Dimension of all 2x2 symmetric matrices?

  • #1
234
6

Main Question or Discussion Point

I think it's 3...

All 2x2 can be written as

[tex]a_1 A_1 + a_2 A_2 + a_3 A_3 + a_4 A_4[/tex]

with

[itex]A_1 =
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
[/itex], [itex]A_2 =
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
[/itex], [itex]A_3 =
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}
[/itex], [itex]A_4 =
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}
[/itex]

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

Answers and Replies

  • #2
Geofleur
Science Advisor
Gold Member
423
176
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.
 
  • Like
Likes Daeho Ro and kostoglotov
  • #3
This is true. It is also interesting to consider the dimension of the antisymmetric matrices, [itex]A^T=-A[/itex].

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

Denote the space of [itex]n \times n[/itex] symmetric matrices as [itex]S_1[/itex] and the space of [itex]n \times n [/itex] antisymmetric matrices as [itex]S_2[/itex].

Then [itex]dim(S_1 \cap S_2) = 0 \Rightarrow dim(S_1 +S_2)=dim(S_1)+dim(S_2)[/itex]

Clearly we cannot gain dimensions by adding together two subsets of the larger set, but we have shown above that we can write any [itex]n \times n [/itex] matrix as a sum of elements in each of these subspaces. So we conclude that [itex]dim(S_1) + dim(S_2)=n^2[/itex] as required.
 
  • Like
Likes Geofleur
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
I think it's 3...

All 2x2 can be written as

[tex]a_1 A_1 + a_2 A_2 + a_3 A_3 + a_4 A_4[/tex]

with

[itex]A_1 =
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
[/itex], [itex]A_2 =
\begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}
[/itex], [itex]A_3 =
\begin{bmatrix}
0 & 0 \\
1 & 0
\end{bmatrix}
[/itex], [itex]A_4 =
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}
[/itex]

And 2x2 Symm = [itex]a_1 A1 + a_2 (A_2 + A_3) + a_4 A_4[/itex], and if we combine [itex]A_2 + A_3[/itex] into a single basis element [itex]A^*[/itex], then [itex]A^*[/itex] is still independent of [itex]A_1[/itex] and [itex]A_4[/itex]...so actually all 2x2 symm matrices should be in a space of dimension 3...and the basis elements are [itex]A_1 \ A_2 \ and \ A^*[/itex]?
Yes, a basis for the space of 2 by 2 symmetric matrices is
[tex]\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}[/tex]
[tex]\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}[/tex] and
[tex]\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}[/tex]
 

Related Threads on Dimension of all 2x2 symmetric matrices?

Replies
2
Views
10K
Replies
4
Views
3K
Replies
5
Views
13K
Replies
1
Views
1K
  • Last Post
Replies
7
Views
5K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
2
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
6
Views
3K
Top