anonymous
  • anonymous
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(\(\vec{u}\), \(\vec{v}\)) ?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
(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.
anonymous
  • anonymous
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?
anonymous
  • anonymous
I believe so, yes...

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UnkleRhaukus
  • UnkleRhaukus
yes yes yes.
anonymous
  • anonymous
And so, if they do not span \(\Re^3\), they are linearly dependent, right?
UnkleRhaukus
  • UnkleRhaukus
yeah three vectors that aren't all linearly independent cannot span \[\mathbb R^3\]
anonymous
  • anonymous
Could "yeah three vectors that aren't all linearly independent cannot span \(\Re^3\)" be the justification of the answer?
UnkleRhaukus
  • UnkleRhaukus
i wouldn't word it like that
anonymous
  • anonymous
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?
UnkleRhaukus
  • UnkleRhaukus
thats is better yes
anonymous
  • anonymous
That's pretty nice :) Thanks! May I ask one more question about linear dependence here?
UnkleRhaukus
  • UnkleRhaukus
sure
anonymous
  • anonymous
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?
UnkleRhaukus
  • UnkleRhaukus
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
anonymous
  • anonymous
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!! :)
anonymous
  • anonymous
If \(\vec{w}\) can be written as a linear combination of \(\vec{u}, \vec{w}\) then they are dependent.
anonymous
  • anonymous
\(\vec{v}\) **
anonymous
  • anonymous
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" :|
UnkleRhaukus
  • UnkleRhaukus
if you find the angle between vectors is π/2 the vectors will be independent
anonymous
  • anonymous
How to find the angle between three vectors??
UnkleRhaukus
  • UnkleRhaukus
\[\theta=\arccos\left(\frac{\vec u \cdot \vec v}{||\vec u||~~||\vec v||}\right)\]
UnkleRhaukus
  • UnkleRhaukus
For multiple vectors you would have to find the angle between each possible pair of vectors
anonymous
  • anonymous
For the case of two vectors, I still can do it. But not for three or more vectors..

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