# Thread: Transformation Matrix, vectorspace, polynomial functions

1. ## Transformation Matrix, vectorspace, polynomial functions

Hello

right now Im studying for my linear algebra exam and I think I pretty much understood the basic concepts but then I tried the following and I am not able to solve it, can someone help me with this:

In the vector space $\mathbb{R}^\mathbb{R}$ is U a subspace with the following polynomials as basis $\mathbf{B}$:
$f_0:x \mapsto 1$
$f_1:x \mapsto x$
$f_2:x \mapsto x^2$
$f_3:x \mapsto x^3$

Additionally there are the following functions:

$p_:0x \mapsto x^3$
$p_1:x \mapsto x^2(1-x)$
$p_2:x \mapsto x(1-x)^2$
$p_3:x \mapsto (1-x)^3$

which form a family $\mathbf{C}:=(p_0,p_1,p_2,p_3)$

a) Calculate the Transformation Matrix $<\mathbf{B}^\ast,\mathbf{C}>$ and the inverse.
b) Show why $\mathbf{C}$ is a basis
c) Calculate the Coordinates $<\mathbf{C},q>$ of $q: x \mapsto -x^3+3x^2-4x+1$
d) For the linear map $g_2: f \mapsto f(2)$calculate the Transformation matrices $$ and $$.

I have no clue how to begin. I think i know how to solve b) and inverting a matrix is not a problem but the rest is not clear to me!
Can you help me, please!

2. Originally Posted by hiddy
Hello

right now Im studying for my linear algebra exam and I think I pretty much understood the basic concepts but then I tried the following and I am not able to solve it, can someone help me with this:

In the vector space $\mathbb{R}^\mathbb{R}$ is U a subspace with the following polynomials as basis $\mathbf{B}$:
$f_0:x \mapsto 1$
$f_1:x \mapsto x$
$f_2:x \mapsto x^2$
$f_3:x \mapsto x^3$

Additionally there are the following functions:

$p_:0x \mapsto x^3$
$p_1:x \mapsto x^2(1-x)$
$p_2:x \mapsto x(1-x)^2$
$p_3:x \mapsto (1-x)^3$

which form a family $\mathbf{C}:=(p_0,p_1,p_2,p_3)$

a) Calculate the Transformation Matrix $<\mathbf{B}^\ast,\mathbf{C}>$ and the inverse.
b) Show why $\mathbf{C}$ is a basis
c) Calculate the Coordinates $<\mathbf{C},q>$ of $q: x \mapsto -x^3+3x^2-4x+1$
d) For the linear map $g_2: f \mapsto f(2)$calculate the Transformation matrices $$ and $$.

I have no clue how to begin. I think i know how to solve b) and inverting a matrix is not a problem but the rest is not clear to me!
Can you help me, please!
What have you tried? Is there a particular part that you're having trouble with? What inner product is this?

3. Originally Posted by Drexel28
What have you tried? Is there a particular part that you're having trouble with? What inner product is this?
Hey

$<\mathbf{B}^\ast,\mathbf{C}>$ this is not a inner product, this is the symbol we use for the transformation matrix ( $T_B^C$).

What i tried is this:

$p_0: x^3$
$p_1:x \mapsto x^2-x^3$
$p_2: x \mapsto x-2x^2+x^3$
$p_3: x \mapsto 1-3x+3x^2-x^3$

therfore: $\begin{matrix}
f: &1 & x & x^2 &x^3 \\
p_0: &0 & 0 & 0 &1 \\
p_1 & 0& 0& 1&-1 \\
p_2 & 0& 1& -2&1 \\
p_3 &1 & -3& 3 &-1
\end{matrix}$

and my transformationmatrix is $\begin{pmatrix}
0 & 0 & 0 &1 \\
0& 0& 1&-1 \\
0& 1& -2&1 \\
1 & -3& 3 &-1
\end{pmatrix}$

Calculating the inverse is not that a problem. Showing that this is a basis I show that the columns are linear independent.
For c) my idea is to multiply the Matrix with the Vector q: $\begin{pmatrix}
1 \\
-4 \\
3 \\
-1 \\
\end{pmatrix}$

And for d) i have no idea!

4. A general method for writing a linear transformation from vector space U to vector space V, given specific bases for U and V, is:]
Apply the linear transformation to each of the basis vector for U in turn. The result will be in V and so can be written as a linear combination of the basis vectors for V. The coeffients in that linear combination will form one column of the matrix.

Here, U and V are the same and you are given two different bases for it so essentially we write the basis "vectors" of the second matrix as linear combinations of the first and use those coefficients as columns of the matrix.

The first basis "vector" is $f_0= 1$ and it is to be transformed into $p_0= x^3= f_3$ so the first column of the matrix representation is $\begin{bmatrix}0 \\ 0 \\ 0 \\ 1\end{bmatrix}$. The second basis "vector" is $f_1= x$ and it is to be transformed into $p_1= x^2(1- x)= x^2- x^3= f_2- f_3$ so the second column of the matrix representation is $\begin{bmatrix}0 \\ 0 \\ 1 \\ -1\end{bmatrix}$. I don't see where you got that "-3". There is no "3" in any of the basis vectors.

To find the "inverse" of that, you can do either of two things:
1) Find the inverse matrix to the one you just found.
2) Do the process in "reverse"- write the basis vectors in the first basis as linear combinations of the vectors in the second basis and use the coefficients as columns of the matrix.

Do it both ways as a check.

5. Originally Posted by hiddy
d) For the linear map $g_2: f \mapsto f(2)$ calculate the Transformation matrices $$ and $$.
Originally Posted by hiddy
my transformation matrix is $\begin{pmatrix}
0 & 0 & 0 &1 \\
0& 0& 1&-1 \\
0& 1& -2&1 \\
1 & -3& 3 &-1
\end{pmatrix}$

Calculating the inverse is not that a problem. Showing that this is a basis I show that the columns are linear independent.
For c) my idea is to multiply the Matrix with the Vector q: $\begin{pmatrix}
1 \\
-4 \\
3 \\
-1 \\
\end{pmatrix}$

And for d) i have no idea!
For d), if $f(x) = ax^3+bx^2+cx+d$ then $f(2) = 8a+4b+2c+d$. So the matrix of the map $g_2$ with respect to B will be $M = \begin{bmatrix}8&4&2&1\end{bmatrix}$.

If $T = \begin{bmatrix}
0 & 0 & 0 &1 \\
0& 0& 1&-1 \\
0& 1& -2&1 \\
1 & -3& 3 &-1
\end{bmatrix}$
is the transformation matrix from B to C then the matrix of $g_2$ with respect to C will be $MT$ (or is it $MT^{-1}$ – I never remember which way round these things go).

6. Originally Posted by HallsofIvy
I don't see where you got that "-3". There is no "3" in any of the basis vectors.
I mixed up rows and columns.

So my results are (in my notation):

a) $<\mathbf{B}^\ast,\mathbf{C}> = \begin{bmatrix}
0 & 0 & 0 &1 \\
0& 0& 1&-3 \\
0& 1& -2&3 \\
1 & -1& 1 &1
\end{bmatrix}$

with the inverse $\begin{bmatrix}
-1 & 1 & 1 &1 \\
3& 2& 1& 0 \\
3& 1& 0 &0 \\
1 & 0& 0 &0
\end{bmatrix}$

this is a basis because they span the whole vector space.

for d) $= M = \begin{bmatrix} 8&4&2&1 \end{bmatrix}$

$=\begin{bmatrix}
0 & 0 & 0 &1 \\
0& 0& 1&-3 \\
0& 1& -2&3 \\
1 & -1& 1 &1
\end{bmatrix} \begin{bmatrix} 8 \\ 4 \\ 2 \\ 1 \end{bmatrix}=$

$=\begin{bmatrix} 1 &-1 & 3 &7 \end{bmatrix}$

for c) calculating the coordinates $<\mathbf{C},q>$ of $q: x \mapsto -x^3+3x^2-4x+1$ do i have to multiply $\begin{bmatrix}
-1 & 1 & 1 &1 \\
3& 2& 1& 0 \\
3& 1& 0 &0 \\
1 & 0& 0 &0
\end{bmatrix} \begin{bmatrix} 1 \\ -4 \\ 3 \\ -1 \end{bmatrix}$
?

Is this now correct?

7. ## Error alert!

For d), I had the B-basis vectors in the wrong order. The matrix M should be $\begin{bmatrix} 1&2&4&8\end{bmatrix}$, not $\begin{bmatrix} 8&4&2&1\end{bmatrix}$, shouldn't it?

Originally Posted by hiddy
I mixed up rows and columns.

So my results are (in my notation):

a) $<\mathbf{B}^\ast,\mathbf{C}> = \begin{bmatrix}
0 & 0 & 0 &1 \\
0& 0& 1&-3 \\
0& 1& -2&3 \\
1 & -1& 1 &1
\end{bmatrix}$

with the inverse $\begin{bmatrix}
-1 & 1 & 1 &1 \\
3& 2& 1& 0 \\
3& 1& 0 &0 \\
1 & 0& 0 &0
\end{bmatrix}$
That looks correct except that the bottom right element in the matrix should be –1, not 1. And in the inverse matrix, the top left element should be 1, not –1.

8. Originally Posted by Opalg
For d), I had the B-basis vectors in the wrong order. The matrix M should be $\begin{bmatrix} 1&2&4&8\end{bmatrix}$
yes! you're right! it should be $\begin{bmatrix} 1&2&4&8\end{bmatrix}$.

What do you say about c)? Correct or not?