# Math Help - linear algebra -vectors

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2. ## Re: linear algebra -vectors

Originally Posted by yeatch
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What are we suppose to do with that list of questions?

3. ## Re: linear algebra -vectors

I need help with the one that don't have a star beside them? I looked at tutorial on YouTube and nothing helps...

4. ## Re: linear algebra -vectors

Have you made any attempt on these at all?

5. ## Re: linear algebra -vectors

Yes of course... I only have five more problems to do....i also review my notes and still have no clue... i went for tutoring and they did not know the material

6. ## Re: linear algebra -vectors

Well let's look at the very first one:

We are supposed to prove that:

$0[x_1,x_2,\dots,x_n] = [0,0,\dots,0]$.

By definition of the scalar product, for any $r \in \Bbb R$, we have:

$r[x_1,x_2,\dots,x_n] = [rx_1,rx_2,\dots,rx_n]$.

Letting $r = 0$ in the above leads to:

$0[x_1,x_2,\dots,x_n] = [0x_1,0x_2,\dots,0x_n]$.

Now, for any $j = 1,2,\dots,n$, is it true that $0x_j = 0$?

7. ## Re: linear algebra -vectors

Omg thank you so much !! yes that is right....what about number 2 a); whats you take on cu and the zero vectors?

8. ## Re: linear algebra -vectors

Ok, suppose $u = [x,y,z]$ (or we could write $u = [x_1,x_2,x_3]$ it's just notation).

By definition, $cu = c[x,y,z] = [cx,cy,cz]$.

If this equals $[0,0,0]$, this means:

$cx = 0$
$cy = 0$
$cz = 0$.

Now if $c = 0$, we have nothing to prove (see answer to number 1 above). So suppose $c \neq 0$. We need to show that in this case $x = y = z = 0$.

Since we are assuming $c \neq 0$, from the 3 equations above, we have:

$x = \dfrac{cx}{c} = \dfrac{0}{c} = 0$

and similarly for $y$ and $z$. Any questions?

9. ## Re: linear algebra -vectors

wow...not at all....i figured out part b ....moving on to number three how can we show that it is associative?

10. ## Re: linear algebra -vectors

I re-looked at my work and made some corrections base on your explained examples...now all i need help with is #three thru 6.

11. ## Re: linear algebra -vectors

3. Just set \displaystyle \begin{align*} \mathbf{x} = \left( x_1, x_2, x_3 \right) , \mathbf{y} = \left( y_1, y_2, y_3 \right) , z = \left( z_1, z_2, z_3 \right) \end{align*} and see if the associative property holds.

4. A vector, say \displaystyle \begin{align*} \mathbf{a} \end{align*}, is a linear combination of two other vectors, say \displaystyle \begin{align*} \mathbf{b} \end{align*} and \displaystyle \begin{align*} \mathbf{c} \end{align*} if it's possible to write \displaystyle \begin{align*} \mathbf{a} = c_1 \mathbf{b} + c_2\mathbf{c} \end{align*}, where \displaystyle \begin{align*} c_1, c_2 \end{align*} are constants.

5. A property of the dot product that will help you: \displaystyle \begin{align*} \mathbf{a} \cdot \left( \mathbf{b} \cdot \mathbf{c} \right) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} \end{align*}.

12. ## Re: linear algebra -vectors

Thanks alot.... Now lastly, do you have any idea how to do #6?

13. ## Re: linear algebra -vectors

Also, you should suspect vector subtraction is NOT associative, since subtraction isn't even associative in the real numbers:

5-(4-3) = 5 - 1 = 4

(5-4)-3 = 1 - 3 = -2.

More abstractly:

a - (b - c) = a - b + c

(a - b) - c = a - b - c, and if these two are to be equal, we get c = -c, which isn't always the case.

We can write any point on the line L as (1+t,2t-1) for some t. Suppose n = (a,b).

Taking the dot product we have n.(1+t,2t-1) = a(1+t) + b(2t-1).

Let's use some specific values for t, to see what happens. At t = 0, we get the point on L, (1,-1). So if this solves n.x = c, for x = (1,-1), we have that c = a - b.

At t = 1, we get the point (2,1) on L, so plugging this into n.x = c gives us c = 2a + b. Since we are using the same "c", this means:

2a + b = a - b, or: a = -2b. So n is of the form (-2b,b).

Suppose we take b = 1. Then n = (-2,1), and c = -3.

Let's verify that this works:

(-2,1).(1+t,2t-1) = -2(1+t) + (2t-1) = -2 - 2t + 2t - 1 = -3.

Why this is called "normal form": note that L is parallel to the line L' = t(1,2), and that n.(1,2) = (-2,1).(1,2) = -2 + 2 = 0, in other words, n is perpendicular (normal) to L.

Let's "turn this problem on its head" by working backwards:

Suppose we have a NON-ZERO vector n = (a,b), and seek all points (x,y) that satisfy (a,b).(x,y) = c, that is:

ax + by = c, or equivalently, by = -ax + c.

We distinguish two cases:

Case 1: b = 0. In this case, then a is non-zero, so we get the (vertical) line:

x = c/a, which can be written as: (x,y) = (c/a,0) + t(0,1). Note that (a,0) and (0,1) are perpendicular.

Case 2: b is non-zero, so y = -(a/b)x + (c/b), which can be written as the line (x,y) = (0,c/b) + t(1,-(a/b)). Note here, as well: (a,b).(1,-(a/b)) = a - a = 0.

In this problem, our line L has a finite slope, so b is not 0.

Since we must have (1,2) parallel to (1,-(a/b)), this means a/b = -2.

Now (1,-1) lies on L, so writing (1,-1) = (0,c/b) + t(1,-(a/b)) = (t,(c-ta)/b)) = (t,(c+2bt)/b)), and solving for t, we see t = 1, so:

-1 = (c+2b)/b = (c/b) + 2, that is: c/b = -3.

Hence the normal form of L is any equation:

b(-2,1).(x,y) = -3b, and cancelling the b's, we get L has the equation: (-2,1).(x,y) = -3.

(Note the normal form isn't unique, we could multiply the constant c and the normal vector n by the same scalar).