# 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

- Zero matrix is the matrix of all 0s.
- \(0^T = 0 \)
- \(0A = A0 = 0 \)

# Matrix Properties

- 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.

# Matrix Transpose

- \((A^T)^T = A \)
- \((A+B)^T = A^T + B^T\)
- \((AB)^T = B^TA^T\)
- \((A^T)^{-1} = (A^{-1})^T\)

# Matrix Inverse

- 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.

### example

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.

# Vector

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:

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

- 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 \(x^{T}y = x \cdot y\)

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