Identity Matrix

Identity matrix is always square (DxD).
The D-dimensional identity matrix is the matrix that has 0s in every cell except the diagonal.
\(I^T = I \)
\(IA = AI = A \)

Zero Matrix

Matrix A is diagonal if all of its off-diagonal elements are zero. That is to say, Identity matrix and zero matrix are both diagonal.
Trace of matrix (tr(A)): \(tr(A) = \sum^D_{d=1} A_{d,d} \)
Determinant: \(det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad – bc \)
Rank: The column rank of A is the dimension of the column space of A, while the row rank of A is the dimension of the row space of A. The rank number is the number of linearly independent rows or columns.

Non-square matrices (m-by-n matrices for which m ≠ n) do not have an inverse. That is to say, all invertible matrix is square matrix.
Commutativity: \(A^{-1}A = AA^{-1} = I\)
Scalar multiplication: \((aA)^{-1} = a^{-1}A^{-1}\)
Transposition: \((A^T)^{-1} = (A^{-1})^T\)
Product: \((AB)^{-1}=B^{-1}A^{-1}\)
Some matrices are square but not invertible. They are singular matrix and its determinant is zero.

There is a way to obtain the inverse matrix thru determinant. Using this, we can find out there is matrix that is square but not invertible.

And it makes sense … look at the numbers: the second row is just double the first row, and does not add any new information.

Suppose we have \(u = \begin{bmatrix}1 \\ 2 \\ 3 \end{bmatrix}\) and \(v = \begin{bmatrix}4 \\ 5 \\ 6 \end{bmatrix}\), the vector inner product is:

\(u^Tv= \begin{bmatrix}1 & 2 & 3 \end{bmatrix}\begin{bmatrix}4 \\ 5 \\ 6 \end{bmatrix} = 1*4 + 2*5 + 3*6 = 32\)

You can multiply two matrices if, and only if, the number of columns in the first matrix equals the number of rows in the second matrix.

matrix multiplication is not commutative(ie. AB != BA)
Given the rules of matrix multiplication, we cannot multiply two vectors when they are both viewed as column matrices. If we try to multiply an n×1 matrix with another n×1 matrix, this product is not defined. To rectify this problem, we can take the transpose of the first vector, turning it into a 1×n row matrix. The interesting part is \(x^{T}y = x \cdot y\)