Check whether a system {$v_1,...,v_m$} of vectors in $\mathbb R^n$ (in $\mathbb R[x]$) is linearly independent.

These are my thoughts:

For {$v_1,...,v_m$} to be linearly independent, prove that:

$\lambda_1v_1 + \lambda_2v_2 + ... + \lambda_mv_m = \theta$ where $\lambda_1, ..., \lambda_m \in F$ and $\lambda_1 +... + \lambda_m = 0$ and $\theta$ represents the null vector ($\underline 0$).

So I'm assuming that each vector is in $\mathbb R^n$ but there are $m$ vectors in this system so I wrote them as a linear combination with scalars. Is this right?

So now how is the best way to do this proof. Can you do it in $\mathbb R$ and $\mathbb R^2$ then generalize or is that normally not allowed?

Will you need to use mathematical induction to generalize? I would just like some advice before I waste my time.

  • $\begingroup$ I think the limitation on the coefficients is that they are not the trivial solutions, namely, at least one of them is different than zero. $\endgroup$
    – Royi
    Commented Apr 4, 2015 at 12:37

1 Answer 1


Easy way to do so is to build a matrix which $ {v}_{1}, {v}_{2}, ... $ are its rows or columns.
Calculate the Determinant of this matrix.

If the determinant is different from zero, the vectors are independent.

Let $ M \in {\mathbb{R}}^{n \times m} $ such that $ M = \begin{bmatrix} \vdots & \vdots & \vdots & \vdots \\ {v}_{1} & {v}_{2} & \cdots & {v}_{m} \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix} $.
Then the following holds, $ \det \left( M \right ) = 0 \Leftrightarrow \left \{ {v}_{1}, {v}_{2}, \cdots {v}_{m} \right \}, \; Linear \, Dependent $.
The intuition behind it comes from the meaning of the determinant, where you can read about at the Wikipedia Article - The Determinant.

This works for $ m = n $.

For the case $ m > n $ they must be dependent.
For the case $ m < n $ you can use the SVD.
If the number of singular values is lower than $ m $ they are dependent.

By the way, this is equivalent of the question whether the equation $ M x = 0 $ has a solution or not.

  • $\begingroup$ Aren't determinants for square matrices only? That means this only works if $m=n$. $\endgroup$
    – Kitegi
    Commented Apr 4, 2015 at 12:28
  • $\begingroup$ Yes. So in your case, if $ m > n $ they must be dependent. for the other case, let me think for few seconds. $\endgroup$
    – Royi
    Commented Apr 4, 2015 at 12:30
  • $\begingroup$ OK, Added a fix. I think the SVD is your friend in this case :-). $\endgroup$
    – Royi
    Commented Apr 4, 2015 at 12:35
  • $\begingroup$ Please could you show me how to use the SVD? $\endgroup$ Commented Apr 4, 2015 at 12:49

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