- anonymous

Show that the vector \(u\) = (1, 2, 3), \(v\) = (0, 1, 2), and \(w\) = (0, 0, 1) span \(\Re ^3\). Are these vectors linearly independent? Please justify your answer.

- chestercat

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- anonymous

it 's not linearly.

- anonymous

Why?

- anonymous

if it was linearly, that should be w-v=v-u. ok??

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## More answers

- anonymous

I don't understand it.. Can you explain?

- anonymous

w-v=((0-0),(0-1),(1-2))=(0,-1,-1)
v-u=((0-1),(1-2),(2-3))=(-1,-1,-1) ok??

- klimenkov

Can you calculate this?
\(\left|\begin{matrix}
1&2&3\\
0&1&2\\
0&0&1
\end{matrix}\right|\)

- anonymous

1

- anonymous

But why did you put the vector horizontally (row vector?!) but not vertically (column vector?!)

- klimenkov

Because nothing will change. Determinant doesn't change if you change the role of rows and columns. You've got, that this determinant is not equal to zero, that means these 3 vectors are linearly independent.

- anonymous

Wow!
I actually did it in this way, was it correct?
Consider \(r\vec{u} + s\vec{v}+t\vec{w}\) for all scalars r, s, t
\[\left[\begin{matrix} 1&2&3 & | &0\\ 0&1&2 & | &0\\ 0&0&1 & | &0\end{matrix} \right]\]
\[ ~ \left[\begin{matrix} 1&0&0 & | &0\\ 0&1&0 & | &0\\ 0&0&1 & | &0\end{matrix} \right]\]
[r, s, t] = [0, 0, 0]
Since we have only [0, 0, 0] is the (trivial) solution, so the three vectors are linearly independent.

- klimenkov

Hm, that is not actually right.

- klimenkov

You need:
\(
r\cdot\left(\begin{matrix}
1\\
2\\
3
\end{matrix}\right)+
s\cdot\left(\begin{matrix}
0\\
1\\
2
\end{matrix}\right)+
t\cdot\left(\begin{matrix}
0\\
0\\
1
\end{matrix}\right)=
\left(\begin{matrix}
0\\
0\\
0
\end{matrix}\right)
\)
And you system of the equations must be:
\(
\left\{
\begin{matrix}
1\cdot r+0\cdot s+0\cdot t=0\\
2\cdot r+1\cdot s+0\cdot t=0\\
3\cdot r+2\cdot s+1\cdot t=0
\end{matrix}
\right.
\)
You've got another system. Understand?

- anonymous

Ahh!!!! So I... actually.... and TOTALLY .....got... that..... wrong :|
Actually, I am confused.
Do I get this part right?
If the three vectors span \(\Re ^n\), and the determinant is not zero, they are linearly independent. If the three vectors span \(\Re ^n\), and the determinant is zero, they are linearly dependent.

- klimenkov

Please, write what \(n\) in \(\Re ^n\) ?

- klimenkov

I have some questions for you. Please, answer them all only by yourself.
1) What is a linear combination?
2) What is a linear independence?
3) What means \(n\) vectors span \(\Re ^n\) ?

- anonymous

1) Linear combination: \(\vec{v}=c_1\vec{v_1} + c_2\vec{v_2} +...+c_n\vec{v_n}\), \(c_1, c_2, ..., c_n\)
2) Linear independence: \(c_1, c_2, ..., c_n\) for not all n being zero
3) n vectors span \(\Re ^n\): the n vectors cannot be expressed in term of the others.
Eg: [1, 2, 3] and [1, 4, 5] => span \(\Re ^2\) but [1, 2, 3] and [2, 4, 6] => span \(\Re ^1\)
Something wrong with last one, I hope you don't mind me checking it now.

- anonymous

1) c1,c2,...,cn => weights (Well, I can only understand it as some scalars)
3) span: Given that
\(\vec{v_1}, \vec{v_2}, ..., \vec{v_p}\) are in \(\Re ^n\)
set of all linear combinations of vectors = subspace of \(\Re ^n\) = spanned by \(\vec{v_1}, \vec{v_2}, ..., \vec{v_p}\)
subspace = (need to be googled)
Consider vector v, and w. If they are both non-zero, non-parallel vectors in \(\Re ^n\)
all possible linear combinations of v and w can be expressed in the form of rv +sw, for all scalars r and s.
all possible linear combinations => filled a plane => plane spanned by v and w.
Sorry, too much irrelevant things here, maybe.

- klimenkov

I think it will be easy to unterstand. \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\) span \(\Re ^n\) if any vector from \(\Re ^n\) can be represented as a linear combination of \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\).

- klimenkov

Now back to your task. Using the system I wrote above you can get, that the vectors are independent. Now you need to show that any vector from \(\Re^3\) can be shown as their linear combination.

- anonymous

"\(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\) span \(\Re ^n\) if any vector from Rn can be represented as a linear combination of \(\vec{v_1}, \vec{v_2}, \ldots, \vec{v_p}\)"
So, in this case, we need to find some vectors \(r\vec{u} + s\vec{v} + t\vec{w}\) from \(\Re ^3\).
Set \(r\vec{u} + s\vec{v} + t\vec{w} = [0, 0, 0]\) to solve r, s, t?!

- anonymous

Not quite. [0, 0, 0] is too trivial...Hmmm... :\

- klimenkov

Take any vector from \(\Re^3\) - \(\vec x=(x_1,x_2,x_3),\, x_i\in \Re\) and show that you can find scalars (weights) - \(r,s,t\) so \(\vec x=r\vec{u} + s\vec{v} + t\vec{w}\).

- klimenkov

But don't take a concrete vector \(\vec x\), like (1,2,3). Solve it in general.

- anonymous

I'm sorry for late reply!
I got \(r = x_1 , \ s = -2x_1 + x_2, \ t=-x_1-2x_2+x_3\)
These are all scalars, I supposed. Aren't they?

- anonymous

- klimenkov

Yes, that's right, but you've put an extra minus in the expression for \(t\). So, you've showed that these three vectors span \(\Re^3\). Congratulations.

- anonymous

:O Thanks!!!
As you may observe, I'm not quite clear with the concepts, right?

- klimenkov

Yes. Actually, it is not a big problem. All you need is to learn honestly.
The biggest part in the solving math problems is to understand the statement of the question. Try to use all the possible sources to understand the problem. In this way you will learn a lot.

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