A Basis for 3x3 Symmetric and Antisymmetric Matrices

The Vector Space of 3x3 Matrices (M)

The set of all real 3x3 matrices forms a vector space M of dimension 9.

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Subspace: Symmetric Matrices (S)

A matrix S is symmetric if S^T = S.

Basis for S (6 matrices):

$$S_1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad S_2 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad S_3 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix}$$ $$S_4 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad S_5 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \quad S_6 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$

• Three diagonal matrices: diag(1,0,0), diag(0,1,0), diag(0,0,1)
• Three off-diagonal matrices: 1 at (i,j) and (j,i) for i < j
• Dimension of S = 6
• Linearly independent, and their combinations span every symmetric 3x3 matrix

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Subspace: Antisymmetric (Skew-Symmetric) Matrices (AS)

A matrix A is antisymmetric (or skew-symmetric) if A^T = -A.

Properties

• Diagonal entries must be zero: a_{ii} = -a_{ii} implies a_{ii} = 0
• Off-diagonal entries satisfy a_{ij} = -a_{ji} for i ≠ j

General form of a 3x3 antisymmetric matrix:

$$A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}$$

Three free parameters (a, b, c) give dimension 3.

Basis for AS (3 matrices):

$$A_1 = \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \quad A_2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix} \quad A_3 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}$$

• Dimension of AS = 3

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S and AS Together Give a Basis for All 3x3 Matrices

Every 3x3 matrix M can be uniquely decomposed into a symmetric part and an antisymmetric part:

$$S = \frac{M + M^T}{2}, \qquad A = \frac{M - M^T}{2}$$

• S is symmetric (S = S^T)
• A is antisymmetric (A = -A^T)
• M = S + A

The combined basis ${S_1, \ldots, S_6, A_1, A_2, A_3}$ has 6 + 3 = 9 elements, matching the dimension of the full space of 3x3 matrices.

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Example: Upper-Triangular All-Ones Matrix

Let U be the 3x3 upper-triangular matrix with all ones on and above the diagonal:

$$U = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$$

Decomposition into S + AS

Symmetric part (S = (U + U^T)/2):

$$U^T = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}, \qquad U + U^T = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{pmatrix}$$ $$S = \frac{U + U^T}{2} = \begin{pmatrix} 1 & \frac12 & \frac12 \\[4pt] \frac12 & 1 & \frac12 \\[4pt] \frac12 & \frac12 & 1 \end{pmatrix}$$

Antisymmetric part (A = (U - U^T)/2):

$$U - U^T = \begin{pmatrix} 0 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & -1 & 0 \end{pmatrix}$$ $$A = \frac{U - U^T}{2} = \begin{pmatrix} 0 & \frac12 & \frac12 \\[4pt] -\frac12 & 0 & \frac12 \\[4pt] -\frac12 & -\frac12 & 0 \end{pmatrix}$$

Check: U = S + A

$$\begin{pmatrix} 1 & \frac12 & \frac12 \\ \frac12 & 1 & \frac12 \\ \frac12 & \frac12 & 1 \end{pmatrix} + \begin{pmatrix} 0 & \frac12 & \frac12 \\ -\frac12 & 0 & \frac12 \\ -\frac12 & -\frac12 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} = U$$

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Summary

Subspace	Dimension	Basis Elements
Symmetric (S)	6	$diag(1,0,0)$, $E_{12}+E_{21}$, $E_{13}+E_{31}$, $diag(0,1,0)$, $E_{23}+E_{32}$, $diag(0,0,1)$
Antisymmetric (AS)	3	$E_{12}-E_{21}$, $E_{13}-E_{31}$, $E_{23}-E_{32}$
All 3x3 matrices (M)	9	Union of S and AS bases