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I want to check if three row vectors are linearly dependent or independent. I have the following row vectors: $u = \left[\begin{array}{r}1 & 3 & -1 & 2\end{array}\right], v=\left[\begin{array}{r}2 & -2 & 5 & 1\end{array}\right], w = \left[\begin{array}{r}1 & -1 & 2 & -1 \end{array}\right]$.

Now, if they are linearly independent it should mean that I would be able to find some non-zero scalars $\lambda_1, \lambda_2, \lambda_3$: such that I am able to satisfy the following equation:

$$\lambda_1u+\lambda_2v+\lambda_3w=0$$

However, I am not really able to understand how to proceed from here:

$$\lambda_1\left[\begin{array}{r}1 & 3 & -1 & 2\end{array}\right]+\lambda_2\left[{\begin{array}{r}2 & -2 & 5 & 1\end{array}}\right]+\lambda_3\left[{\begin{array}{r}1 & -1 & 2 & -1\end{array}}\right] = 0$$

Please note that my textbook has not yet talked about rank or determinants. Any hints on how to solve this without those? Should I solve a linear system? Should I build a matrix using these row vectors as rows?

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  • $\begingroup$ Hint: Letting $A = \left [ a_{ij} \right ]$, the vectors $a_{j}$’s are linearly independent if $Ax= 0 \Rightarrow x=0$. $\endgroup$ – OGC Oct 1 '18 at 8:48
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You are always able to find such scalars: just take $\lambda_1=\lambda_2=\lambda_3=0$. The question is: is there some other solution? If there is, the vectors are linearly dependent. Otherwise, they are linearly independent.

So, solve the system$$\left\{\begin{array}{l}\lambda_1+2\lambda_2+\lambda_3=0\\3\lambda_1-2\lambda_2-\lambda_3=0\\-\lambda_1+5\lambda_2+2\lambda_3=0\\2\lambda_1+\lambda_2-\lambda_3=0\end{array}\right.$$Is there some solution besides the null solution or not?

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  • $\begingroup$ Thanks. So do we need to put the row vectors as columns? I thought there was a difference between row vectors and column vectors. $\endgroup$ – Cesare Oct 1 '18 at 8:45
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    $\begingroup$ I did not think of them as rows. All I did was$$\lambda_1\begin{bmatrix}1&3&-1&2\end{bmatrix}+\lambda_2\begin{bmatrix}2&-2&5&1\end{bmatrix}+\lambda_3\begin{bmatrix}1&-1&2&-1\end{bmatrix}=\begin{bmatrix}\lambda_1+2\lambda_2+\lambda_3&3\lambda_1-2\lambda_2-\lambda_3&-\lambda_1+5\lambda_2+2\lambda_3&2\lambda_1+\lambda_2-\lambda_3\end{bmatrix}.$$ $\endgroup$ – José Carlos Santos Oct 1 '18 at 8:49
  • $\begingroup$ Wow, this is beautiful, thanks. $\endgroup$ – Cesare Oct 1 '18 at 8:50
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    $\begingroup$ @Cesare I'm glad I could help. $\endgroup$ – José Carlos Santos Oct 1 '18 at 8:51
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A much quicker and easier approach is to form a matrix from these vectors and see if it has full rank. In this case, if these vectors are linearly independent, the rank has to be $3$. It turns out that the rank is indeed $3$, so the given vectors are linearly independent.

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