Dimension of all 2x2 symmetric matrices?

[kostoglotov]

I think it's 3...

All 2x2 can be written as

a_1 A_1 + a_2 A_2 + a_3 A_3 + a_4 A_4

with

A_1 =<br /> \begin{bmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; 0<br /> \end{bmatrix}<br />, A_2 =<br /> \begin{bmatrix}<br /> 0 &amp; 1 \\<br /> 0 &amp; 0<br /> \end{bmatrix}<br />, A_3 =<br /> \begin{bmatrix}<br /> 0 &amp; 0 \\<br /> 1 &amp; 0<br /> \end{bmatrix}<br />, A_4 =<br /> \begin{bmatrix}<br /> 0 &amp; 0 \\<br /> 0 &amp; 1<br /> \end{bmatrix}<br />

And 2x2 Symm = a_1 A1 + a_2 (A_2 + A_3) + a_4 A_4, and if we combine A_2 + A_3 into a single basis element A^*, then A^* is still independent of A_1 and A_4...so actually all 2x2 symm matrices should be in a space of dimension 3...and the basis elements are A_1 \ A_2 \ and \ A^*?

 

[Geofleur]

Yes, but note that the title says "diagonal matrices", which aren't the same as symmetric matrices. The space of 2$\times$2 diagonal matrices has dimension 2.

 

[Jack Davies]

This is true. It is also interesting to consider the dimension of the antisymmetric matrices, A^T=-A.

In general for the space of n \times n matrices, you can write A=\frac{1}{2} (A+A^T)+\frac{1}{2}(A-A^T) for any matrix A (i.e 'decompose' into symmetric and antisymmetric parts). Furthermore, the sum of the dimensions of these two spaces always adds to n^2:

Denote the space of n \times n symmetric matrices as S_1 and the space of n \times n antisymmetric matrices as S_2.

Then dim(S_1 \cap S_2) = 0 \Rightarrow dim(S_1 +S_2)=dim(S_1)+dim(S_2)

Clearly we cannot gain dimensions by adding together two subsets of the larger set, but we have shown above that we can write any n \times n matrix as a sum of elements in each of these subspaces. So we conclude that dim(S_1) + dim(S_2)=n^2 as required.

 

[HallsofIvy]

I think it's 3...

All 2x2 can be written as

a_1 A_1 + a_2 A_2 + a_3 A_3 + a_4 A_4

with

A_1 =<br /> \begin{bmatrix}<br /> 1 &amp; 0 \\<br /> 0 &amp; 0<br /> \end{bmatrix}<br />, A_2 =<br /> \begin{bmatrix}<br /> 0 &amp; 1 \\<br /> 0 &amp; 0<br /> \end{bmatrix}<br />, A_3 =<br /> \begin{bmatrix}<br /> 0 &amp; 0 \\<br /> 1 &amp; 0<br /> \end{bmatrix}<br />, A_4 =<br /> \begin{bmatrix}<br /> 0 &amp; 0 \\<br /> 0 &amp; 1<br /> \end{bmatrix}<br />

And 2x2 Symm = a_1 A1 + a_2 (A_2 + A_3) + a_4 A_4, and if we combine A_2 + A_3 into a single basis element A^*, then A^* is still independent of A_1 and A_4...so actually all 2x2 symm matrices should be in a space of dimension 3...and the basis elements are A_1 \ A_2 \ and \ A^*?

Yes, a basis for the space of 2 by 2 symmetric matrices is
\begin{bmatrix}1 &amp; 0 \\ 0 &amp; 0 \end{bmatrix}
\begin{bmatrix}0 &amp; 1 \\ 1 &amp; 0 \end{bmatrix} and
\begin{bmatrix}0 &amp; 0 \\ 0 &amp; 1 \end{bmatrix}