Linear maping

[mammothrob]

T: P3 ----> P4

\(\displaystyle \begin{array}{l}
T(ax^3 + bx^2 + cx + d) = (3x + 2)(ax^3 + bx^2 + cx + d) \\
B = \{ x^3 ,x^2 ,x,1\} \\
B' = \{ x^4 ,x^3 ,x^2 ,x,1\} \\
\end{array}\)


Find the matrix for T relative to the bases B and B'.

Now, by evaluating each element of B through my trans formation i get
4 new polynomials.

I used their co efficients as entries in the columns of my matrix.


3000
2300
0230
0023
0002


My question is this... in problems where I am mapping say

R2--->R3


I have to either use a transiton matricie, or use linear combinations of
T(B) with B' to get my "matrix for T relative to B and B' "columns?

Is

3000
2300
0230
0023
0002


really my matrix for T relative to B and B' or is more work needed?

 

[pka]

T: P3 ----> P4\(\displaystyle \begin{array}{l}
T(ax^3 + bx^2 + cx + d) = (3x + 2)(ax^3 + bx^2 + cx + d) \\
B = \{ x^3 ,x^2 ,x,1\} \\
B' = \{ x^4 ,x^3 ,x^2 ,x,1\} \\
\end{array}\)
Find the matrix for T relative to the bases B and B'.
I used their co efficients as entries in the columns of my matrix.
3000
2300
0230
0023
0002

My question is this... in problems where I am mapping say
R2--->R3
I have to either use a transiton matricie, or use linear combinations of
T(B) with B' to get my "matrix for T relative to B and B' "columns? Is
3000
2300
0230
0023
0002

I really don’t understand your point!
Your work on the matrix for T is correct.
However, to map \(\displaystyle R^2 \mapsto R^3\) ones needs a \(\displaystyle 3 \times 2\) matrix. But that is not the example you gave? So what is your question?

 

[mammothrob]

T:R2 -----> R3

\(\displaystyle \begin{array}{l}
T(x,y) = (x + y,x,y) \\
B = \{ (1, - 1),(0,1)\} \\
B' = \{ (1,1,0),(0,1,1)(1,0,1)\} \\
\end{array}\)


The matrix for T relative to B and B' is

1 0
0 0
-1 1


To get that it was a more involved process to get that matrix.

Involving linear combinations and solving of systems.

Where as my other problem maping to polynomials, I was just able to read the coeffecients off of T(B)s.

Does that make since?

Just seemed to easy, like im missing some steps.

 

[daon]

When you say "Relative to B and B'," what exactly do you mean? Do you mean "The transformation from R2 to R3 using the Bases B (for R2), B' (for R3)?"

 

[mammothrob]

The matrix A is the matrix relative to B and B'.

I mean if you Multiply A with coordinate matrix relaitve to B you get a coordinate matrix relative to B'

 

[pka]

To get that it was a more involved process to get that matrix. Involving linear combinations and solving of systems. Where as my other problem maping to polynomials, I was just able to read the coeffecients off of T(B)s.
Just seemed to easy, like im missing some steps.

Is this it?
Yes the first is quite easy because both spaces are using a standard basis.
The second problem is using a non-standard basis. That makes it much more difficult.

 

[daon]

The matrix A is the matrix relative to B and B'.

I mean if you Multiply A with coordinate matrix relaitve to B you get a coordinate matrix relative to B'

Okay.. the way you said it confused me, but that is probably more due to the fact that I had a different teacher.

Anyway, for the nonstandard...

In your second post, the example which I assume is correct, you are using a 3x2 matrix as a representation of T. This means that it takes a vector v from a 3-dimentional vector space to a 2-dimentional vector space.

So, when you are not using the standard basis, you do need to do a row reduction. You would be reducing the basis you are going TO (as columns) augmented with the the matrix representing T acording to the basis you are coming FROM.

i.e. [ B1 B2 ... Bm | T(A1) T(A2) ... T(An) ], Where B is the basis you are going to and A is the basis you are comming from. This is why the standard basis makes things so much nicer, since no row reduction is needed if going TO a standard basis.

You would row reduce the LHS to the standard I_m identity matrix, but be sure to make the same changes to the RHS augmented matrix. The resulting matrix should be m x n, and that will be the matrix you are looking for.

There are other ways, but it can be more difficult.

 

[mammothrob]

yesss...

I see it now. I was missing the concept between standard and non-standard bases. Much appreciated.