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We have 4 5x7 matrices $A_1, A_2, A_3, A_4$ of rank 1, 2, 3, 4 that are in row echelon form. Is it true that the set of these matrices is necessarily linearly independent?

I would think that the echelon form condition ensures that their nonzero rows are ordered in such a way that by doing linear combinations of these matrices you'd at best get 4 matrices of rank 1 with 1 nonzero row and zeroes everywhere else. E.g. say you can't zero out the 4th row of matrix $A_4$ using the other three. Which would suggest they are LI. Is that correct reasoning? Thanks for any help.

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  • $\begingroup$ As vectors in the $35$-dimensional space of matrices? $\endgroup$
    – pancini
    Commented Jan 1, 2021 at 22:29
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    $\begingroup$ Yes, this reasoning is correct. $\endgroup$ Commented Jan 1, 2021 at 22:32

1 Answer 1

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Essentially, yes. The form of each matrix is something like \begin{align*} A_1 &= \left( \begin{matrix} 1 & ? & ? & ? & ? & ? & ? \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0& 0 & 0 & 0 \end{matrix} \right) \\ A_2 &= \left( \begin{matrix} 1 & ? & ? & ? & ? & ? & ? \\ 0 & 1 & ? & ? & ? & ? & ? \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0& 0 & 0 & 0 \end{matrix} \right) \\ A_3 &= \left( \begin{matrix} 1 & ? & ? & ? & ? & ? & ? \\ 0 & 1 & ? & ? & ? & ? & ? \\ 0 & 0 & 1 & ? & ? & ? & ? \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0& 0 & 0 & 0 \end{matrix} \right) \\ A_4 &= \left( \begin{matrix} 1 & ? & ? & ? & ? & ? & ? \\ 0 & 1 & ? & ? & ? & ? & ? \\ 0 & 0 & 1 & ? & ? & ? & ? \\ 0 & 0 & 0 & 1 & ? & ? & ? \\ 0 & 0 & 0 & 0& 0 & 0 & 0 \end{matrix} \right) \\ \end{align*}

where $?$ can be any scalar from your field. (I could use variables, but that would be a pain. Note that the $?$ may be distinct from each other and do not represent any particular quantity.) Take $\alpha_i$ to be four scalars. Then

\begin{align*} &\text{the 4rd row of} \left( \sum_{i=1}^4 \alpha_i A_i \right)\\ &= \left(\begin{matrix} 0 & 0 & 0 & \alpha_4 & \alpha_4 ? & \alpha_4 ? & \alpha_4 ? \\ \end{matrix} \right) \end{align*}

Obviously, this is a zero vector iff $\alpha_4 = 0$. Then consider

\begin{align*} &\text{the 3rd row of} \left( \sum_{i=1}^4 \alpha_i A_i \right)\\ &= \left(\begin{matrix} 0 & 0 & \alpha_3 + \alpha_4 & \alpha_3 ? + \alpha_4 ?& \alpha_3 ? + \alpha_4 ? & \alpha_3 ? + \alpha_4 ? & \alpha_3 ? + \alpha_4 ? \\ \end{matrix} \right) \end{align*}

Since $\alpha_4 = 0$ though,

\begin{align*} &\text{the 3rd row of} \left( \sum_{i=1}^4 \alpha_i A_i \right)\\ &= \left(\begin{matrix} 0 & 0 & \alpha_3 & \alpha_3 ? & \alpha_3 ? & \alpha_3 ? & \alpha_3 ? \\ \end{matrix} \right) \end{align*}

so, once again, this is only a zero vector iff $\alpha_3 = 0$. You can continue this to show that

$$\sum_{i=1}^4 \alpha_i A_i = \mathcal O \implies \alpha_i = 0 \; \forall i \in \{1,2,3,4\}$$

where $\mathcal O$ is the $5 \times 7$ all-zeroes matrix, the identity of the vector space $M_{5 \times 7}(F)$. Thus, the set of $A_i$ is linearly independent.

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    $\begingroup$ Even less pain would be to just use indices, like $\sum_{i=1}^4 \alpha_i A_i=0$ implies $\sum_{i=1}^4 \alpha_i A_i(j,k)$. As we have $A_i(4,4)=\delta_{4,i}$ we get $\alpha_4=0$. Evaluating at $(j,k)=(1,1), (2,2), (3,3)$ we get $\alpha_i=0$. Surely your answer is more instructive the way you wrote it. Just had to laugh when you were writing about the pain :D $\endgroup$ Commented Jan 1, 2021 at 22:46

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