Linear Algebra Lecture 3

Linear Algebra Lecture 3

1.Matrix multiplication in 4 ways

2.Inverse of A, AB, AT

3.Gauss-Jordan idea

Matrix multiplication

A[⋯⋯]B[⋯⋯]=C[⋯⋯]

Standard rule

Cij is from row i of A and column j of B.

Cij=Ai⋅Bj=∑nk=1aikbkj

When are we allowed to multiply matrices?

If they are square, they’ve got to be the same size.

If they are rectangular, not the same size.

Am×nBn×p=Cm×p

Multiply by column

Am×n[⋯⋯]Bn×p[⋮]=Cm×p[⋯⋯]

Multiply a matrix by column one tells the first column of the answer. And multiply matrix A times each column of matrix B can get the columns of the answer.

Columns of C are combinations of columns of A.

Multiply by row

Am×n[⋯]Bn×p⎡⎣⎢⋯⋯⋯⎤⎦⎥=Cm×p[⋯]

Rows of C are combinations of rows of B.

Column times row

When row times a column, it gave a number.

If I multiply a column of A times a row of B, I get a full-size matrix.

Am×1B1×p=Cm×p

⎡⎣⎢234⎤⎦⎥[16]=⎡⎣⎢234121824⎤⎦⎥

Times all columns and rows at a time.

AB=∑(ColsofA)×(RowsofB)

⎡⎣⎢234789⎤⎦⎥[1060]=⎡⎣⎢234⎤⎦⎥[16]+⎡⎣⎢789⎤⎦⎥[00]

Block Multiplication

We can also cut the matrix into blocks and do the multiplication by blocks.

Chop a matrix A into four square blocks, and B is a same size square also.

A[⋯⋯⋯⋯]B[⋯⋯⋯⋯]→A[A1A3A2A4]B[B1B3B2B4]=[A1B1+A2B3A3B1+A4B3A1B2+A2B4A3B2+A4B4]

Inverses

Inverses of square matrices

If you know a square matrix A, is it invertible or not? If it is invertible, then there is some other matrix, call it A inverse.

For square matrices, a left inverse is also a right inverse. If I can find a matrix on the left that gets the identity, then also that matrix on the right will produce that identity.

A−1A=I=AA−1

For rectangular matrices, we’ll see a left inverse isn’t a right inverse.

If A inverse exists?

These matrices are called invertible(可逆) or non-singular(非奇异).

If singular case, no inverse.

A=[1326]

A square matrix won’t have an inverse because I can find a vector X with Ax=0. The matrix can’t have an inverse if some combination of columns gives nothing.

Because, I could take Ax=0, I could multiply by A inverse. AA−1=I and 0×A−1=0, so x=0.

So our conclusion will be that non-invertible matrices, singular matrices, some combinations of their columns gives the zero column, there is no way A inverse can recover.

How to compute A inverse?

A[1327]A−1[abcd]=I[1001]

Gauss-Jordan idea solves two equations at once.

⎧⎩⎨⎪⎪⎪⎪⎪⎪[1327][ab]=[10][1327][cd]=[01]

Use augmented matrix, add the identity at the right-hand side of the matrix, then do elimination.

A[1237]I1001→(UpperTriangularForm)[1031]1−201→(UpwardElimination)I[1001]7−2−31

E[AI]=[IA−1]

Because EA=I tells us E=A−1