Confused about the dimension of a span of a set of vectors ls

Question

The question is:

What is the dimension of the following subspace of $\mathbb{R^5}$? $$span\left( \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ -1 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0\\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ -1 \\ 1\\ -1\\ 0 \end{pmatrix} \right)$$

I don't understand how to go about doing this. Do in reduce the matrix or what?

Thanks!

Answer 1

You construct a linearly independent set (a basis) from those given six vectors. Let's denote them by $e_1,\dots,e_6$, from left to right.

Since $e_1$ is not zero, choose it as the first vector in the basis. Then it is easy to see that $e_1$ and $e_2$ are linearly independent ($e_1$ has second component $0$, whilst $e_2$ has it $1$). So now we have two linearly independent vectors.

Clearly $e_3=0$ is not linearly independent from $e_1$ and $e_2$, so we don't choose it. Also since $e_1 = -e_4$, the set $\{e_1,e_2,e_4\}$ would be linearly dependent, same for $e_6 = -e_2$.

You can check that you cannot write $e_5$ as a linear combination of $e_1$ and $e_2$. Thus $\{e_1,e_2,e_5\}$ is a maximal linearly independent set of the six vectors, and thus forms a basis for your subspace.