Subspace theorem problem (matrix)

[NewtonianAlch]

Homework Statement

Suppose A is a fixed matrix in M. Apply the subspace theorem to show that

S = {x $\in$ ℝ : Ax = 0}

The Attempt at a Solution

Zero vector for x = <0,0,0>

A*<0,0,0> = 0

Therefore zero vector is in ℝ and S is non-empty.

Addition:

For u & v $\in$ S

u + v = 0 + 0 = 0

A*(u+v) = 0 => A(0) = 0

S is closed under addition

Multiplication:

λ $\in$ R

λ*A*u = λ*0 = 0

Therefore closed under multiplication and by subspace theorem, if a subspace of S.

Is this correct?

 

[Mark44]

Homework Statement

Suppose A is a fixed matrix in M. Apply the subspace theorem to show that

S = {x $\in$ ℝ : Ax = 0}

I can't tell what space x is in. It just shows up as a box in my browser. Is it R³?

Also, what is it you're supposed to show. All you have above is a definition for S. Are you supposed to show that S is a subspace of R³ (or whatever the space is)?

The Attempt at a Solution

Zero vector for x = <0,0,0>

A*<0,0,0> = 0

Therefore zero vector is in ℝ and S is non-empty.

Addition:

For u & v $\in$ S

u + v = 0 + 0 = 0

No, you're not given any information that allows you to conclude that u and v are zero vectors or that they add to 0.

A*(u+v) = 0 => A(0) = 0

No, for the reason given above. You need to use the properties of matrix multiplication.

S is closed under addition

Multiplication:

λ $\in$ R

λ*A*u = λ*0 = 0

No, because you can't say that u is necessarily equal to 0.

Therefore closed under multiplication and by subspace theorem, if a subspace of S.

Is this correct?

 

[CompuChip]

I assume that instead of ℝ you mean Rⁿ (Mark: it shows up in my browser as a blackboard $\mathbb{R}$).

It looks almost correct to me, except for this line:
u + v = 0 + 0 = 0
You are taking u and v arbitrarily in S then they are not necessarily equal to zero. You may want to use the linearity of matrix multiplication here.

No, because you can't say that u is necessarily equal to 0.

Actually he used that Au = 0 there, so that is correct.

 

[Mark44]

I assume that instead of ℝ you mean Rⁿ (Mark: it shows up in my browser as a blackboard $\mathbb{R}$).

The OP is using vectors in R³, though.

It looks almost correct to me, except for this line:
u + v = 0 + 0 = 0
You are taking u and v arbitrarily in S then they are not necessarily equal to zero. You may want to use the linearity of matrix multiplication here.


Actually he used that Au = 0 there, so that is correct.

Right. I mistakenly thought the OP replaced u with 0, but it was Au being replaced. I didn't catch that.

 

[HallsofIvy]

Homework Statement

Suppose A is a fixed matrix in M. Apply the subspace theorem to show that

S = {x $\in$ ℝ : Ax = 0}

You mean "show that S is a subspace of"- I guess R³ although you said ℝ.

The Attempt at a Solution

Zero vector for x = <0,0,0>

A*<0,0,0> = 0

Therefore zero vector is in ℝ and S is non-empty.

Yes, that is correct.

Addition:

For u & v $\in$ S

u + v = 0 + 0 = 0

No. What you have written here is that u= v= 0 and that is not necessarily true.
However, what you may have meant to write may have been

A*(u+v) = 0 => A(0) = 0

Not quite. You need A(u+ v)= Au+ Av= 0+ 0= 0.

S is closed under addition

Multiplication:

λ $\in$ R

λ*A*u = λ*0 = 0

Again, not quite. You want $A(\lambda u)= \lambda A(u)= \lambda(0)= 0$

Therefore closed under multiplication and by subspace theorem, if a subspace of S.

Is this correct?

 

[NewtonianAlch]

That should have been an R^n yes! I mistakenly left that out, sorry about the confusion!

 

[NewtonianAlch]

Thanks HallsofIvy, I guess I need to expand and be more explicit in each step of the way. I just assumed I didn't have to show that commutative, or was it associative law of multiplication. But I understand what you mean!

 

[NewtonianAlch]

I'm not too sure what's wrong with the addition aspect.

Given that I want to essentially prove that the matrix A multiplied by an addition of vectors would equal zero A*(u + v) = 0

So would it suffice to say that the above statement is true, and u + v = 0 but that neither u or v has to equal 0?

 

[HallsofIvy]

But it isn't true that u+ v= 0. What is true is that Au= 0 and Av= 0. That's completely different.

 

[Deveno]

for emphasis, u in S does NOT mean u is 0. it just means that Au = 0.

of course, the 0-vector 0 is ONE of the vectors in S, but an arbitrary vector in S need not be the 0-vector (it's hard to say for sure without knowing more about the matrix (or linear map) A).

what you need to show, is that:

if u,v is in S, so Au = 0 and Av = 0,

then A(u+v) = 0.

next, you need to show that if c is in R, and u is in S, that cu is also in S

(which entails showing that A(cu) = 0).