College Mathematics Linear Algebra and Analytic Geometry

It is to find his extremely linear independent part group.

Necessity.

If a1....an and b1....bn equivalent, then they can be represented linearly with each other.

Either x is taken from la, then x can be changed by a1....An linear representation. Since A1....

An can be made by B1....bn is expressed linearly, so x can also be represented by b1....bn is expressed linearly, so x also belongs to lb.

So LA is included in LB.

In the same way, if X belongs to LB, then X also belongs to LA. So LB is included in LA.

Since LA and LB are subsets of each other, LA=LB.

Sufficiency. If la=lb, then, take x from la, and x also belongs to lb.

Since A1....An belongs to LA, then A1....An also belongs to LB.

Since A1....An belongs to LB, and each AI can be defined by B1....bn is linear, so vector group a can be represented linearly by b.

In the same way, the demonstrable vector group b can also be represented linearly by a.

So A is equivalent to B.

1+x+。。x n-1 is solved as i i=1,2,3., n then n=1 because ( 1+x+. x^n-1）（x-1）=x^n -1

f1(x n)+xf2(x n)+xn-2fn-1(x^n)=(1+x+..x^n-1)g(x)

Bring in the above equation with i.

f1（1）+ω1f2(1)+.n-2fn-1（1）=0

。f1（1）+ωn-1f2(1)+.n-1^n-2fn-1（1）=0

Examining the above system of linear equations, the coefficient determinant is the Van der Mon determinant, so the coefficient determinant is not 0, i.e., there is only a zero solution.

So f1(1)=f2(1)=..=fn-1(1)=0

That is, the sum of the coefficients is 0

This is the definition ,,, definition,。。 Where's your question?

A is irrelevant, the rank is k, and the rank of b is also k, so it is also irrelevant.

The reverse is wrong, and the counter-example is given.

a:(1,0)

b;(1,0,0,0)

If a is not related above, then B is not related, which is correct.

Here's a counter-example:

a(0,0）

b(0,0,1,0)

It's the same question as the one just now, right?

Necessity. Set A1....An is a maximally irrelevant group, then any A can be represented linearly by it. Counter-evidence. Let for a certain a, there are two linear combinations such that .

a=k1 a1+..kn an=u1 a1+..un an, ki and ui are not exactly equal, then.

k1-u1)a1+..kn-un)an=0, where a certain (ki-ui) is not equal to 0, then ai can be expressed linearly by its banquet covector, and a1....An is extremely Li Yinda has nothing to do with the group contradiction.

Sufficiency. Since each A can be represented by A1....An linear representation, a1....An must contain the largest irrelevant group of the vector group omega. Just prove A1....An is linearly independent. Anti-Harassment Evidence Law.

If correlated, an AI can be linearly represented by the rest of the vectors. Establish.

a=k1 a1+..0 ai+..kn an

then a= ai+ a1+.0 ai+..kn an)

k1 a1+..ai+..kn an, which is the only contradiction with a linear representation.