MIT18.06 Linear Algebra (11)

lec11

matrix space

SS : space of 3x3 symmetric matrix : 6 dimension

UU: space of 3x3 upper triangular matrix : 6 dimension

SUS\cup U is not a matrix space.

$S\cap U isamatrixspace,is a matrix space, \dim(S\cap U)=3$

$S+U :anyof: any of S +anyof**+** any of U$ = all 3x3’s.

dim(S+U)=9\dim(S+U)=9

dim(S+U)+dim(SU)=dim(S)+dim(U)=12\dim(S + U) + \dim(S \cap U) = \dim(S) + \dim(U) = 12

solution to:

d2ydx2+y=0\frac{d^2y}{dx^2}+y=0

is

y=c1sinx+c2cosxy=c_1\sin x+c_2\cos x

yy is a vector space, too.

All rank 1 matrix : A=UVTA=UVT where U,VU,V are column vectors.

All matrix can be a combination of rank one matrices.

All rank 4 matrices is not a matrix space. rank(A+B)rank(A)+rank(B)rank(A + B) \le rank(A) + rank(B)

$S =all= all V inin \mathbb{R}^4 withwith v_1+v_2 + v_3 + v_4 = 0$

$S isavectorspace,is a vector space, \dim(S) = 3$

$S isnullspacetois null space to A = \begin{bmatrix}1 & 1 & 1 & 1\end{bmatrix} i.e. i.e.  AV = 0 $ $ S A = \begin{bmatrix}1 & 1 & 1 & 1\end{bmatrix} 的空空间,即的空空间,即 AV = 0$ ,kms

dimN(A)=nr\dim N(A) = n − r

e.g. A=[1111],r=1,n=4,dimN(A)=41=3A = \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}, r = 1, n = 4, \dim N(A) = 4 - 1 = 3

A=[1(pivot)111(free varibles)]A = \begin{bmatrix}1 \text{(pivot)} & | & 1 & 1 & 1 \text{(free varibles)}\end{bmatrix}

Basis for SS i.e. Special solution for N(A)N(A) :

[1100],[1010],[1001]\begin{bmatrix} -1\\1\\0\\0 \end{bmatrix}, \begin{bmatrix} -1\\0\\1\\0 \end{bmatrix}, \begin{bmatrix} -1\\0\\0\\1 \end{bmatrix}

N(AT)=0,dimN(AT)=0N(AT)={0}, \dim N(AT) =0

small world graph

Graph

Graph = {nodes, edges}


MIT18.06 Linear Algebra (11)
https://yzzzf.xyz/2022/09/27/MIT16.08-lec11/
Author
Zifan Ying
Posted on
September 27, 2022
Licensed under