Getting to row echelon form: \left[\begin{array}{rrr} 1 & 1 & 5 \\ 0 & -3 & -9 \end{array}\right] \overrightarrow{E_{2}(-1 / 3)}\left[\begin{array}{lll} 1 & 1 & 5 \\ 0 & 1 & 3 \end{array}\right] \overrightarrow{E_{12}(-1)}\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 1 & 3 \end{array}\right] .
\begin{aligned} z & =2 \\ x+y+z & =2 \\ 2 x+2 y+4 z & =8 \end{aligned}
Augmented matrix:
\left[\begin{array}{llll} 0 & 0 & 1 & 2 \\ 1 & 1 & 1 & 2 \\ 2 & 2 & 4 & 8 \end{array}\right] \stackrel{E_{12}}{\longrightarrow}\left[\begin{array}{llll} (1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ 2 & 2 & 4 & 8 \end{array}\right] \xrightarrow[E_{31}(-2)]{\longrightarrow}\left[\begin{array}{rlll} 1 & 1 & 1 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right]
We keep on going… \begin{aligned} & {\left[\begin{array}{rrrr} (1) & 1 & 1 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 1 & 2 \end{array}\right] \xrightarrow[E_{2}(1 / 2)]{\longrightarrow}\left[\begin{array}{rrrr} 1 & 1 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 1 & 2 \end{array}\right]} \\ & \overrightarrow{E_{32}(-1)}\left[\begin{array}{rrrr} (1) & 1 & 1 & 2 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] \xrightarrow[E_{12}(-1)]{\longrightarrow}\left[\begin{array}{rrrr} (1) & 1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] . \end{aligned}
There’s still no information on y.
\begin{aligned} & x=-y \\ & z=2 \\ & y \text { is free. } \end{aligned}
To rotate a vector by \theta:
=\left[\begin{array}{rr} \cos \theta &-\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{l} r \cos \phi \\ r \sin \phi \end{array}\right]=\left[\begin{array}{rr} \cos \theta&-\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]
Scaling:
A = \left[\begin{array}{ll} z_1 & 0 \\ 0 & z_2 \end{array}\right]
Shearing: adding a constant shear factor times one coordinate to another coordinate of the point. A = \left[\begin{array}{ll} 1 & s_2 \\ s_1 & 1 \end{array}\right]
Adjacency matrix: A square matrix whose (i, j) th entry is the number of edges going from vertex i to vertex j
Incidence matrix: A matrix whose rows correspond to vertices and columns correspond to edges. The (i, j) th entry is 1 if vertex i is the tail of edge j, -1 if it is the head, and 0 otherwise.
A distribution vector is a vector whose entries are nonnegative and sum to 1.
A stochastic matrix is a square matrix whose columns are distribution vectors.
A Markov chain is a discrete dynamical system whose initial state \mathbf{x}^{(0)} is a distribution vector and whose transition matrix A is stochastic, i.e., each column of A is a distribution vector.
From HW2: put ths difference equation in matrix form:
y_{k+2}-y_{k+1}-y_{k}=0
Steps: 1. Make two equations. Solve for y_{k+2}: y_{k+2}=y_{k+1}+y_{k}. Also of course y_{k+1}=y_{k+1}. 2. Define the vector \mathbf{y}^{(k)}=\left[\begin{array}{l} y_{k} \\ y_{k+1} \end{array}\right] 3. Put the two equations in matrix form: . . .
\left[\begin{array}{l}y_{k+1} \\ y_{k+2}\end{array}\right]=\left[\begin{array}{ll}0 & 1 \\ 1 & 1\end{array}\right]\left[\begin{array}{c}y_{k} \\ y_{k+1}\end{array}\right]
Reaction-diffusion:
a_{x,t+1} = a_{x,t} + dt\left( \frac{D_{a}}{dx^{2}}(a_{x+1,t} + a_{x-1,t} - 2a_{x,t}) \right)
Heat in a rod: \begin{equation*} -y_{i-1}+2 y_{i}-y_{i+1}=\frac{h^{2}}{K} f\left(x_{i}\right) \end{equation*}
The determinant of a square n \times n matrix A=\left[a_{i j}\right], \operatorname{det} A, is defined recursively:
If n=1 then \operatorname{det} A=a_{11};
otherwise,
then
\begin{aligned} \operatorname{det} A & =\sum_{k=1}^{n} a_{k 1}(-1)^{k+1} M_{k 1}(A) \\ & =a_{11} M_{11}(A)-a_{21} M_{21}(A)+\cdots+(-1)^{n+1} a_{n 1} M_{n 1}(A) \end{aligned}
The column space of the m \times n matrix A is the subspace \mathcal{C}(A) of \mathbb{R}^{m} spanned by the columns of A.
The row space of the m \times n matrix A is the subspace \mathcal{R}(A) of \mathbb{R}^{n} spanned by the transposes of the rows of A
A basis for the column space of A is the set of pivot columns of A. (Find these by row reducing A and choosing the columns with leading 1s)
The null space of the m \times n matrix A is the subset \mathcal{N}(A) of \mathbb{R}^{n}
\mathcal{N}(A)=\left\{\mathbf{x} \in \mathbb{R}^{n} \mid A \mathbf{x}=\mathbf{0}\right\}
\mathcal{N}(A) is just the solution set to A \mathbf{x}=\mathbf{0}
For example, if A is invertible, A \mathbf{x}=\mathbf{0} has only the trivial solution \mathbf{x}=\mathbf{0}
so \mathcal{N}(A) is just \left\{\mathbf{0}\right\}.
A is invertible exactly if \mathcal{N}(A)=\{\mathbf{0}\}
Given an m \times n matrix A.
Compute the reduced row echelon form R of A.
Use R to find the general solution to the homogeneous system A \mathbf{x}=0.
\mathbf{x}=x_{i_{1}} \mathbf{w}_{1}+x_{i_{2}} \mathbf{w}_{2}+\cdots+x_{i_{n-r}} \mathbf{w}_{n-r}
where x_{i_{1}}, x_{i_{2}}, \ldots, x_{i_{n-r}} are the free variables.
To find the least squares solution to A \mathbf{x}=\mathbf{b}, we minimize the squared error \left\|A \mathbf{x}-\mathbf{b}\right\|^{2} by solving the Normal Equations for \mathbf{x}: \begin{aligned} \mathbf{A}^{T} \mathbf{A} \mathbf{x} &=\mathbf{A}^{T} \mathbf{b} \end{aligned}
If A is an m \times n full-column-rank matrix, then A=Q R, where the columns of the m \times n matrix Q are orthonormal vectors and the n \times n matrix R is upper triangular with nonzero diagonal entries.
A=\left[\mathbf{w}_{1}, \mathbf{w}_{2}, \mathbf{w}_{3}\right]=\left[\mathbf{q}_{1}, \mathbf{q}_{2}, \mathbf{q}_{3}\right]\left[\begin{array}{ccc} 1 & \frac{\mathbf{q}_{1} \cdot \mathbf{w}_{2}}{\mathbf{q}_{1} \cdot \mathbf{q}_{1}} & \frac{\mathbf{q}_{1} \cdot \mathbf{w}_{3}}{\mathbf{q}_{1} \cdot \mathbf{q}_{1}} \\ 0 & 1 & \frac{\mathbf{q}_{2} \cdot \mathbf{w}_{3}}{\mathbf{q}_{2} \cdot \mathbf{q}_{2}} \\ 0 & 0 & 1 \end{array}\right]
If A is a square n \times n matrix, the equation \operatorname{det}(\lambda I-A)=0 is called the characteristic equation of A
The eigenvalues of A are the roots of the characteristic equation.
For each scalar \lambda in (1), use the null space algorithm to find a basis of the eigenspace \mathcal{N}(\lambda I-A).
The pseudoinverse of A is A^{+}=V S^+ U^{T}
S^{+} is the matrix with the reciprocals of the non-zero singular values on the diagonal, and zeros elsewhere.
Can find least squares solutions:
\begin{aligned} A x & =b \\ x & \approx A^{+} b \end{aligned}