How to tell if a columns of matrix are linear dependent?

$\begingroup$

How can it be seen if the following matrix is linear dependent?

Let $A= \begin{bmatrix} 0 & -3 & 9& \\ 2&1& 7 \\ -1& 4 &-5 \\ 1&-4&-2 \end{bmatrix} $

First operation I perfomed was switch r1 to r4 and -2r1+r2

$A= \begin{bmatrix} 1 & -4 & -2& \\ 0&9& -3 \\ -1& 4 &-5 \\ 0&-3&9 \end{bmatrix} $

Next performed was $r1+r3$ to get

Let $A= \begin{bmatrix} 1& -4 & -2& \\ 0&9& -3\\ 0& 0 &-7 \\ 0&-3&9 \end{bmatrix} $

next done was$-1/3r2+r4$ and got

Let $A= \begin{bmatrix} 1& -4 & -2& \\ 0&9& -3\\ 0& 0 &-7 \\ 0&0&-8 \end{bmatrix} $

Thus it is shown because there no free variable it is linear independent. (I think)

My question is is it also linear indepedent because the vectors are not multiples of the first vector.

$\endgroup$ 3

2 Answers

$\begingroup$

To answer your question

My question is is it also linear indepedent because the vectors are not multiples of the first vector.

No. Just because the second and third columns are not multiples of the first, it does not mean they are linearly independent. Take for example the matrix

$$\begin{pmatrix} 1 & 1 & 1\\1&2&3\end{pmatrix}$$

None of the columns are multiples of the others, but the columns do form a linearly dependent set. You know this without any real work, since $3$ vectors in $\mathbb{R}^2$ cannot form a linearly independent set.

$\endgroup$ 3 $\begingroup$

You are right that after row reducing and finding that there are no free variables (because every column has a pivot), then all of the columns are linearly independent.

By knowing the set of three vectors is linearly independent, we know that the third column vector cannot be written as a linear combination of the first column vector and the second column vector. That is, there do not exist $c_{1}$, $c_{2} \in \mathbb{R}$ such that $v_{3} = c_{1}v_{1} + c_{2}v_{2}$. ($v_{i}$ is my notation for the $i$-th column vector.)

Similarly, we know $v_{2}$ cannot be written as a linear combination of $v_{1}$ and $v_{3}$.

We also know $v_{1}$ cannot be written as a linear combination of $v_{2}$ and $v_{3}$.

That is what is meant by the $3$ vectors being linearly independent. You can't write any one of them as a linear combination of the others.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

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