How is this a basis for the vector space of symmetric 2x2 matrices?

Question

*EDIT, I posted the incorrect set of matrices. Unsure how to indicate this,but I've correct the question now. \begin{bmatrix}0&1\\1&1\end{bmatrix} \begin{bmatrix}1&1\\1&0\end{bmatrix} \begin{bmatrix}1&2\\2&-3\end{bmatrix}

This comes from a MC question that has 6 sets of three 2x2 matrices. 5 are basis for the vector space of symmetric 2x2 matrices and one is not. According to the answer, this one should be a basis. However, I'm having difficulty seeing how.

My understanding is that if it's a basis for symmetric 2x2 matrices, then there must exist c1, c2, c3 such that: c1[M1] + c2[M2] +c2[M3]= any symmetric 2x matrix, ie. \begin{bmatrix}1&0\\0&1\end{bmatrix} But, I can't find any such constants that would make this set of matrices equal to that matrix.

Answer 1

You are given three matrices,

$ A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$

$ B = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$

$ C = \begin{bmatrix} -2 & -2 \\ -2 & 1 \end{bmatrix}$

A very quick inspection gives $C = A - 3B$ so that $1A - 3B - 1C = 0$

Thus the three matrices are not linearly independent.

The vector space of symmetric 2 x 2 matrices has dimension 3, ie three linearly independent matrices are needed to form a basis.

The standard basis is defined by$ M = \begin{bmatrix} x & y \\ y & z \end{bmatrix} = x\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + y\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} + z\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$

Clearly the given $A,B,C$ cannot be equivalent, having only two independent matrices.

So, why not?

Look for: Copying error? Did you mis-write or mis-type something?

Sign error?

Misunderstood question?

Is one of the other answers correct so this one then could be wrong?

And then after all other routes are exhausted, yes texts sometimes contain typos.

Answer 2

Any linear combination of the three matrices above will produce a $2 \times 2$ matrix with first three identical entries. So these matrices can not span the space of all $2 \times 2$ symmetric matrices, so they do not form a basis.