MIT18.06 Linear Algebra (14)

lec14

orthogonal vectors（正交）

$x^\top y=0$

$\| x\| ^2+\| y\| ^2=\| x+y\| ^2$

Subspace

Subspace S is orthogonal to subspace T

means: every vector in S is orthogonal to every vector in T.

row space is orthogonal to null space.

null space and row space are orthogonal complement in $\mathbb{R}^n$

that means: null space contains all vectors $\bot$ row space.

solve $A^\top A \hat{x}=A^\top B$

from $Ax=B$ to $A^\top A \hat{x} = A^\top B$ , to find “best” solution $\hat{x}$ .

1. $N(A^\top A)=N(A)$
2. $rank(A^\top A)= rank(A)$
3. $A^\top A$ is invertible iff A has independent columns. (upd@0429: equivalent to 1. and 2. )

proof of $N(A^\top A)=N(A)$

Prove $\operatorname{rank}A^TA=\operatorname{rank}A$ for any $A\in M_{m \times n}$