# Linear Algebra

A lot of content summarized from Mark Schmidt’s notes on Linear Algebra

## # Notation

Generally column major

- Scalar (1,1): $\alpha$
- Column Vector (m, 1): $\begin{bmatrix}a_1 \\ a_2 \end{bmatrix}$
- Row Vector (1, n): $\begin{bmatrix}a_1 & a_2\end{bmatrix}$
- Matrix (m, n): $\begin{bmatrix}a_{1,1} & a_{2,1} \\ a_{1,2} & a_{2,2}\end{bmatrix}$

## # Operations

### # Transpose

$(A^T)*{ij} = (A)*{ji}$

A matrix is **symmetric** if $A = A^T$

### # Vector Addition

Associative (brackets don’t matter) and commutative (order independent)

$$a + b = \begin{bmatrix}a_1 \\ a_2 \end{bmatrix} + \begin{bmatrix}b_1 \\ b_2 \end{bmatrix} = \begin{bmatrix}a_1 + b_1 \\ a_2 + b_2 \end{bmatrix}$$

### # Scalar Multiplication

Associative (brackets don’t matter) and commutative (order independent)

$$\alpha b = \alpha \begin{bmatrix}b_1 \\ b_2\end{bmatrix} = \begin{bmatrix}\alpha b_1 \\ \alpha b_2\end{bmatrix}$$

### # Inner Product

Between two vectors of the same length, multiply each element together to get a scalar result

$$a^Tb = \sum_{i = 1}^{n}a_ib_i = \gamma$$

A specific version of this is the **dot product** which can be expressed as the inner product between two vectors, $a \cdot b = a^Tb$

- Commutative: $a^Tb = b^Ta$
- Distributive across addition: $a^T(b+c) = a^Tb + a^Tc$

### # Outer Product

Between two vectors of the same length, create a matrix multiplying each combination of elements in each vector.

Given two vectors $u = \begin{bmatrix}u_1 \\ u_2 \\ \vdots \\ u_m \end{bmatrix}$ and $v = \begin{bmatrix}v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix}$,

$$u \otimes v = A = \begin{bmatrix}u_1v_1 & u_1v_2 & \dots & u_1v_n\\ u_2v_1 & u_2v_2 & \dots & u_2v_n \\ \vdots & \vdots & \ddots & \vdots \\ u_mv_1 & u_mv_2 & \dots & u_mv_n \end{bmatrix}$$

The resulting matrix $A$ is always rank-1.

### # Multiplication

In general, we can multiply matrices A and B when the number of columns in A matches the number of rows in B

If A is (m, n) and B is (n, p), then AB is (m, p)

- Associative: $A(BC) = (AB)C$
- Distributive across addition: $A(B + C) = AB + AC$
- Generally not commutative: $AB \neq BA$
- Transposing reverses order: $(AB)^T = B^TA^T$
- Matrix powers don’t change order: $(AB)^2 = ABAB$
- Matrix-vector multiplication always yields a vector: $x^TAy = x^T(Ay) = (Ay)^Tx = y^TA^Tx$

## # Properties

### # Vector Norm

A scalar measure of a vector’s length

- $\Vert x \Vert \geq 0$
- Euclidean Norm: $\Vert x \Vert_2 = \sqrt{\sum_{i=1}^n x_i^2}$ (this is also called the L-2 Norm)
- $\Vert x \Vert_2^2 = x^Tx$

### # Rank

- The dimension of the vector space generated (or spanned) by its columns.
- This corresponds to the number of linearly independent columns of A.
- This minimal set of vectors that span a space is called a basis

### # Orthogonal

If for some set of vectors $q$:

- $q_i^Tq_j = 0$, we call $q_i$ and $q_j$ orthogonal
- $q_i^Tq_j = 1$, we call $q_i$ and $q_j$ orthonormal

Inner product of square orthogonal matrices is the identity matrix: $Q^TQ = I = QQ^T$

### # Linear Dependence

A vector is linearly dependent on a set of vectors if it can be written as a linear combination of them

$$c = \alpha_1 b_1 + \alpha_2 b_2 + \dots + \alpha_n b_n$$

A set of vectors is linearly dependent if and only if the zero vector can be written as a non-trivial combination of any of the vectors.

A matrix with fully independent columns has **full column rank**. If this is the case, $Ax = 0$ implies that $x = 0$

## # Special Matrices

### # Identity Matrix

1’s on the diagonal and 0’s otherwise. $I_n$ denotes an (n,n) identity matrix.

Multiplication by the identity matrix yields the original matrix. Columns of the identity matrix are called elementary vectors.

### # Diagonal Matrix

$$D = \begin{bmatrix}d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix}$$

## # Spaces

### # Range (Column-space)

Subspace spanned by the columns of a matrix.

A system $Ax=b$ is solvable if and only if b is in $A$’s column-space

### # Subspace

A non-empty subset of vector space that is closed under addition and scalar multiplication

Possible spaces of $\mathbb{R}^3$

- 0 Vector
- Any line or plane through the origin
- All of $\mathbb{R}^3$

We say that the vectors generate or span the subspace when you can reach any point in the subspace through a linear combination of the vectors.

## # Matrices as transformation

Viewing $Ax = T(x)$

A linear transformation can’t move the origin. But, if there are linearly dependent columns, there are non-zero vectors that *can* be transformed to zero. The set of vectors that can be transformed to 0 is called the null-space.

Null space: $\mathcal N (A)$ is all $x$ such that $Ax = 0$

### # Fundamental Theorem of Linear Algebra

- $r$ is the dimension of the column-space which is the same as the dimension of the row-space
- The null-space is orthogonal to the row-space

## # Inverses

We can find the inverses if and only if A is square and doesn’t have null-space outside of the zero vector (otherwise we either lose information to the null-space or can’t get to all vectors)

If the inverse exists, it is a unique matrix such that $A^{-1}A = I = AA^{-1}$

Some identities

- $(A^{-1})^T = (A^T)^{-1}$
- $(\gamma A)^{-1} = \gamma^{-1} A^{-1}$
- Assuming both $A^{-1}$ and $B^{-1}$ exist, $(AB)^{-1} = B^{-1}A^{-1}$

Special inverses of diagonal matrices

$$D=\begin{bmatrix}d_1 & 0 & 0 \\ 0 & d_2 & 0 \\ 0 & 0 & d_3 \end{bmatrix}$$

$$D^{^-1}=\begin{bmatrix}1/d_1 & 0 & 0 \\ 0 & 1/d_2 & 0 \\ 0 & 0 & 1/d_3 \end{bmatrix}$$

## # Solving Linear Equations

Given A and b, we want to solve for x in $Ax = b$

Say, $\begin{bmatrix}2 & -1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}1 \\ 5\end{bmatrix}$.

We can interpret this multiple ways:

- By rows: $x$ is the intersection of the hyperplanes $2x-y=1$ and $x+y=5$
- By columns: $x$ is the linear combination that yields the RHS in $x\begin{bmatrix}2 \\ 5\end{bmatrix} + y \begin{bmatrix}-1 \\ 1\end{bmatrix} = \begin{bmatrix}1 \\ 5 \end{bmatrix}$
- Transformation

$Ax=b$ generally has a solution when $b$ is in the column-space of A. It has a single unique solution if the columns of A are linearly independent.

If $Ax=b$ has as solution we say it is consistent.

Basically, $x = A^{-1}b$

We can solve using Gaussian Elimination