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

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

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

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

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
$$\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}$$
$$\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$$ and
$$\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}$$