# Need help, test coming soon!

• Sep 23rd 2008, 06:43 PM
aeubz
Need help, test coming soon!
Need explanations!

My test is coming up real soon!

Need help with these questions.

Sorry for the rudeness (Doh) ... tests make me... nervous...

Describe all vectors v = [x,y] that are orthogonal to u = [a,b].

And I really really need help understanding "span" in linear algebra. How do you determine if a vector is the span of a matrix?

prob) determine if the vector b is in the span of the columns of the matrix A.
A =
[1 2]
[3 4]
b =
[5]
[6]

prob) show that R^2 =
span
([1] , [1])
([1] , [-1])
• Sep 23rd 2008, 07:14 PM
Chris L T521
Hello (Hi)

Since I don't know much about span, I'll take the easy one (Wink)

Quote:

Originally Posted by aeubz
Describe all vectors v = [x,y] that are orthogonal to u = [a,b].

Take the dot product of $\displaystyle \bold v$ and $\displaystyle \bold u$

$\displaystyle \bold u\cdot\bold v=ax+by$

The vectors are orthogonal when $\displaystyle \bold u\cdot\bold v=0$

So , we see that $\displaystyle ax+by=0\implies by=-ax$

Now, we need to manipulate the vector $\displaystyle \bold v$:

$\displaystyle \left[\begin{array}{c}x\\y\end{array}\right]={\color{red}\frac{b}{b}}\left[\begin{array}{c}x\\y\end{array}\right]=\frac{1}{b}\left[\begin{array}{c}bx\\by\end{array}\right]$

Since, $\displaystyle by=-ax$, $\displaystyle \bold v=\frac{1}{b}\left[\begin{array}{c}bx\\-ax\end{array}\right]=\frac{x}{b}\left[\begin{array}{c}b\\-a\end{array}\right]$

Since $\displaystyle \frac{x}{b}$ is another constant, let's denote it by $\displaystyle k$

So all the possible vectors will take on the form $\displaystyle \color{red}\boxed{\bold v=k\left[\begin{array}{c}b\\-a\end{array}\right];~k,a,b\in\mathbb{R}}$.

Does this make sense? :D

--Chris