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Let $A$ be the set of $2 \times 2$ real, symmetric matrices.

Consider the set of matrices $\{\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \}$. I am curious for an explanation for why this is not the basis for $A$.

I know that the dimension of $A$ is 3, and the basis that is more consistent with this dimension is $\{\begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \}$. This basis satisfies the definition of a basis.

However, it is not clear to me why the first set does not satisfy the definition of a basis (i.e. it spans $A$ and is linearly independent). The set spans the space of all $2 \times 2$ matrices, so it certainly spans $A$. So, I am guessing there is some reason why the set is no linearly independent.

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    $\begingroup$ The set of matrices you have given is a basis for all $2 \times 2$ matrices. $\endgroup$ – copper.hat Nov 20 '17 at 20:37
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Note that not all the matrices in the first basis are elements of $A$, as they are not symmetric. You can't have basis elements that are not even part of the space they supposedly span. That set is certainly a basis. It's just not a basis for $A$, but rather for the whole space of $2\times 2$ matrices.

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