# Linear algebra questions

• March 16th 2009, 09:58 AM
GreenandGold
Linear algebra questions
I have this 2x3 matrix, M:
|203|
|151|

the problem that i am having is that all the entries have bars over them. Can any one explain the bar notation. I've ever seen it before. Thanks!!!!

The problem that I'm given is to find an L:V->W such that relative to B and B', the matrix M is associated with L.

I think it would help if I knew what the bar notation ment.
• March 17th 2009, 03:50 AM
Showcase_22
I think you can do it this way:

The first thing to notice is that dim(V)=3 and dim(W)=2. Hence:

$\begin{pmatrix}
{2}&{0}&{3}\\
{1}&{5}&{1}
\end{pmatrix}\begin{pmatrix}
{1}\\
{0}\\
{0}
\end{pmatrix}=\begin{pmatrix}
{2}\\
{1}
\end{pmatrix}$

Repeating the process for all the standard basis of a 3D space gives the vectors which span W to be $\begin{pmatrix}
{2}\\
{1}
\end{pmatrix},\begin{pmatrix}
{0}\\
{5}
\end{pmatrix}$
and $\begin{pmatrix}
{3}\\
{1}
\end{pmatrix}$
.

However these vectors just span. If we have three vectors which span, and the vector space only has a dimension of 2, one of the vectors must be a linear combination of the others.

The vector $\begin{pmatrix}
{0}\\
{5}
\end{pmatrix}$
is a linear combination of 15 $\begin{pmatrix}
{2}\\
{1}
\end{pmatrix}$
vectors and -10 $\begin{pmatrix}
{3}\\
{1}
\end{pmatrix}$
vectors (these were found using simultaneous equations).

Therefore, upon sifting, we get:

$B'=\begin{pmatrix}
{2}\\
{1}
\end{pmatrix},\begin{pmatrix}
{0}\\
{5}
\end{pmatrix}$
where $B=\begin{pmatrix}
{1}\\
{0}\\
{0}
\end{pmatrix},\begin{pmatrix}
{0}\\
{1}\\
{0}
\end{pmatrix},\begin{pmatrix}
{0}\\
{0}\\
{1}
\end{pmatrix}$
.
• March 17th 2009, 07:04 AM
GreenandGold
I without question I understood all of that thank you. However are you saying that the function L is the elementary basis vector? In other words are you saying L(x)=(1,0,0)

What if B={(0,1,1)(1,1,1),(0,0,1)} and B'={(0,1),(1,0)}. Then would I just take a basis vector from B or B' and use the inverse matrix m^-1 to solve for x? Hopefully im thinking of this right.
• March 17th 2009, 10:22 AM
Showcase_22
Quote:

Originally Posted by GreenandGold
I without question I understood all of that thank you.

No problem =p

Quote:

However are you saying that the function L is the elementary basis vector? In other words are you saying L(x)=(1,0,0)
I haven't heard of an elementary basis vector before (is it the same as the standard basis?).

I'm saying that:

L(1,0,0)=(2,1)
L(0,1,0)=(0,5)
L(0,0,1)=(3,1)= $\frac{3}{2}$(2,1)- $\frac{1}{10}$(0,5)

L is a mapping from V to W so it maps the basis vectors of V (which I made to be the standard basis B={(1,0,0),(0,1,0),(0,0,1)}) to the vectors that i've written above (which is the basis B'={(2,1),(0,5)}).

Quote:

What if B={(0,1,1)(1,1,1),(0,0,1)} and B'={(0,1),(1,0)}. Then would I just take a basis vector from B or B' and use the inverse matrix m^-1 to solve for x? Hopefully im thinking of this right.
Let's see:

From the matrix M, (0,1,1) has to be equal to 2(0,1)+(1,0)=(1,2) (by reading the first column).

(0,1,1) $\neq$ (1,2) so those sets for B and B' cannot be used.

However, I think it works better if you fix B'={(0,1),(1,0)} and treat B as a set of unknown vectors, say B={A,B,C} where each vector is of the form $(x,y,z)$.

The big problem here arises from the fact that M is not invertible (since it isn't a square matrix). If the matrix was square I think the method you devised would work pretty well! =p
• March 19th 2009, 03:56 AM
GreenandGold
thanks that really cleared up much of my confusion...