3
$\begingroup$

Let A = ($a_{ij}$) be an mxn matrix. If the set of row vectors of A is linearly independent, is the set column vectors too? What happens if the row vectors are linearly dependent. Does it affect the linear dependence of the column vectors?

I believe that since the row rank = column rank and m and n are not equal, the linear dependence and independence of the set of row vectors and column vectors should not depend on each other. I'm not sure how to build up a solid argument though.

$\endgroup$
1
$\begingroup$

You can build an intuition for that by using the number of pivots (in other words "rank")

Consider the following matrix:

\begin{equation*} A_{5,4} = \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,4} \\ \vdots & \vdots & \ddots & \vdots \\ a_{5,1} & a_{5,2} & \cdots & a_{5,4} \end{pmatrix} \end{equation*}

Assuming it can be transformed to reduced row echelon form, we get the following (this is just a concrete example for that sake of understanding):

\begin{equation*} A'_{5,4} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \end{equation*}

we have $4$ pivots, $4$ columns, and $5$ rows, the number of pivots is enough for the columns to be linearly independent but it's not enough for the rows to be so, one row won't have a pivot.

if in the systems of equations $A\vec x=0$ one row doesn't have a pivot then we don't have a unique solution such that $\vec x=0$ and by definition the rows aren't linearly dependent.

This applies to any non-square matrix (number of rows $\neq $ number of columns), I chose a $5 X 4$ one just as a concrete example.

To directly answer your questions:

What happens if the row vectors are linearly dependent. Does it affect the linear dependence of the column vectors?

No, this is clearly shown in the above example, rows are linearly depedent and columns aren't.

If the set of row vectors of A is linearly independent, is the set column vectors too?

No, this can be shown in a similar fashion where we have number of columns bigger than the number of rows.

It's worth noting that if you are dealing with a square matrix the linear dependence/independence of the rows and columns are related (one imply the other) as a pivot for a column will be a pivot for the row.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This means that linear dependence in case of a square matrix is related but not in the case where m is not equal to n, right? $\endgroup$ – Just another person May 28 at 10:30
  • $\begingroup$ Yes. Since each pivot for a column is a pivot for a row, example: a $3x3$ matrix, if we have 3 columns all with pivots we can conclude the columns are linearly independent, but we also have 3 rows, so these pivots are enough to make the rows linearly independent as well $\endgroup$ – Sergio May 28 at 10:35
0
$\begingroup$

The determinant of a matrix $A$ is the same as that of its transpose $A^T$. And since invertibility is equivalent to a non zero determinant, the linear independence of the column vectors is equivalent to that of the row vectors.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I was looking for an algebraic argument. This just seems like an abstract intuition. When we prove that the row rank is equal to the column rank, we use a whole mathematical process rather than just using invertibility. $\endgroup$ – Just another person May 28 at 9:46
  • $\begingroup$ It may appear abstract now, but at a certain point all these equivalences become second nature and part of the same intuition. You'll get used to it. I'm not saying this to suggest it isn't worth it to build algebraic arguments (I reckon at this stage it probably is for you anyway), it's just that after a while all of them become just background noise and not really that necessary. $\endgroup$ – Matheus Andrade May 28 at 9:52
  • $\begingroup$ I agree with you, these intuitions will become a second nature At this point though, they're hard to explain and build mathematical arguments with. $\endgroup$ – Just another person May 28 at 9:57
  • $\begingroup$ I understand where you're coming from. $\endgroup$ – Matheus Andrade May 28 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.