Linear Transformations Question

Hi all, have no idea how to go about doing this!

In this question let β = {1, x, x^{2}, x^{3}} be the standard basis for http://upload.wikimedia.org/wikipedi...4ec0b06eb1.png_{[x]. }Let Φ: http://upload.wikimedia.org/wikipedi...4ec0b06eb1.png_{3}[x] http://upload.wikimedia.org/wikipedi...3eafbb4972.png http://upload.wikimedia.org/wikipedi...4ec0b06eb1.png_{3}[x] be the linear transformation defined by :

Φ : p(x) http://upload.wikimedia.org/wikipedi...78bbf9bac6.png (x^{2})(d^{2}p/dx^{2})+2(x+1)dp/dx

a) Find matrix [Φ]^{β}_{β of }Φ relative to the basis β

b) Consider the different basis γ = {1, 1+x, 1+3x+3x^{2}, 1+6x+15x^{2}+15x} for http://upload.wikimedia.org/wikipedi...4ec0b06eb1.png_{3}[x]. Find the change-of-basis matrices [id]^{β}_{γ and } [id]^{γ}_{β.
c) use your answer to part b to calculate }[Φ]^{γ}_{γ and observe that it's diagonal.
d) use answer to part c to calculate }Φ http://upload.wikimedia.org/wikipedi...ed408f6a5c.png Φ http://upload.wikimedia.org/wikipedi...ed408f6a5c.png ...http://upload.wikimedia.org/wikipedi...ed408f6a5c.png Φ(q) (10 times), where q = 15 +15x +15x^{2} + 15^{3}. (I think it may be best to use coordinate representation of q in the basis γ here. )

Re: Linear Transformations Question

To find the matrix of a given linear transformation, L, from V to V, given an ordered basis $\displaystyle \{a_n\}$ for V, apply L to each of the basis vectors in turn, and write the result as a linear combination of the basis vectors. The coefficients of that linear combination form one column of the matrix. Here, the second basis "vector" is the function "x". Applying the given linear transformation to that gives $\displaystyle 2(x+ 1)= 2x+ 2= 2(1)+ 2(x)+ 0(x^2)+ 0(x^3)$ because the first derivative is 1 and the second derivative is 0. The second column of the matrix representation is

$\displaystyle \begin{bmatrix}2 \\ 2 \\ 0 \\ 0 \end{bmatrix}$

Do the same thing, applying L to 1, $\displaystyle x^2$, $\displaystyle x^3$, to find the other three columns.

To find the matrices $\displaystyle [id]^\beta_\gamma$ and $\displaystyle [id]^\gamma_\beta$ do the same thing with the identity transformation. For example, the identity map takes the second basis vector in [tex]\beta[tex], "x", to x which we need to write as a linear combination in the basis [tex]\gamma[tex]: $\displaystyle x= (-1)(1)+ (1)(x+ 1)+ 0(1+ 3x+ 3x^2)+ 0(1+ 6x+ 15x^2+ 15x^3)$ so the second column of $\displaystyle [id]^\beta_\gamma$ is

$\displaystyle \begin{bmatrix}-1 \\ 1\\ 0\\ 0 \end{bmatrix}$

Do the same thing, the other way around, to find the matrix $\displaystyle [id]^\gamma_\beta$ or find the inverse of that function.

To find the matrix $\displaystyle [\phi]^\gamma_\gamma$, you can **either**

1) Do the same as in (a): Applying the linear transformation to 1+ x gives $\displaystyle (x+ 1)= 0(1)+ 2(x+ 1)+ 0(1+3x+ 3x^2)+ 0(1+ 6x+ 15x^2+ 15x^3)$ so the second column of $\displaystyle [\phi]^\gamma_\gamma$ is

$\displaystyle \begin{bmatrix}0 \\ 2 \\ 0 \\ 0\end{bmatrix}$

or

2) Multiply: $\displaystyle [\phi]^\gamma_\gamma= [id]^\beta_\gamma[\phi]^\beta_\beta[id]^\gamma_\beta$ which converts from basis $\displaystyle \gamma$ to basis $\displaystyle \beta$, applies the linear transformation in that basis, then converts back.

Of course, applying the linear transformation "n" times is the same as multiplying by the corresponding matrix n times. And because $\displaystyle [\phi]^\gamma_\gamma$ is diagonal, it is easy to take powers:

$\displaystyle \begin{bmatrix}a & 0 & 0 & 0 \\ 0 & b & 0 & 0\\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & d\end{bmatrix}^n= \begin{bmatrix}a^n & 0 & 0 & 0 \\ 0 & b^n & 0 & 0 \\ 0 & 0 & c^n & 0 \\ 0 & 0 & 0 & d^n\end{bmatrix}$

Re: Linear Transformations Question

To find the matrix of a given linear transformation, L, from V to V, given an ordered basis $\displaystyle \{a_n\}$ for V, apply L to each of the basis vectors in turn, and write the result as a linear combination of the basis vectors. The coefficients of that linear combination form one column of the matrix. Here, the second basis "vector" is the function "x". Applying the given linear transformation to that gives [itex]2(x+ 1)= 2x+ 2= 2(1)+ 2(x)+ 0(x^2)+ 0(x^3)[/itex] because the first derivative is 1 and the second derivative is 0. The second column of the matrix representation is

$\displaystyle \begin{bmatrix}2 \\ 2 \\ 0 \\ 0 \end{bmatrix}$

Do the same thing, applying L to 1, [itex]x^2[/itex], [itex]x^3[/itex], [itex] to find the other three columns.

To find the matrices [itex][id]^\beta_\gamma[/itex] and [itex][id]^\gamma_\beta[/itex] do the same thing with the identity transformation. For example, the identity map takes the second basis vector in [itex]\beta[itex], "x", to x which we need to write as a linear combination in the basis [itex]\gamma[itex]: [itex]x= (-1)(1)+ (1)(x+ 1)+ 0(1+ 3x+ 3x^2)+ 0(1+ 6x+ 15x^2+ 15x^3)[/itex] so the second column of [itex][id]^\beta_\gamma[/itex] is

$\displaystyle \begin{bmatrix}-1 \\ 1\\ 0 \ 0 \end{bmatrix}$

Do the same thing, the other way around, to find the matrix [itex][id]^\gamma_\beta[/itex] or find the inverse of that function.

To find the matrix [itex][\phi]^\gamma_\gamma[/itex], you can **either**

1) Do the same as in (a): Applying the linear transformation to 1+ x gives [itex](x+ 1)= 0(1)+ 2(x+ 1)+ 0(1+3x+ 3x^2)+ 0(1+ 6x+ 15x^2+ 15x^3)[/itex] so the second column of [itex][\phi]^\gamma_\gamma[/itex] is

$\displaystyle \begin{bmatrix}0 \\ 2 \\ 0 \\ 0\end{bmatrix}$

or

2) Multiply: [itex][\phi]^\gamma_\gamma= [id]^\beta_\gamma[\phi]^\beta_\beta[id]^\gamma_\beta[/itex] which converts from basis [itex]\gamma[/itex] to basis [itex]\beta[/itex], applies the linear transformation in that basis, then converts back.

Of course, applying the linear transformation "n" times is the same as multiplying by the corresponding matrix n times. And because [itex][\phi]^\gamma_\gamma[/itex] is diagonal, it is easy to take powers:

$\displaystyle \begin{bmatrix}a & 0 & 0 & 0 \\ 0 & b & 0 & 0\\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & d\end{bmatrix}^n= \begin{bmatrix}a^n & 0 & 0 & 0 \\ 0 & b^n & 0 & 0 \\ 0 & 0 & c^n & 0 \\ 0 & 0 & 0 & d^n\end{bmatrix}$