# Matrix

tags
Linear Algebra, Math

We star with a set of equations with nine coefficients, three unknowns, and three right-hand sides.

\begin{align*} 2u + v + w &= 5 \\\
4u - 6v &=-2 \\\
-2u + 7v + 2w &=9 \end{align*}

We represent the right hand side with a column vector (or a 1 by 3 matrix). The unknowns of the system are represented as a column vector as well.

$x=\left[\begin{array}{l}{u} \ {v} \ {w}\end{array}\right]$

We define a square 3 by 3 matrix of the coefficients

$A=\left[\begin{array}{ccc} {2} & {1} & {1} \\\ {4} & {-6} & {0} \\\ {-2} & {7} & {2} \end{array}\right]$

While we present a square matrix here, matrices can be rectangular generally with $$m$$ rows and $$n$$ columns called an $$m$$ by $$n$$ matrix.

We can then define the equation as the matrix form $$Ax = b$$

\begin{align*} A x &= b \\ \left[\begin{array}{ccc}{2} & {1} & {1} \\ {4} & {-6} & {0} \\ {-2} & {7} & {2}\end{array}\right]\left[\begin{array}{c}{u} \\ {v} \\ {w}\end{array}\right]&=\left[\begin{array}{c}{5} \\ {-2} \\ {9}\end{array}\right] \end{align*}.

## Operations

This is straightforward. We add each coefficient of the lhs to the corresponding coefficient in the rhs.

\begin{align*} \left[\begin{array}{cc} {2} & {1} \\\
{3} & {0} \\\
{0} & {4} \end{array}\right] + \left[\begin{array}{cc} {1} & {2} \\\
{-3} & {1} \\\
{1} & {2} \end{array}\right] = \left[\begin{array}{ll} {3} & {3} \\\
{0} & {1} \\\
{1} & {6} \end{array}\right] \end{align*}

### Product

The first straightforward product is the scalar product. This is straightforward and follows from the definition of addition

$2\mathbf{A} = \mathbf{A} + \mathbf{A}; \{\mathbf{A} \in \mathcal{R}^{m \times n}\}$

We can also define a product between two matrices $$\mathbf{A}: n \times m$$ and $$\mathbf{B}: m \times k$$.

\begin{align*} \left[\begin{array}{ccc} {A_{11}} & {A_{12}} & {A_{13}} \\\
{A_{21}} & {A_{22}} & {A_{23}} \end{array}\right] \times \left[\begin{array}{cc} {B_{11}} & {B_{12}} \\\
{B_{21}} & {B_{22}} \\\
{B_{31}} & {B_{32}} \end{array}\right] &= \left[\begin{array}{ll} {\sum_{i=1}^3 A_{1i} B_{i1}} & {\sum_{i=1}^3 A_{1i} B_{i2}} \\\
{\sum_{i=1}^3 A_{2i} B_{i1}} & {\sum_{i=1}^3 A_{2i} B_{i2}} \end{array}\right] \end{align*}

The Hadamard product is often used in machine learning. It is quite simply the element-wise produce between two matrices: $$\mathcal{A}, \mathcal{B}: n \times m$$.

\begin{align*} \left[\begin{array}{cc} {A_{11}} & {A_{12}} \\\
{A_{21}} & {A_{22}} \\\
{A_{31}} & {A_{32}} \end{array}\right] \odot \left[\begin{array}{cc} {B_{11}} & {B_{12}} \\\
{B_{21}} & {B_{22}} \\\
{B_{31}} & {B_{32}} \end{array}\right] &= \left[\begin{array}{ll} {A_{11}B_{11}} & {A_{12}B_{12}} \\\
{A_{21}B_{21}} & {A_{22}B_{22}} \\\
{A_{31}B_{31}} & {A_{32}B_{32}} \end{array}\right] \\\
\left[\begin{array}{cc} {2} & {1} \\\
{3} & {0} \\\
{0} & {4} \end{array}\right] \odot \left[\begin{array}{cc} {1} & {2} \\\
{-3} & {1} \\\
{1} & {2} \end{array}\right] &= \left[\begin{array}{ll} {2} & {2} \\\
{-9} & {0} \\\
{0} & {8} \end{array}\right] \end{align*}

• Kronecker Product

### Transpose

Transpose mirrors a matrix along its diagonal starting from the top left corner. $[\mathbf{A}^\top]_{ji} = \mathbf{A}_{ij}$.

### Inverse

The inverse of a matrix follows from the inverse of a scalar (i.e. $$\alpha \alpha^{-1} = 1$$). For a matrix we instead want the product of a matrix and its inverse to be the identity matrix:

$\mathbf{A}\mathbf{A}^{-1} = \mathbf{I}$

The inverse for a square matrix exists iff its determinant is not zero $$\text{det}(\mathbf{A}) = |\mathbf{A}| \neq 0$$. If this condition is met, we can calculate the inverse using the adjugate

$\mathbf{A} = \frac{1}{|\mathbf{A}|} \text{adj}(\mathbf{A})$.

See below for the adjugate of a matrix.

• Moore-penrose Inverse

### Trace

The trace of a matrix is the sum of arguments on the main diagonal.

$\text{tr}(\mathbf{A}) = \sum_{i=1}^n a_{ii}$

• Properties:

• $$\text{tr}(\mathbf{A} + \mathbf{B}) = \text{tr}(\mathbf{A}) + \text{tr}(\mathbf{B})$$
• $$\text{tr}(\mathbf{A}) = \text{tr}(\mathbf{A}^\top)$$
• $$\text{tr}(\mathbf{A}^{\top} \mathbf{B}) = \text{tr}(\mathbf{A} \mathbf{B}^\top) = \text{tr}(\mathbf{B}^{\top} \mathbf{A}) = \text{tr}(\mathbf{B} \mathbf{A}^{\top}) =\sum_{i, j} \mathbf{A}_{i j} \mathbf{B}_{i j}$$