Linear Algebra

matrix types

orthogonal (rotating)

diagonal (scaling)

definition

symmetric (lucky coincidence)

linear map ( $T_D:\vec{v}\mapsto D\vec{v}$ )

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has $n$ eigenvectors which are the standard basis of $\mathbb{R}^n$

definition

square matrix whose columns of entries are unit vectors perpendicular to each other

each eigenvector $\vec{v}_i$ is scaled by a factor of the top-left diagonal entry $d_i$ (eigenvalue)

linear map ($T_O:\vec{v}\mapsto O\vec{v}$)

determinant is plus-minus one

the determinant is the product of the top-left diagonal entries

definition

$n\times n$ matrix whose entries are symmetric over the top-left diagonal

linear map ($T_S: \vec{v}\to S\vec{v}$)

decomposition

linear homomorphism

there exists a linear map

$T_D$ which scales each basis vector $\vec{b}_i$ in $\mathbb{R}^n$ with the eigenvalue $\lambda_i$

has $n$ eigenvectors perpendicular to each other

each eigenvector $\vec{v}_i$ is scaled by the eigenvalue $\lambda_i$

composition

every matrix

linear map ($T_A:\vec{v}\to A\vec{v}$)

$T_O$ which rotates each
basis vector $\vec{b}_i$ in $\mathbb{R}^n$ to the eigenvector $\vec{v}_i$

rotate the eigenvectors to the standard basis of $\mathbb{R}^n$, scale them, and rotate them back ($T_S=T_O\circ T_D\circ\mathrm{inv}(T_O)$).

decomposition

there exists a linear map

$T_{A_{m\times n}}$ which maps each vector from $\mathbb{R}^n$ to $\mathbb{R}^m$

$T_{D_{m\times m}}$ which scales each basis vector $b_i$ with the square root of the nonzero eigenvalue $\lambda_i$ of $T_S$

$T_{O_{n\times n}}$ which rotates each
basis vector $\vec{b}_i$ in $\mathbb{R}^n$ to the eigenvector $\vec{v}_i$ of $T_{S_{n\times n}}$

vector space

important axioms

definition

$T_{O_{n\times n}}$ which rotates each
basis vector $\vec{b}_i$ in $\mathbb{R}^n$ to the eigenvector $\vec{v}_i$ of $T_{S_{m\times m}}$

scalar-field compability

vector addition distributive

scalar addition distributive

A vector space is denoted ($V, \oplus, \odot$), which is a set $V$ together with vector addition $\oplus$ and scalar multiplication $\odot$ over a field $(F,+,\cdot)$ .

$(a\cdot b)\odot v=a\odot(b\odot v)$

$a\odot(u\oplus v)=a\odot u\oplus a\odot v$

$(a+b)\odot v=a\odot v\oplus b\odot v$

definition

A linear homomorphism is a vector function $f:V\to W$ such that given two vectors $u$ and $v$ in $V$ then

examples of vector spaces

axiom of extensionality

vector space of n-tuples of field numbers with componentwise addition and multiplication

basis

Smallest subset which spans the vector set

coordinates

$\phi:(\lambda_1,\ldots,\lambda_n)\in F^n\mapsto V\ni \lambda_1\odot b_1\oplus\cdots\oplus\lambda_n\odot b_n$

Axiom of choice implies every vector space has a basis

$B=\{(1,\ldots,0),\ldots,(0,\ldots,1)\}$

$[x]_B=\begin{bmatrix}x_1 \\ \vdots \\ x_n \end{bmatrix}$

definition

Only basis such that coordinate vector of $x$ has the same elements as $x$ itself.

$m$-by-$n$ matrix to vector function ($F^n\to F^m$)

General matrix

2D rotation matrix

Diagonal matrix

$f(a,b)=(ac, bd)$

$f(a,b)=(a\cos\theta\mp b\sin\theta, a\sin\theta\pm b\cos\theta)$

$\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6\end{bmatrix}$

Gauss-Jordan Elimination

Resources

https://math.stackexchange.com/questions/1936784/why-do-elementary-row-operations-change-the-column-space-without-changing-the-so

Operations

Transpose of matrix

Covector

Takes vector as input and outputs a number, like a measuring device

A covector is also linear

You going to the neighbours house is the linear homomorphism and you taking the measurement device home is its transpose. You measure the same weight home as at your neighbour.

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Scale

Pivot

Swap

Intuition

Every invertible matrix $F$ can be expressed as a matrix product of a sequence of elementary matrices: $F=E_1E_2E_1E_3E_1\ldots$

n-tuples of real numbers $\Leftrightarrow x=(x_1,\ldots,x_n)\in F^n$

It is impossible to specify a set without being able to tell what its elements are.

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Let $(V, \oplus,\odot)$ and $(W,\diamond,\circ)$ be vector spaces over the same field $(F,+,\cdot)$.

$f(k\odot v)=k\circ f(v)\in W$

$f(u\oplus v)=f(u)\diamond f(v)\in W$

zero vector space

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span

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QR decomposition

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$f:(a,b)\in R^2\mapsto R^3\ni (a+ 2b,3a+ 4b,5a + 6b)$

change of basis

$f(\phi_A(a, b))=f(a\vec{u}+b\vec{v})=af(\vec{u})+bf(\vec{v})$

$af(u_1,u_2)+bf(v_1,v_2)=a(u_1+ 2u_2,3u_1+ 4u_2,5u_1 + 6u_2)$
$+b(v_1+ 2v_2,3v_1+ 4v_2,5v_1 + 6v_2)$

Let $B=\{\vec{u},\vec{v}\}$ be the ordered set of basis vectors to $\mathbb{R}^2$, where $\vec{u}=(u_1,u_2)$ and $\vec{v}=(v_1,v_2)$

$af(\vec{u})+bf(\vec{v})=af(u_1,u_2)+bf(v_1,v_2)$

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$\begin{bmatrix} c & 0 \\ 0 & d\end{bmatrix}$

$\begin{bmatrix}\cos\theta & \mp\sin\theta \\ \sin\theta & \pm\cos\theta\end{bmatrix}$