Linear Algebra Lecture 11

Linear Algebra Lecture 11

1. Matrix space

2. Bases of new vector spaces

3. Rank one matrices

Matrix space

All 3×33×3 matrices, every 3 by 3 matrix is one of vectors. They are vectors in my vector space, can add matrices, and multiply them by scalar numbers. And can take combination of matrices.

Subspace of the matrix space

All symmetric matices (对称矩阵)

All upper triangular matrices

All diagonal matrices (对角矩阵)

Basis for M = all 3×33×3 matrices

The most obvious basis would be the matrix

⎡⎣⎢100000000⎤⎦⎥,⎡⎣⎢000100000⎤⎦⎥,⎡⎣⎢000000100⎤⎦⎥,...,⎡⎣⎢000000001⎤⎦⎥[100000000],[010000000],[001000000],...,[000000001]

Our space is practically the same as nine dimensional space, just the nine number are written in a square instead of in a column.

Dimension and basis for subspace

Subspace	Basis	Dimension
Original	⎡⎣⎢100000000⎤⎦⎥,⎡⎣⎢000100000⎤⎦⎥,...,⎡⎣⎢000000001⎤⎦⎥[100000000],[010000000],...,[000000001]	9
Symmetric	⎡⎣⎢100000000⎤⎦⎥,⎡⎣⎢000010000⎤⎦⎥,⎡⎣⎢000000001⎤⎦⎥,⎡⎣⎢010100000⎤⎦⎥,⎡⎣⎢000001010⎤⎦⎥,⎡⎣⎢001000100⎤⎦⎥[100000000],[000010000],[000000001],[010100000],[000001010],[001000100]	6
Upper triangular	⎡⎣⎢100000000⎤⎦⎥,⎡⎣⎢000010000⎤⎦⎥,⎡⎣⎢000000001⎤⎦⎥,⎡⎣⎢000100000⎤⎦⎥,⎡⎣⎢000000010⎤⎦⎥,⎡⎣⎢000000100⎤⎦⎥[100000000],[000010000],[000000001],[010000000],[000001000],[001000000]	6
Diagonal	⎡⎣⎢100000000⎤⎦⎥,⎡⎣⎢000010000⎤⎦⎥,⎡⎣⎢000000001⎤⎦⎥[100000000],[000010000],[000000001]	3

Intersection : S∩US∩U = symmetric and upper triangular = diagonal

dim(S∩U)=3dim(S∩U)=3

Sum : S+US+U = any element of S + any element of U = all 3×33×3 matrices

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

dim(S)+dim(U)=dim(S∩U)+dim(S+U)dim(S)+dim(U)=dim(S∩U)+dim(S+U)

Solution spaces

A vector space doesn’t have vectors in it, it’s come from differential equations.

All solutions of d2ydx2+y=0d2ydx2+y=0

All combinations y=c1cosx+c2sinxy=c1cosx+c2sinx

dimension and basis

basis : cosxcosx and sinxsinx

dim(solutionspace)dim(solutionspace) = 2 (because we have a second order equation)

Rank one matrices

Example 1: A=[1248510]A=[1452810]

The dimension of column space and row space of AA is one.

dimC(A)=rank=dimC(AT)=1dimC(A)=rank=dimC(AT)=1

A=[1248510]=[12][145]A=[1452810]=[12][145]

Every rank one matrix has the form one column times one row. A=uvTA=uvT

Rank one matrices are like building blocks, can produce every matrix.

Subset of rank one matrices is not a subspace, because when adding two rank one matrices, the rank may be two.

Example 2:

In R4R4, v=⎡⎣⎢⎢⎢v1v2v3v4⎤⎦⎥⎥⎥v=[v1v2v3v4] , SS = all vv in R4R4 with v1+v2+v3+v4=0v1+v2+v3+v4=0.

SS = nullspace of A=[1111]A=[1111]

RankofA=r=1RankofA=r=1

dimN(A)=n−r=4−1=3dimN(A)=n−r=4−1=3

Basis for SS , set 0 and 1 for free variable :

⎡⎣⎢⎢⎢−1100⎤⎦⎥⎥⎥,⎡⎣⎢⎢⎢−1010⎤⎦⎥⎥⎥,⎡⎣⎢⎢⎢−1001⎤⎦⎥⎥⎥[−1100],[−1010],[−1001]

C(A)=R1,dimC(A)=1C(A)=R1,dimC(A)=1

N(AT)={0}N(AT)={0}

The null space has dimension three, the row space has dimension one,

The column space has dimension one, the left null space has dimension zero.

{3+1=4=n1+0=1=m{3+1=4=n1+0=1=m

The smallest subspace is empty set, that has only one point, that point is zero dimensional, and the basis is empty.