Linear independence (1 Viewer)

Sy123 · May 4, 2014

$\\ $If$ \ \vec{v}_1 , \vec{v}_2 , \vec{v}_3 \ \text{are linear independent} \\ \\ $and$ \ \ a\vec{v}_1 + b\vec{v}_2 + c\vec{v}_3 = x\vec{u}_1 + y\vec{u}_2 + z\vec{v}_3 \\ \\ $then this implies$ \ \vec{u}_1 , \vec{u}_2 , \vec{u}_3 \ \text{is linear indepdendent}$

Is this true? I am very strongly inclined to say yes, but I feel as though I'm missing something and just want to make sure

seanieg89 · May 4, 2014

Sy123 said:
$\\ $If$ \ \vec{v}_1 , \vec{v}_2 , \vec{v}_3 \ \text{are linear independent} \\ \\ $and$ \ \ a\vec{v}_1 + b\vec{v}_2 + c\vec{v}_3 = x\vec{u}_1 + y\vec{u}_2 + z\vec{v}_3 \\ \\ $then this implies$ \ \vec{u}_1 , \vec{u}_2 , \vec{u}_3 \ \text{is linear indepdendent}$

Is this true? I am very strongly inclined to say yes, but I feel as though I'm missing something and just want to make sure

What are a,b,c,x,y,z? just some fixed constants? If this is the case the statement certainly isn't true.

Or are you saying something more like "If every linear combination of the v_i's can be written as a linear combination of the u_i's, and the v_i's are linearly independent then so are the u_i's"?

The latter statement is true by a dimensionality argument.

Drongoski · May 4, 2014

$\vec {v}_1, \vec{v}_2 $ and $ \vec {v}_3$ are linearly indepenent means in effect: none of them can be derived from the other 2 by way of a linear combination of the other two.

Equivalently, $x_1 \vex{v}_1 + x_2 \vec {v}_2 + x_3 \vec {v}_3 = \vec{0} \implies x_1, x_2 $ and $ x_3 = 0$

Sy123 · May 4, 2014

seanieg89 said:
What are a,b,c,x,y,z? just some fixed constants? If this is the case the statement certainly isn't true.

Or are you saying something more like "If every linear combination of the v_i's can be written as a linear combination of the u_i's, and the v_i's are linearly independent then so are the u_i's"?

The latter statement is true by a dimensionality argument.

Yep just fixed constants, thank you

Sy123 · May 4, 2014

So what if I altered my statement:

$\\ $Using the above conditions, the statement$ \\ \\ x \vec{u_1} + y\vec{u_2} + z \vec{u_3} = 0 \\ $implies \\ \alpha \vec{v_1} + \beta \vec{v_2} + \gamma \vec{v}_3 = 0 \\ \text{Which implies} \ \ \alpha = \beta = \gamma = 0$

seanieg89 · May 4, 2014

Sy123 said:
Yep just fixed constants, thank you

Yeah so the answer is just no then.

A really silly/trivial counterexample is as follows:

The standard basis (e_1,e_2,e_3) of R^3 is linearly independent, and 1e_1+0e_2+0e_3 can be written as 1e_1+0e_1+0e_1. (But e_1,e_1,e_1 is obviously not linearly dependent.)

seanieg89 · May 4, 2014

Sy123 said:
So what if I altered my statement:

$\\ $Using the above conditions, the statement$ \\ \\ x \vec{u_1} + y\vec{u_2} + z \vec{u_3} = 0 \\ $implies \\ \alpha \vec{v_1} + \beta \vec{v_2} + \gamma \vec{v}_3 = 0 \\ \text{Which implies} \ \ \alpha = \beta = \gamma = 0$

Huh? Can you please elaborate here? Are you changing the question or your answer?

Sy123 · May 5, 2014

seanieg89 said:
Huh? Can you please elaborate here? Are you changing the question or your answer?

I'm just changing the question, seeing what I can get with the initial conditions (of v's being linear independent)

seanieg89 · May 5, 2014

Sy123 said:
I'm just changing the question, seeing what I can get with the initial conditions (of v's being linear independent)

Not much if the 6 scalars are still just some particular constants. Eg this equation is trivially true with all constants 0, no matter what the u_j's are.

Davo_01 · May 5, 2014

Sy123 said:
So what if I altered my statement:

$\\ $Using the above conditions, the statement$ \\ \\ x \vec{u_1} + y\vec{u_2} + z \vec{u_3} = 0 \\ $implies \\ \alpha \vec{v_1} + \beta \vec{v_2} + \gamma \vec{v}_3 = 0 \\ \text{Which implies} \ \ \alpha = \beta = \gamma = 0$

Is this question ur asking related to the method ur trying to use to prove 3d in math1902 assignment?

SpiralFlex · May 5, 2014

GoldyOrNugget · May 6, 2014

SpiralFlex said:
$Do you mean something like this\?$

$Suppose we have a set$ \;S = \{v_{1}, v_{2}, ..., v_{n}\} \;$of linearly independent vectors and a set$\; T = \{w_{1},....,w_{m}\}\; $where for each$ \;T\; $is a linear combination of elements of$\; S \;$with$\; m > n. \; $Then$\; T \;$is linearly independent.$

I think that should read "T is linearly dependent".

Linear independence (1 Viewer)

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