Two questions involving linear independence

[kant]

How do you know if this:
| 0_-8_5|
|3_-7_4 |
|-1_5_-4|
| 1_-3_2|

a linearly independent set?


The answer at the back of the book say that it is independent, but obvious there are free variable in this matrix , thus imply a nontrival solution for AX=0, so it must be depend.

Let:

v( sub 1) =

|1 |
|-3|
|2 |

v( sub 2 ) =

|-3 |
|9 |
|-6 |

v( sub 3)

|5|
|-7|
|h |


for what value of h is v( sub 3) in Span{ V( sub 2), v( sub2)}?

for what of h is {v(s1), v(s2), v(s3) } linear independent?

 

[EnumaElish]

for what of h is {v(s1), v(s2), v(s3) } linear independent?

There is no such value of h because v1 and v2 are already dependent.

 

[kant]

There is no such value of h because v1 and v2 are already dependent.

The anwer at the back say "all h".

 

[EnumaElish]

Let's see. v1, v2, v3 are lin. indep. iff "a v1 + b v2 + c v3 = 0 implies a = b = c = 0. " See http://planetmath.org/encyclopedia/LinearIndependence.html .

a (1,-3,2) + b (-3,9,-6) + c (5,-7,h) = 0

a - 3b + 5c = 0
-3a + 9b - 7c = 0
2a -6b + hc = 0

a = 3b - 5c
-3(3b - 5c) + 9b - 7c = 0
-9b + 15c + 9b - 7c = 0
15c - 7c = 0 ===> c = 0 ===> a = 3b
2a - 6b = 0
6b - 6b = 0 for all b

So a v1 + b v2 + c v3 = 0 is true for all a, all b and c = 0, i.e. it doesn't imply a = b = c = 0 (because it works with any a and b as long as c alone is zero). Therefore the vectors are not linearly independent for any h.

 

[EnumaElish]

Your original matrix is 4x3, so the row vectors have to be dependent. But the column vectors may be independent.

 

[kant]

Let's see. v1, v2, v3 are lin. indep. iff "a v1 + b v2 + c v3 = 0 implies a = b = c = 0. " See http://planetmath.org/encyclopedia/LinearIndependence.html .

a (1,-3,2) + b (-3,9,-6) + c (5,-7,h) = 0

a - 3b + 5c = 0
-3a + 9b - 7c = 0
2a -6b + hc = 0

a = 3b - 5c
-3(3b - 5c) + 9b - 7c = 0
-9b + 15c + 9b - 7c = 0
15c - 7c = 0 ===> c = 0 ===> a = 3b
2a - 6b = 0
6b - 6b = 0 for all b

So a v1 + b v2 + c v3 = 0 is true for all a, all b and c = 0, i.e. it doesn't imply a = b = c = 0 (because it works with any a and b as long as c alone is zero). Therefore the vectors are not linearly independent for any h.

Hmm... From my point of view:

The question of weather the three vector are depend or not can be resolved by solving

|1_-3_5_0 |
|-3_9_-7_0
|2_-6_h_0 |


or

|1_-3_ 5_0|
|0_0_8_0_ |
|0_0_h-10_0|

The answer at the back stated that the three vector are dependent for "all h". I simple do not see that is the case, since if h is 10, then there is a free variable. In other word, this imply a nontrival solution for X, thus the three vector must be depend. It is not the case when h is not 10. Can you tell me what is wrong with my reasoning?

 

[EnumaElish]

Sorry, I am a slow thinker. What do you mean by "solving
|1_-3_5_0 |
|-3_9_-7_0
|2_-6_h_0 |"?

Also how do you get
|1_-3_ 5_0|
|0_0_8_0_ |
|0_0_h-10_0|
from
|1_-3_5_0 |
|-3_9_-7_0
|2_-6_h_0 |?

The answer at the back stated that the three vector are dependent for "all h". I simple do not see that is the case

We seem to arrive at the same conclusion with different methods. I am trying to understand the method you have used. Please clarify if you need me to comment on it.

 

[VietDao29]

Okay,
Lets prove that a = b = c = 0 is not the only solotion to $$a v_1 + b v_2 + c v_3 = 0$$
Matrix A:
$$A : = \left[ \begin{array}{ccc} 1 & -3 & 5 \\ -3 & 9 & -7 \\ 2 & -6 & h \end{array} \right]$$
You will always have:
$$\det{A} = 0 \mbox{, } \forall h$$
Therefore a = b = c = 0 is not the only solution to $$a v_1 + b v_2 + c v_3 = 0$$, there fore for all h, the 3 vectors v1, v2, v3 are dependent.
h = 10 is one case that the 3 vectors are dependent, you can try another h like h = 1, pi, ...
Viet Dao,

 

[HallsofIvy]

Your terminology is very confusing: you talk about a "linearly independent set" but what you show is not a set at all: it looks like either a determinant or a matrix. What set of vectors are you talking about?

You then refer to the vector (1, -3, 2) which I presume you take from the bottom row. But then you talk about (-3, 9, -6) which I don't see any where in the "matrix" and which is clearly not independent of (1, -3, 2). (-3, 9, -6)= (-3)(1, -3, 2).

One common way of determining whether a set of vectors is independent is to write a matrix having those vectors as columns and then row reduce. Is that what is intended? Certainly the vectors (0, 3, -1, 1), (-8, -7, 5, -3), and (5, 4,-4, 2) are independent.

 

[VietDao29]

I am trying to prove that:
$$\left\{ \begin{array}{ccc} 1a - 3b + 5c & = & 0 \\ -3a + 9b - 7c & = & 0 \\ 2a - 6b + hc & = & 0 \end{array} \right$$(1)
can have other solutions apart from a = b = c = 0 (this solution is obvious).
(1) have only one set of solution iff det(A) <> 0.
But because det(A) = 0, for all h. Then (1) can have other solutions, and therefore the 3 vectors are dependent. Am I missing something?
Viet Dao,

 

[HallsofIvy]

I am trying to prove that:
$$\left\{ \begin{array}{ccc} 1a - 3b + 5c & = & 0 \\ -3a + 9b - 7c & = & 0 \\ 2a - 6b + hc & = & 0 \end{array} \right$$(1)
can have other solutions apart from a = b = c = 0 (this solution is obvious).
(1) have only one set of solution iff det(A) <> 0.
But because det(A) = 0, for all h. Then (1) can have other solutions, and therefore the 3 vectors are dependent. Am I missing something?
Viet Dao,

Am I missing something? What does this have to do with the original question?

 

[VietDao29]

Err, maybe it's true that my terminology's very limited as I am not a native English speaker. And I self-study some maths (I'm going to grade 11th this year). So it may be true that I am wrong...
Anyway, I learn that a group of vectors {v1, v2,..., vn} is linear independent is the only solution to:
a1v1 + a2v2 + ... + anvn = 0 is ai = 0 for i = 1 .. n
And I try to prove that there exists a, b, c (a ^ 2 + b ^ 2 + c ^ 2 <> 0) such that:
$$a \left[ \begin{array}{c} 1 \\ -3 \\ 2 \end{array} \right] + b \left[ \begin{array}{c} -3 \\ 9 \\ -6 \end{array} \right] + c \left[ \begin{array}{c} 5 \\ -7 \\ h \end{array} \right] = \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right]$$
-----------------
So in order to find h that makes the 3 vectors independent, I find h such that det(A) <> 0.
Because in the book states that:
$$\left\{ \begin{array}{ccc} a_1x_1 + a_2x_2 + ... + a_nx_n & = & 0 \\ b_1x_1 + b_2x_2 + ... + b_nx_n & = & 0 \\ ... \\ h_1x_1 + h2_x_2 + ... + h_nx_n & = & 0 \end{array} \right$$ (1)
Say :
$$A : = \left[ \begin{array}{ccc} a_1 & ... & a_n \\ \vdots & \ddots & \vdots \\ h_1 & ... & h_n \end{array} \right]$$
From the book, it states : (1) has 1 set of solution iff det(A) <> 0.
And (1) has many sets of solution iff det(A) = 0.
And because det(A) = 0 for all h. So I conclude that there must exist another set of solution to (1) apart from a = b = c = 0.
And because of that the 3 vectors are dependent, no mater what h is chosen.
By the way, the original question is : For what of h is {v(s1), v(s2), v(s3) } linear independent?
So am I... wrong ?
Viet Dao,