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

*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.

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.

• I actually did post the wrong set of matrices! Wasn't paying enough attention, my bad. I've corrected it now. Could you perhaps attempt the question again? – 3.14Pie Mar 20 '17 at 2:30
• Well, now, no problem. Test the three revised matrices for independence and they will pass. Show that you can express any symmetric matrix $\begin{bmatrix} x & y \\ y & z \end{bmatrix}$ as a linear combination $c_1D + c_2E + c_3F$ uniquely where $D, E, F$ are your given matrices, by solving for the constants in terms of $x, y, z$ if you need more detail. – victoria Mar 20 '17 at 2:38
• By the way, the erasing mania gives me a huge headache. Your question as it was made sense and I spent the time and trouble to answer it. Now you have erased the first half and put in the corrected matrices, but you have left the second half which does not connect to the new matrices, and now my carefully crafted answer plus the other answer and comments look insane. How about instead we try to learn from our mistakes, leave the original problem as it was, put a row of stars below and then add : Correction:... Wouldn't that be a lot clearer? – victoria Mar 20 '17 at 2:42
• Ah thanks for the tip! Sorry, I'll definitely do that in the future. – 3.14Pie Mar 20 '17 at 2:52
• You could do it now.... I hate my answers looking insane because the questioner has shifted ground. – victoria Mar 20 '17 at 2:58

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.