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I have a set of three matrices and want to know whether or not they form a basis for $ℝ^{3\times3}$

Any help is appreciated

the matrices: \begin{bmatrix}1&0&0\\0&1&0\\0 &0&1\end{bmatrix}\begin{bmatrix}0&1&0\\0&0&1\\0 &0&0\end{bmatrix}\begin{bmatrix}0&0&1\\0&0&0\\0 &0&0\end{bmatrix}

I know I have to show they are linearly independent and span the vector space, but I really don't get how I should do that

Thanks in advance

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    What is the dimension of $\mathbb{R}^{3\times 3}$? – 2017-01-09
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    I think 3, right? – 2017-01-09
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    You have 9 entries in each matrix, so you should expect that the dimension will be 9, and in particular 3 matrices cannot span the entire space. – 2017-01-09
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    @Amaluena I guess you were thinking of each column or row as a vector, no? – 2017-01-09
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    i may be wrong but if see more precisely 1st vector of your basis spans a three dimensional vector space(say V) and 2nd vector spans 2 dimensional vector space possibly contained in V and 3rd spans 1 dimensional vector space also contained in V if one can prove 1st basis vector is l.c of other two – 2017-01-11

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Denote $C_1,C_2$ and $C_3$ the $3 \times 3$ matrices from your question. Is it true that $$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \in \text{span}(C_1,C_2,C_3)?$$

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    No, they don't, I see now they never span the entire space – 2017-01-09