Question

Concept-checking questions

Consider vectors \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\), if \(\vec{w}\) is a linear combination of \(\vec{u}\) and \(\vec{v}\), does that mean that
i) \(\vec{w}\) is linearly dependent on \(\vec{u}\) and \(\vec{v}\) ?
ii) sp(\(\vec{u}\), \(\vec{v}\), \(\vec{w}\)) = sp(\(\vec{u}\), \(\vec{v}\)) ?

Answer 1

(i) follows straight from the definition of linear dependence :-)
For (ii), recognize that the linear span of a set of vectors is the set of all its finite linear combinations; if \(\vec{u},\vec{v},\vec{w}\) are linearly dependent, i.e. \(\vec{w}\) is a linear combination of \(\vec{u},\vec{v}\), then it should be immediately apparent that the spanning sets are identical.

Answer 2

If I can show \(\vec{u}=(u_1, u_2, u_3)\), \(\vec{v}=(v_1, v_2, v_3)\), and \(\vec{w}=(w_1, w_2, w_3)\) span \(\Re^3\), can I immediately draw a conclusion that the vectors are linearly independent?

Answer 3

I believe so, yes...

Answer 4

yes yes yes.

Answer 5

And so, if they do not span \(\Re^3\), they are linearly dependent, right?

Answer 6

yeah three vectors that aren't all linearly independent cannot span \[\mathbb R^3\]

Answer 7

Could "yeah three vectors that aren't all linearly independent cannot span \(\Re^3\)" be the justification of the answer?

Answer 8

i wouldn't word it like that

Answer 9

Since it cannot span \(\Re^3\), that means one of the vectors is a linear combination of the other vectors, therefore, the vectors are linearly dependent.
^ Could that be one?

Answer 10

thats is better yes

Answer 11

That's pretty nice :) Thanks!
May I ask one more question about linear dependence here?

Answer 12

sure

Answer 13

Checking the linear dependence of vectors, say, \(\vec{u} =(u_1,u_2,u_3), \vec{v} =(v_1,v_2,v_3),\) and \(\vec{w} =(w_1,w_2,w_3)\)
Method I)
\[\left[\begin{matrix}u_1 & v_1 & | & w_1 \\ u_2 & v_2 & | & w_2 \\u_3 & v_3 & | &w_3\end{matrix}\right]\]
Solve it, if it is
a) consistent (that is it has solution), then it is linearly dependent
b) inconsistent (that is it has no solution), the it is linearly independent

Method II)
Find the determinant of \[\left[\begin{matrix}u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\u_3 & v_3 & w_3\end{matrix}\right]\]
If the determinant is
a) equal to 0, it is linearly dependent
b) not equal to 0, it is linearly independent

Is that correct?
Is there any other way to check the linear dependence?

Answer 14

for the first method
if (0,0,0) is the only solution the the vectors are independent

the second method is right,

and im sure there are other methods but i can think of them

Answer 15

I think for the first method, it's actually testing if \(\vec{w}\) is a linear combination of \(\vec{u}\) and \(\vec{v}\)
Please leave a comment here if you can think of any of them. That would be really helpful!

Thank you for your help!! :)

Answer 16

If \(\vec{w}\) can be written as a linear combination of \(\vec{u}, \vec{w}\) then they are dependent.

Answer 17

\(\vec{v}\) **

Answer 18

And that's why
"if it is
a) consistent (that is it has solution), then it is linearly dependent
b) inconsistent (that is it has no solution), the it is linearly independent" :|

Answer 19

if you find the angle between vectors is π/2 the vectors will be independent

Answer 20

How to find the angle between three vectors??

Answer 21

\[\theta=\arccos\left(\frac{\vec u \cdot \vec v}{||\vec u||~~||\vec v||}\right)\]

Answer 22

For multiple vectors you would have to find the angle between each possible pair of vectors

Answer 23

For the case of two vectors, I still can do it. But not for three or more vectors..