How to Tell if these vectors linearly independent?

$$v_1 = (3,1,0,1)$$ $$v_2 = (5,2,0,1)$$ $$v_3 = (1,0,-1,2)$$

i've used matrix reduction and found the reduction form as: \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{bmatrix}

some source said that if there is a row of zero in the reduced matrix, the vectors would be dependently linear and the other source say otherwise, so which is true?

  • 2
    $\begingroup$ Just by inspection...if $a\vec v_1+b\vec v_2+c\vec v_3=0$ then $c=0$ simply by looking at the third coordinate. And since the last two coordinates of $\vec v_1$ and $\vec v_2$ are the same we must have $a=-b$, but it is easy to see that this does not work. $\endgroup$ – lulu Oct 15 '17 at 17:29

Suppose you put them in a matrix, $A= \begin{bmatrix} v_1 & v_2 & v_3\end{bmatrix}$ and reduce them to row echelon form.

As we can see from the matrix that it has $3$ pivot columns, i.e., it is full rank, hence it is linearly independent.


Suppose you put them as $A= \begin{bmatrix} v_1^T \\ v_2^T \\v_3^T \end{bmatrix}$, and reduce them, to be linearly independent, you should not have any zero rows.

  • $\begingroup$ This answer is correct. I hope the OP understands why it says that there is no nontrivial linear combination of the vectors that gives the $0$ vector rather than relying on the rule for "zero rows". $\endgroup$ – Ethan Bolker Oct 15 '17 at 17:29
  • $\begingroup$ Thank you very much, for your answer! If we put them as the vector's transpose there is no zero rows, perhaps that is what my source implied. $\endgroup$ – Refo Ilmiya Oct 15 '17 at 18:34

You can tell that immediately, considering the third component of the given vectors.
There is no way to linearly combine two $0$'s to produce $-1$, or $0$ and $-1$ to produce $0$ !

  • $\begingroup$ sorry, I saw just now that this was also in lulu's comment $\endgroup$ – G Cab Oct 15 '17 at 17:45

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