Download Linear Algebra Lecture #3

Linear Algebra Lecture #3
[齋藤ｘｘ]は「齋藤正彦著『線型代数入門』東京大学出版会のｘｘページ参照」の意味。
[語句]
augmented matrix ここでは行列 A と単位行列を並べたもの[齋藤 53 例 2：この方法で逆行列を求
めることを講義では Gauss-Jordan idea と呼んでいる]
block multiplication 区分けを使った掛け算[齋藤 38]
determinant 行列式[齋藤 78]
dot product 内積(a・b で表す。“a dot b”と読む；= inner product)
entry(行列の)成分[齋藤 31]
Gauss ガウス
linearly dependent 線型従属[齋藤 99]
linearly independent 線型独立[齋藤 99]
Jordan ジョルダン（フランスの数学者；Jordan canonical form ジョルダンの標準形）
left inverse 左逆元(e を単位元とすると、a の左逆元とは、xa=e を満たす x のこと)
Lo and behold, the inverse will show up here.
rectangular matrix 長方形の行列(正方行列でない行列のこと)
right inverse 右逆元(e を単位元とすると、a の右逆元とは、ax=e を満たす x のこと；正方行列で
は右逆元と左逆元は一致する。[齋藤 88 (5)])
square matrix 正方行列[齋藤 40]
summation formula ここではシグマ記号を使って表すことを指す。
transpose 転置行列(この講義では A の転置行列を表すのに AT という記号(A transpose と読む)を使
っている) [齋藤 37]
Multiplication and Inverse Matrices
Overview of the lecture
matrix multiplication (four ways)
Inverse of A, AB and AT
Gauss-Jordan way of finding A-1
How to multiply matrices
First Way: regular way
[ m×n ] [ n×p ] = [ m×p ]
A
B=
C
n
the (i, j)-entry of C = cij = (row i of A)・(column j of B) = ∑ aik bkj
k=1
“the sum from k equals 1 to n of a sub ik times b sub kj”
Second Way: column way (matrix times column vector)
[ m×n ] [ n×p ] = [ m×p ]
A
B=
C
A×(column 1 of B) = column 1 of C
A×(column 2 of B) = column 2 of C
A×(column p of B) = column p of C
from lecture 2
???
3
? ? ? × 4 = 3×(column 1) + 4×(column 2) + 5×(column 3)
???
5
3×3 3×1 3×1
(matrix×column vector= linear combination of columns of the matrix)
column k of C = A×(column k of B) = linear combination of columns of A
Third Way: row way (row vector times matrix)
(row 1 of A)×B = row 1 of C
(row 2 of A)×B = row 2 of C
(row m of A)×B = row m of C
from lecture 2
???
[1 2 7]×? ? ? = 1×(row 1) + 2×(row 2) + 7×(row 3)
???
1×3 3×3 1×3
row×matrix = linear combination of rows of the matrix
row k of C = (row k of A)×B = linear combination of rows of B
Fourth Way: column times row
(column of A)×(row of B) =
m×1
1×p
2
2 12
3 ×[1 6] = 3 18
4
4 24
m×p
the row of the product is a multiple of the row vector
the column of the product is a multiple of the column vector
n
A B = ∑ (column k of A)×(row k of B)
k=1
block multiplication (multiplication by blocks)
Inverse Matrix
A: square matrix
Is it invertible (non-singular) or not? If the inverse matrix (A-1) exists,
A-1A = I (left inverse) AA-1 = I (right inverse)
For square matrices, the left inverses are also right inverses.
the case where there is no inverse (singular case)
A = 1 3 Why is A singular? (The determinant is zero.)
26
Suppose
1 3 ? ? =1 0
Column 1 of the product 1 is linear combination of 1 and 3
26 ? ? 01
0
2
6
This is impossible.
The square matrix A is not invertible iff there is a non-zero x with Ax = 0.
1 3 3= 0
2 6 -1 0
Why is A not invertible? Because A-1Ax = 0, hence x = 0. (contradiction!)
How to find the inverse?
13 a c =10
27 bd 01
A
A-1
I
13 a =1
27 b 0
13 c =0
27 d 1
Finding the inverse is like solving two systems. In general, A×(column j of A-1) = column j of I
Gauss-Jordan way of finding A-1 (solve n systems of equations at once)
Do eliminations and get A to I. Use elimination upwards.
1 3 1 0  1 3 1 0  1 0 7 -3
2701
0 1 -2 1
0 1 -2 1
A I
I
A-1
Why did we get A-1?
E [A I] = [I X]
E = the product of all the elimination matrices
E A = I. So E = A-1
X = E I = E = A-1