# Are three matrices linearly independent and form a basis of $M_2(\mathbb R)$?

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 cs 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.

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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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So I can rewrite the matrices like this: $$A = (-1, 1, -1, 1)$$ and so on for the others. How does that simplify the problem? I will get: $$c_1(-1, 1, -1, 1) + c_2(1, 1, -1, -1) + c_3(-1, 1, 1, -1) = 0$$ Will that give me four equations? –  MadScone Dec 9 '11 at 16:52
Well, it doesn't simplify the problem, but you said you know how to check whether vectors were linearly independent, so I was explaining how to think of it in terms of what you're already familiar with. So yes, you get the four equations \begin{align*}-c_1+c_2-c_3&=0\\ c_1+c_2+c_3&=0\\ -c_1-c_2+c_3&=0\\ c_1-c_2-c_3&=0\end{align*} –  Zev Chonoles Dec 9 '11 at 17:56
Ok thanks, that's all I really needed to know I suppose. I solved the question like this: $$\operatorname{rref}\left[\begin{array}{ccc|c} -1&1&-1&0 \\ 1&1&1&0 \\ -1&-1&1&0 \\ 1&-1&-1&0 \end{array}\right] = \left[\begin{array}{ccc|c} 1&0&0&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \end{array}\right]$$ so the matrices are linearly independent. Assuming that's right, my question has been answered. –  MadScone Dec 9 '11 at 19:41
@MadScone: I've fixed up the formatting on your comment above. And that looks right to me :) –  Zev Chonoles Dec 9 '11 at 19:58

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 }

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