Symmetric matrices

$A = A^T$ is a symmetric matrix. $A = -A^T$ is an antisymmetric matrix. All symmetric and antisymmetric matrices are square.

Symmetric

The sufficient condition is that $A = A^T$.

If a matrix is symmetric, this means that

1. Can be decomposed into $U\Sigma U^T$ where $U$ is the the eigenbasis (spectral theorem)

2. Has orthogonal eigenbasis (spectral theorem)

3. Can be factored into $VV^T$. If $\Sigma$ is non-negative, the $V$ is real (derived from the factorization)

4. Inverse is also symmetric

Note that symmetry does not guarentee invertibility as the eigenbasis may have eigenvalues of $0$.

Side note: if a matrix is not symmetric, it can still be decomposed through SVD.

Sums of symmetric matrices

$A + A^T$ is symmetric and $A - A^T$ is antisymmetric. Furthermore, because $A = \frac{1}{2} (A + A^T) + \frac{1}{2}(A - A^T)$, we can express any matrix $A$ as the sum of a symmetric and an antisymmetric matrix!!

Transpose

Transpose is just flipping rows and columns. Theoretically it deals with dual spaces but practically it's just the flip

These properties are important

• ${A^T}^T = A$

• $(AB)^T = B^TA^T$

• $(A + B)^T = A^T + B^T$

Other properties

A matrix is symmetric if $A^T = A$ and slew-symmetric if $A^T = -A$

If you transpose a rectangular matrix, think of it as rotating around the major diagonal

$(AB)^T_{i,j} = B^TA^T$

• Proof

  $(AB)^T_{ij} = (AB)_{ji}$

  This is the product of the jth row of $A$ with the ith column of $B$, which means the jth colum of $A^T$ with the ith row of $B^T$, which is just $(B^TA^T)_{ij}$

$(Ax)^T = x^TA^T = x_1a_1^T +... + x_na_n^T$ where $a_k$ is a column in the original matrix $A$

Proof : this is really simple. It's just using the transpose rule with the definition of a matrix-vector product