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I have the set of matrices $ \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix} $ $ \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} $ $ \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix} $ $ \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix} $ and I'm asked to check if said set is a basis of $ \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} $, that is, any 2x2 matrix. I know that in this case it is really simple to do it like $$\pmatrix{a&b\cr c&d\cr}=\alpha\pmatrix{1&0\cr0&0\cr}+\beta\pmatrix{0&1\cr0&0\cr}+\gamma\pmatrix{0&0\cr1&0\cr}+\delta\pmatrix{0&0\cr0&1\cr}$$

But not all cases are that simple, so I was wondering if I could do it another way: When I'm asked to check if a set of vectors is a basis of a vector space $R^n$ I just see if said vectors are linearly independent, given that if I have n linearly independent vectors $\{(v1,...,vn),...,(u1,...un)\}$ they'll be a basis for $R^n$. Can I use the same test and say that if

$$\pmatrix{0&0\cr 0&0\cr}=\alpha\pmatrix{1&0\cr0&0\cr}+\beta\pmatrix{0&1\cr0&0\cr}+\gamma\pmatrix{0&0\cr1&0\cr}+\delta\pmatrix{0&0\cr0&1\cr}$$

Has only one solution $\alpha=\beta=\gamma=\delta=0$ then they are all linearly independent, and since it's all 2x2 it'll be a basis for all 2x2 matrices? Does that make any sense?

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Yes, this is the right approach. The matrices span the entire space, and the last equation shoes that they are linearly independent. – copper.hat May 1 '13 at 18:55
Your solution works, but it also requires knowing that the $2\times 2$ matrix space is four dimensional. – vadim123 May 1 '13 at 18:55

Another way, if you know anything about determinants, is to represent each matrix as a column of a $4\times 4$ matrix and compute the determinant. For instance, you can represent a $2\times 2$ matrix $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ as the column of a $4\times 4$ matrix via $$ \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}. $$ Do this for all matrices in the set you want to check for linearly independence and then compute the determinant. If the determinant is not zero, then the set of matrices is linearly independent. Note that this is essentially the same thing as what you're doing, just in determinant form.

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$\mathbb{R}^{N\times N}$ as a linear space (with addition between elements and multiplication by scalars) is no different than $\mathbb{R}^{2N}$ endowed with these same operations. There is a bijective mapping between elements and operations in these two spaces, so any tricks you know about vectors is $\mathbb{R}^4$ are applicable to $\mathbb{R}^{2\times 2}$

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