linear algebra 1

Transpose

$\boldsymbol{(AB)}^T = \boldsymbol{B}^T \cdot \boldsymbol{A}^T$

Element-wise op, Hadamard product

Dot product

• projection = Mechanical work is the dot product of force and displacement vectors.

• output is scalar.

Matrix multiplication is not commutative.

$\boldsymbol{A} \cdot \boldsymbol{B} \stackrel{?}{=} \boldsymbol{B} \cdot \boldsymbol{A}$

Identity Matrix

Matrix inverse

$\boldsymbol{A} \cdot \boldsymbol{A} ^{-1} = \boldsymbol{I}$

For $\boldsymbol{A} ^ {-1}$ to be exist, $\boldsymbol{Ax} = \boldsymbol{b}$ must have exactly one solution.

Necessary & sufficient conditions to find inverse matrix = A is square, m = n = rank(A)

Rank

• linearly independent columns

• vector space generated by its columns

Norms

• mapping vectors to non-negative values.

$L ^ 2$ norm, $||x||$ = Euclidean distance

squared $L ^ 2$norm = better computationally, but increase slowly near the origin.

$L ^ 1$norm = good for zero & nonzero is important.

$L ^ \infty$norm = absolute value of the element with the largest magnitude in the vector.