Problem on linear independence and matrices

  • Thread starter vince89
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  • #1
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Can I ask for some help?

Suppose that {v1,v2...vn} is a linearly independent set of vectors and A is a singular matrix.
Prove or disprove: The set {Av1, Av2, ...Avn} is linearly independent.
 

Answers and Replies

  • #2
How can you relate two different basis?
 
  • #3
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the vectors are the same. and they're not bases, they're just sets of vectors
 
  • #4
But the vetors of the first set are linearly independent, so they form a basis. If the second set has linearly independent members it should also be a base. Correct?
 
  • #5
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but doesnt the singular matrix change the property of linear independence?
 
  • #6
So the members of the second set can not be independent! :smile:
 
  • #7
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hmm...i dont quite get it. can you write a rough proof. :) thanks a lot!
 
  • #8
Ok!
The first set forms a basis since the vectors are linearly independent.

Suppose now that the members of the 2nd set are linearly independent, thus they also form a basis. Two basis are related by a non-singular matrix, but in our case they related by a singular one. Thus the members of the 2nd are not linearly independent.

What about that?
 
  • #9
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how does the singular matrix change the linear independence of the basis?
 
  • #10
Let me show it with equations.

Call the vectors of the 2nd set [itex]\bar{v}^\alpha[/itex], then

[tex] \bar{v}^\alpha=A^\alpha_\beta\,v^\beta[/tex]

In order for [itex]\bar{v}^\alpha[/itex] to be linearly independent, it must hold that

[tex]\lambda_\alpha \, \bar{v}^\alpha=0\Rightarrow \lambda_{\alpha}=0 \quad \forall \alpha[/tex]

Now we have

[tex]\lambda_\alpha \, \bar{v}^\alpha=0\Rightarrow \lambda_\alpha \, A^\alpha_\beta\,v^\beta=0 \Rightarrow \lambda_\alpha \, A^\alpha_\beta=0 [/tex]

The last equality hols because [itex]v^\alpha[/tex] are linearly independent.

This is a [itex]n\times n[/itex] homogeneous system for the unknows [itex]\lambda_\alpha [/itex]. In order for that to have only the trivial solution it must hold [itex]det(A)\neq 0[/itex]. But [itex]det(A)= 0[/itex] so there is a solution for [itex]\lambda_\alpha [/itex] besides the trivial one. Thus the vectors [itex]\bar{v}^\alpha[/itex] are dependent.
 
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  • #11
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if there is a trivial solution, what does it imply? :) thanks btw.
 
  • #12
HallsofIvy
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How about just trying for a counter example? What is the simplest singular linear transformation you know?
 
  • #13
Trivial solution means [tex]\lambda^\alpha=0[/tex], which would cause [tex]\bar{v}^\alpha[/tex] to be independent
 
  • #14
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Hello,

I hope I am not spoiling the fun, but I think things are getting confused.

Anything could happen here. Since we are not requiring that {v_1, ..., v_n} is a basis, here are two examples. Take any set of linearly independent vectors and let A = 0. Then of course their images are not linearly independent. On the other hand, take A to be some nonzero matrix, and let v_1 be any vector not in its kernel. Then {v_1} is linearly independent, and so is {Av_1}.
 
  • #15
Hello,

I hope I am not spoiling the fun, but I think things are getting confused.

Anything could happen here. Since we are not requiring that {v_1, ..., v_n} is a basis, here are two examples. Take any set of linearly independent vectors and let A = 0. Then of course their images are not linearly independent. On the other hand, take A to be some nonzero matrix, and let v_1 be any vector not in its kernel. Then {v_1} is linearly independent, and so is {Av_1}.

The OP was:

Suppose that {v1,v2...vn} is a linearly independent set of vectors and A is a singular matrix.

Since the [itex]v_i[/itex] are linearly independent, then they form a basis. Assuming of course, the dimension of the vector space is n. :smile:
 
  • #16
HallsofIvy
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All you really need is a 'counterexample'. If {v1, v2, ..., vn} is a set of independent vectors, and A is the linear transformation that takes every v into 0, what can you say about {Av1, Av2, ..., Avn}?
 
  • #17
mathwonk
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whats your definition of singular? if it means the columns are not independent, you are done. at least assuming you know the basic theory of dimension.
 
  • #18
HallsofIvy
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The OP was:



Since the [itex]v_i[/itex] are linearly independent, then they form a basis. Assuming of course, the dimension of the vector space is n. :smile:
Yes, and masnevets' point was that there is no reason to assume that! In any case, his point was an extension of what I said: Suppose {v1, v2, ...} is a set of independent vectors and A is the zero operator (Av= 0 for all v).
 

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