Question

1

Answer 1

is this the full question?

Answer 2

\[\left\{ \left(\begin{matrix}1 \\ 0\end{matrix}\right),\left(\begin{matrix}2 \\ 0\end{matrix}\right),\left(\begin{matrix}6 \\ 0\end{matrix}\right) \right\}\]

Answer 3

Can you write any of the vectors in the set as a linear combination of the vectors in the set?

In other words, for non-zero constants \(k_1,k_2,k_3\), can you express some vector in the given set \(v\) like this:
\[k_1\binom10+k_2\binom20+k_3\binom60=v~~?\]

Answer 4

okay, but why

Answer 5

Do you know what the definition of linear independence is? It's straight from the definition.

Also, just noticed a slight error: \(v\) should be the zero vector, not any vector in the set...

Answer 6

yes, i know that you getta set them all equal to 0 vector right?

Answer 7

Right, but you have to be able to do that without having all the constants be zero.

The vector equation can be written as a system of equations:
\[\begin{cases}k_1+2k_2+6k_3=0\\
0+0+0=0\end{cases}\]
If you can find at least one set of solutions to this system (not \(k_1=k_2=k_3=0\)), then you're done, and you say the set is indeed linearly independent.

Answer 8

isn't it if the only thing you can find is k1=k2=k3=0, then it is independent, if you can find something else then it is not?

Answer 9

Yes, you're right. I meant *dependent* at the end of that sentence >.< This is what happens when you don't do linear algebra for a while.

Answer 10

hahaha. Ya this subject is easy but it is messy compared to my other math classes. Well the problem I am having is I know how to set the stuff in the boxes, but still get stuck in solving them

Answer 11

Try fixing one of the constants as zero, say \(k_3=0\). You'll be able to find a solution for one variable in terms of the other. If there's at least one non-zero solution for the remaining variables, you're done.

Answer 12

1 0 | 0
2 0 | 0
6 0 | 0
6k1-k3 gave me 1 0 , 2 0 , 0 0
not sure if I am going right way

Answer 13

Whenever I think of linearly independent or dependent all it really means is if I'm given 2 vectors, can I get everywhere in a 2D space? If I'm given 3 vectors can I get everywhere in a 3D space? And of course, if I'm given n-vectors, can I get everywhere in an n-dimensional space?

Don't get hung up on the concept of what "the" fourth dimension is or anything like that. In linear algebra by "space" they don't necessarily mean physical space, a 4D space could mean you have red, blue, green, and orange balls that you're keeping in baskets and surely the amount of green balls is linearly independent of red balls?

Answer 14

Suppose \(k_3=0\), then you have
\[k_1+2k_2=0~~\implies~~k_2=-2k_2\]
Does this have any non-zero solutions?

Answer 15

no it doesn't. 0 0 0 is only possible

Answer 16

Typo, should be \(k_1=-2k_2\)...

What if \(k_1=1\) ? Then surely \(k_2=-\dfrac{1}{2}\), and neither are zero.

Answer 17

ahhh i see. So they are linearly dependent right?

Answer 18

Right! We've found at least one set of nontrivial solutions for the constants, so we satisfy the definition for linear dependence.

Answer 19

what does ' non trivial ' mean

Answer 20

Non-zero

Answer 21

oh thanks a lot

Answer 22

For \(x+y=0\), the trivial solution \(x=y=0\). Think of "trivial solution" as "easiest numbers that make the equation true."

Answer 23

You're welcome!