# Linear Algebra

## Lecture 1: The geometry of linear equations

### Linear equations as matrices

$2x - y = 0 \\ -x + 2y = 3 \\ \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix} \\ Ax = b$

### Row picture

• Each row in $$Ax = b$$ is a line.
• Intersection of the lines is the solution.

### Column picture

Instead of $\begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix} \\$ we think of $x \begin{bmatrix} 2 \\ -1 \end{bmatrix} + y \begin{bmatrix} -1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix}$

We want to find a linear weighting of $$[2, -1]^T$$ and $$[-1, 2]^T$$ that reaches $$[0, 3]^T$$.

## Lecture 2: Elimination with Matrices

### Identity matrix

$I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

### Permutation matrix

$P_{1,2} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$ #### Rows v. columns - $$P_{1,2}A$$ will swap the 1st and 2nd rows of $$A$$. - $$AP_{1,2}$$ will swap the 1st and 2nd columns of $$A$$.

### Elimination algorithm

Give an $$n \times n$$ matrix:

• Take the first pivot $$(1, 1)$$.
• Set $$(1, 2), ..., (1, n)$$ to zero by subtracting the first row.
• Repeat with the next pivots $$(2, 2), ..., (n, n)$$

The resulting matrix is the upper triangular matrix $$U$$.

#### Elimination matrices

We can define the subtractions in the elimination algorithm with elimination matrices.

For example, to subtract 4 row 1s from 2, we use: $E_{1,2} = \begin{bmatrix} 1 & 0 & 0 \\ -4 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

Thus, the elimination algorithm is repeating this step: $E_{2,3}(E_{1,3}(E_{1,2}A)) = U \\ EA = U$