本文为《Linear algebra and its applications》的读书笔记

Linearly independent sets

The definition of linear independence in a vector space is just the same as in R n \mathbb R^n Rn.

The main difference between linear dependence in R n \mathbb R^n Rn and in a general vector space is that when the vectors are not n n n-tuples, the homogeneous equation
usually cannot be written as a system of n n n linear equations. That is, the vectors cannot be made into the columns of a matrix A A A in order to study the equation A x = 0 A\boldsymbol x = \boldsymbol 0 Ax=0. We must rely instead on the definition of linear dependence and on Theorem 4.

EXAMPLE 1

Let p 1 ( t ) = 1 , p 2 ( t ) = t \boldsymbol p_1(t)= 1, \boldsymbol p_2(t)= t p1(t)=1,p2(t)=t , and p 3 ( t ) = 4 − t \boldsymbol p_3(t)= 4 - t p3(t)=4−t. Then { p 1 , p 2 , p 3 } \{\boldsymbol p_1,\boldsymbol p_2,\boldsymbol p_3\} {p1,p2,p3} is linearly dependent in P \mathbb P P because p 3 = 4 p 1 − p 2 \boldsymbol p_3 = 4\boldsymbol p_1 - \boldsymbol p_2 p3=4p1−p2.

基 / 极大线性无关组 / 极小生成集

Observe that when H ≠ V H \neq V H=V , condition ( i i ) (ii) (ii) includes the requirement that each of the vectors { b 1 , . . . , b p } \{\boldsymbol b_1,...,\boldsymbol b_p\} {b1,...,bp} must belong to H H H

EXAMPLE 3

Let A A A be an invertible n × n n \times n n×n matrix. Then the columns of A A A form a basis for R n \mathbb R^n Rn.

EXAMPLE 4

Let { e 1 , . . . , e n } \{\boldsymbol e_1,...,\boldsymbol e_n\} {e1,...,en} be the columns of the n × n n \times n n×n identity matrix, I n I_n In. The set { e 1 , . . . , e n } \{\boldsymbol e_1,...,\boldsymbol e_n\} {e1,...,en} is called the standard basis (标准基) for R n \mathbb R^n Rn (Figure 1).

EXAMPLE 6

Let S = { 1 , t , t 2 , . . . , t n } S = \{1, t, t^2,..., t^n\} S={1,t,t2,...,tn}. S S S is is called the standard basis for P n \mathbb P^n Pn.

The Spanning Set Theorem

生成集定理

As we will see, a basis is an “efficient” spanning set that contains no unnecessary vectors.
In fact, a basis can be constructed from a spanning set by discarding unneeded vectors.

Two Views of a Basis

When the Spanning Set Theorem is used, the deletion of vectors from a spanning set must stop when the set becomes linearly independent. Thus a basis is a spanning set that is as small as possible.
A basis is also a linearly independent set that is as large as possible. If S S S is a basis for V V V, and if S S S is enlarged by one vector—say, w \boldsymbol w w—from V V V , then the new set cannot be linearly independent, because S S S spans V V V , and w \boldsymbol w w is therefore a linear combination of the elements in S S S.

Bases for N u l A Nul\ A Nul A and C o l A Col\ A Col A

We already know how to find vectors that span the null space of a matrix A A A. The discussion in Section 4.2 pointed out that our method always produces a linearly independent set when N u l A Nul\ A Nul A contains nonzero vectors. So, in this case, that method produces a basis for N u l A Nul\ A Nul A.
The next two examples describe a simple algorithm for finding a basis for the column space.

EXAMPLE 8

Find a basis for C o l B Col\ B Col B, where

SOLUTION

Each nonpivot column of B B B is a linear combination of the pivot columns. By the Spanning Set Theorem, we may discard b 2 \boldsymbol b_2 b2 and b 4 \boldsymbol b_4 b4, and S = { b 1 , b 3 , b 5 } S=\{\boldsymbol b_1,\boldsymbol b_3,\boldsymbol b_5\} S={b1,b3,b5} will still span C o l B Col\ B Col B. S S S is a basis for C o l B Col\ B Col B.

What about a matrix A A A that is not in reduced echelon form? Recall that any linear dependence relationship among the columns of A A A can be expressed in the form A x = 0 A\boldsymbol x =\boldsymbol 0 Ax=0, where x \boldsymbol x x is a column of weights. When A A A is row reduced to a matrix B B B, the columns of B B B are often totally different from the columns of A A A. However, the equations A x = 0 A\boldsymbol x =\boldsymbol 0 Ax=0 and B x = 0 B\boldsymbol x =\boldsymbol 0 Bx=0 have exactly the same set of solutions. Then the vector equations
also have the same set of solutions.
That is, the columns of A A A have exactly the same linear dependence relationships as the columns of B B B.

EXAMPLE 9

It can be shown that the matrix
is row equivalent to the matrix B B B in Example 8. Find a basis for C o l A Col\ A Col A.

SOLUTION

In Example 8 we saw that
so we can expect that
Thus { a 1 , a 3 , a 5 } \{\boldsymbol a_1,\boldsymbol a_3,\boldsymbol a_5\} {a1,a3,a5} is a basis for C o l A Col\ A Col A. The columns we have used for this basis are the pivot columns of A A A.

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Chapter 4 (Vector Spaces): Linearly independent sets； Bases (基)

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