n = 0$. Suppose we have a row of all zeros, i.e. $v_1 = [0,0,\ldots,0]$. Then, pick $\alpha_1 = 1$ and $\alpha_i = 0$ for $i = 2,\ldots,m$. It is easy to see that $\alpha_1v_1+\cdots+\alpha_nv_n = 0$, so that set of vectors is linearly dependent. $\endgroup$
    – JimmyK4542
    Commented Aug 30, 2014 at 2:24
  • $\begingroup$ Does that mean every non-square matrix has row vectors which are linearly dependent? $\endgroup$
    – Krishna Deshmukh
    Commented Jan 20, 2019 at 10:16

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