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I know how to prove whether or not vectors are linearly independent, but can't apply the same thing to matrices it seems. Given three 2x2 matrices, for example:

$A = \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix}$ $B = \begin {bmatrix} 1&1 \\\\ -1&-1 \ \end{bmatrix}$ $C = \begin {bmatrix} -1&1 \\\\ 1&-1 \ \end{bmatrix}$

I want to test whether or not these are linearly dependent. So with vectors I would do something like:

$ c_1A + c_2B + c_3C = 0$

Where the $c_i$'s are some scalar constants, and prove that the only solution of that is when $c_1 = c_2 = c_3 = 0$

So how do I go about solving this:

$ c_1 \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix} + c_2 \begin {bmatrix} 1&1 \\\\ -1&-1 \ \end{bmatrix} + c_3 \begin {bmatrix} -1&1 \\\\ 1&-1 \ \end{bmatrix} = 0$

Or I am going about this completely the wrong way?

Any help would be hugely appreciated.

2 Answers 2

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You're going about it exactly the right way. EDIT: As David Mitra points out, you have to prove that $c_1=c_2=c_3=0$, not just that $c_1+c_2+c_3=0$.

In fact, you can just think of the matrices as being vectors of length 4: $\begin{pmatrix}a & b \\ c& d\end{pmatrix}\mapsto (a,b,c,d)$ and use your knowledge about the linear independence of vectors.

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    @MadScone: I've fixed up the formatting on your comment above. And that looks right to me :)2011-12-09
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It's perfectly fine (except that, if you want to prove independence, you need to show $c_1=c_2=c_3=0$, not that their sum is 0). Next, do the matrix arithmetic on the left hand side:

$ c_1 \begin {bmatrix} -1&1 \\\\ -1&1 \ \end{bmatrix} + c_2 \begin {bmatrix} 1&1 \\\\ -1&-1 \ \end{bmatrix} + c_3 \begin {bmatrix} -1&1 \\\\ 1&-1 \ \end{bmatrix} = \begin {bmatrix} -c_1+c_2-c_3&c_1+c_2+c_3\\\\ -c_1-c_2+c_3 & c_1-c_2-c_3 \ \end{bmatrix} ={\bf 0}. $

Since a matrix is the zero matrix if and only if each of its components is 0, you get the system of equations $\eqalign{ -c_1+c_2-c_3&=0\cr c_1+c_2+c_3&=0 \cr -c_1-c_2+c_3&=0 \cr c_1-c_2-c_3&=0 } $