Linearly Independent Sets: Subsets of A Are Also Independent

[_Bd_]

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 don't know if I should have more math
. . .
the back of the book just says Proof.

 

[fzero]

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 don't 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?

 

[_Bd_]

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 I am 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 that's what I am thinking but I don't know if there is any more math and less text involved
?

 

[HallsofIvy]

My question then, is "what definition of 'independent' are you using?"

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

 

[JonF]

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

 

[_Bd_]

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 definition of linear independence that says: if {v₀, v₁ ,…, v_n } are linear independent iff a₀v₀ + a₁v₁ + ... + a_nv_n = 0 implies a₀, a₁ ,…, a_n = 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?

 

[JonF]

Make sure you note that in the definition I gave you it’s for any a_n'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 ,…, v_n } be a set of linear independent vectors. Suppose {v_p0, v_p1 ,…, v_pk } is a subset of vectors and it is linear dependent. If it is linearly dependent… thus we reach a contradiction. I'll let you fill in the rest.

 

[Mark44]

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 + ... + c_nv_n = 0 is c₁ = c₂ = ... = c_n = 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 = ... = c_n = 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.