Linear dependance

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Homework Statement



Show that if A = {v_1, v_2, v_3. . .v_n} is a linearly independent set then any subset of A is also linearly independent

Homework Equations



proof.


The Attempt at a Solution



so if A is a set of vectors
[v_1 ]
[v_2 ]
[v_3 ]
[. . . ]
[v_n ]
for it to be independent then the determinant must NOT be zero and so far I know that det are just for square matrices.
therefore A must be a square matrix with a det not = to zero.
therefore it must be possible to reduce it to the identity matrix
therefore since the Identity matrix has a leading 1 on different columns/rows you can NOT write any of them as a linear combination of any other ones, therefore any subset would be independent also ??

Im not sure about this since its mostly text and I dont know if I should have more math
. . .
the back of the book just says Proof.
 
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  • #2
fzero
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you can NOT write any of them as a linear combination of any other ones, therefore any subset would be independent also ??

Im not sure about this since its mostly text and I dont know if I should have more math
. . .
the back of the book just says Proof.

Focus on this last bit. If a subset were linearly dependent, you could write an equation for one of the vectors as a linear combination of the others. Is that consistent with A being linearly independent? What assumption did we make that must be false?
 
  • #3
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Focus on this last bit. If a subset were linearly dependent, you could write an equation for one of the vectors as a linear combination of the others. Is that consistent with A being linearly independent? What assumption did we make that must be false?

thats kind of what i pointed. . . or Im not sure where you're getting at
say it reduces to

[1 0 0 0 ]
[0 1 0 0 ]
[0 0 1 0 ]
[0 0 0 1 ]

so for example vector 1 (1 0 0 0) can NOT be written as any combination of the other 3 vectors. . .because none of them have any coefficients in that column. . . so even if you were to only take a subset
[ 1 0 0 0 ]
[ 0 0 0 1 ]
[ 0 0 1 0 ]
then neither of those can be written as combinations of the others (therefore they are indepenent?)
or thats what Im thinking but I dont know if there is any more math and less text involved
?
 
  • #4
HallsofIvy
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My question then, is "what definition of 'independent' are you using?"

The definition that appears in most texts has nothing to do with matrices!
 
  • #5
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You should be able to show this pretty straight forward by using the defintion of linear independence that says: if {v0, v1 ,…, vn } are linear independent iff a0v0 + a1v1 + ... + anvn = 0 implies a0, a1 ,…, an = 0. Try a proof by contradiction.
 
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  • #6
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My question then, is "what definition of 'independent' are you using?"

The definition that appears in most texts has nothing to do with matrices!

The chapter is on vector spaces . . .



You should be able to show this pretty straight forward by using the defintion of linear independence that says: if {v0, v1 ,…, vn } are linear independent iff a0v0 + a1v1 + ... + anvn = 0 implies a0, a1 ,…, an = 0. Try a proof by contradiction.

ok so . . .then c1v1 + c2v2. . .cnvn = 0 then the set is independent if c1, c2. . .cn are zero.

if c1,c2. . .cn are NOT zero then the set contains at least 1 c_iv_i which is NOT equal to zero and therefore does not satisfy the independence theorem . . .
is that cool?
 
  • #7
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Make sure you note that in the definition I gave you it’s for any an's have to = 0, not just for one particular group

I'd formal it up a bit, but that's the general idea. Start with something like let {v0, v1 ,…, vn } be a set of linear independent vectors. Suppose {vp0, vp1 ,…, vpk } is a subset of vectors and it is linear dependant. If it is linearly dependant… thus we reach a contradiction. I'll let you fill in the rest.
 
  • #8
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ok so . . .then c1v1 + c2v2. . .cnvn = 0 then the set is independent if c1, c2. . .cn are zero.

Make sure you understand the subtlety of this definition, which is that the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0 is c1 = c2 = ... = cn = 0.

I can always set any linear combination of vectors to 0, whether they are linearly dependent or linearly independent, and come up with the solution c1 = c2 = ... = cn = 0. The key difference is that for linearly independent vectors, this is the only one solution. It will be one of many solutions for a set of linearly dependent vectors.

For example, the vectors v1 = <2, 1, 1> and v2 = <4, 2, 2> are obviously linearly dependent.

The equation c1v1 + c2v2 = 0 has a solution c1 = c2 = 0, even though these vectors are linearly dependent, but there are many other solutions, such as c1 = 2 and c2 = -1.
 

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