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RolyPoly
Group Title
Conceptchecking 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(\(\vec{u}\), \(\vec{v}\)) ?
 one year ago
 one year ago
RolyPoly Group Title
Conceptchecking 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(\(\vec{u}\), \(\vec{v}\)) ?
 one year ago
 one year ago

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oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
I believe so, yes...
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
yes yes yes.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
And so, if they do not span \(\Re^3\), they are linearly dependent, right?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
yeah three vectors that aren't all linearly independent cannot span \[\mathbb R^3\]
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
Could "yeah three vectors that aren't all linearly independent cannot span \(\Re^3\)" be the justification of the answer?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i wouldn't word it like that
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
thats is better yes
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
That's pretty nice :) Thanks! May I ask one more question about linear dependence here?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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!! :)
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
If \(\vec{w}\) can be written as a linear combination of \(\vec{u}, \vec{w}\) then they are dependent.
 one year ago

oldrin.bataku Group TitleBest ResponseYou've already chosen the best response.1
\(\vec{v}\) **
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
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" :
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
if you find the angle between vectors is π/2 the vectors will be independent
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
How to find the angle between three vectors??
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\theta=\arccos\left(\frac{\vec u \cdot \vec v}{\vec u~~\vec v}\right)\]
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
For multiple vectors you would have to find the angle between each possible pair of vectors
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.0
For the case of two vectors, I still can do it. But not for three or more vectors..
 one year ago
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