Linear algebra: Find the matrix of linear transformation

  • #1
208
1

Homework Statement


Check if [itex]L(p)(x)=(1+4x)p(x)+(x-x^2)p'(x)-(x^2+x^3)p''(x)[/itex] is a linear transformation on [itex]\mathbb{R_2}[x][/itex]. If [itex]L(p)(x)[/itex] is a linear transformation, find it's matrix in standard basis and check if [itex]L(p)(x)[/itex] is invertible. If [itex]L(p)(x)[/itex] is invertible, find the function rule of it's inverse.

Homework Equations


-Linear transformations

The Attempt at a Solution



[itex]L(p)(x)[/itex] is a linear transformation if
[itex]L(z(x)+w(x))=L(z)(x)+L(w)(x) \forall z,w\in\mathbb{R_2}[x][/itex] and
[itex]L(\alpha z(x))=\alpha L(z)(x)\forall z\in \mathbb{R_2}[x],\forall\alpha\in\mathbb{R}[/itex]

Let [itex]p(x)=a+bx+cx^2\Rightarrow L(p)(x)=a+(4a+2b)x+(3b+c)x^2[/itex].
Let [itex]z(x)=p+qx+rx^2[/itex] and [itex]w(x)=\alpha+\beta x+\gamma x^2[/itex].

[itex]L(z(x))=p+(4p+2q)+(3q+r)x^2[/itex] and [itex]L(w(x))=\alpha+(4\alpha+2\beta)+(3\beta+\gamma)x^2[/itex] then [itex]L(z(x))+L(w(x))=(p+\alpha)+(4p+2q+4\alpha+2\beta)x+(3q+r+3\beta+\gamma)x^2[/itex]
[itex]L(z(x)+w(x))=L((p+\alpha)+(q+\beta)x+(r+\gamma)x^2)[/itex].
From this [itex]\Rightarrow L(z(x)+w(x))=L(z)(x)+L(w)(x)[/itex].

[itex]L(\alpha z(x))=L(\alpha p+\alpha qx+\alpha rx^2)=\alpha p+(4\alpha p+2\alpha q)x+(3\alpha q+\alpha r)x^2[/itex].
[itex]\alpha L(z)(x)=\alpha(p+(4p+2q)x+(3q+r)x^2)[/itex].
From this [itex]\Rightarrow L(\alpha z(x))=\alpha L(z)(x)[/itex] and [itex]L(p)(x)[/itex] is a linear transformation.

Standard basis for [itex]\mathbb{R_2}[x][/itex] is [itex]\mathcal{B}=\{1,x,x^2\}[/itex].
[itex][L]_{\mathcal{B}}=
\begin{bmatrix}
1 & 0 & 0 \\
4 & 2 & 0 \\
0 & 3 & 1 \\
\end{bmatrix}
[/itex].

Is the matrix [itex][L]_{\mathcal{B}}[/itex] correct?
If not, how to construct it?

[itex]L(p)(x)[/itex] is invertible because [itex]\det [L]_{\mathcal{B}}\neq 0[/itex].
[itex][L]^{-1}_{\mathcal{B}}=
\begin{bmatrix}
1 & 0 & 0 \\
-2 & 1/2 & 0 \\
6 & -3/2 & 1 \\
\end{bmatrix}
[/itex]

Function rule of [itex]L^{-1}(p)(x)[/itex] can be read from [itex][L]^{-1}_{\mathcal{B}}\Rightarrow L^{-1}(p)(x)=a+(-2a+\frac{1}{2}b)x+(6a-\frac{3}{2}b+c)x^2[/itex].
Is the function rule of [itex]L^{-1}(p)(x)[/itex] correct?
If not, how to construct it?

Also, how to check if [itex]L(p)(x)[/itex] is bijective?
 

Answers and Replies

  • #2
535
72
Once you proved linearity of ##L##, then it is bijective iff ##\det [L]_{\cal B} \neq 0 ##
You can check your answer by verifying that ##[L]_{\cal B}[L]_{L({\cal B})}^{-1} = I_3##.

If a linear transformation is bijective, then its inverse is linear. So when ##[L]_{L({\cal B})}^{-1}## is known, you know the coordinates on ##L({\cal B})## of ##L^{-1}##, and you can re-assemble your function
 
Last edited:
  • #3
208
1
Once you proved linearity of ##L##, then it is bijective iff ##\det [L]_{\cal B} \neq 0 ##
You can check your answer by verifying that ##[L]_{\cal B}[L]_{\cal B}^{-1} = I_3##.

If a linear transformation is bijective, then its inverse is linear. So when ##[L]_{\cal B}^{-1}## is known, you know the coordinates on ##{\cal B}## of ##L^{-1}##, and you can re-assemble your function

Are matrices of [itex]L[/itex] and [itex]L^{-1}[/itex] correct?
 
  • #4
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72
##L## is correct, and ##L^{-1}## is correct if ##LL^{-1} = I_3##
 
  • #5
208
1
##L## is correct, and ##L^{-1}## is correct if ##LL^{-1} = I_3##

Could you explain the method for finding the matrix of linear transformation that involves polynomials?
 
  • #6
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72
First of all, your results are correct, I didn't tell you anything because you could check the answer by yourself.

Could you explain the method for finding the matrix of linear transformation that involves polynomials?

Yes, you do exactly as usual. If you take the standard basis of ## \mathbb{R}_2[X]##, which is ##{\cal B} = (1,X,X^2)##, you are required to find a decomposition on ##{\cal B}## of ##L(1), L(X), L(X^2)## and assemble their coordinates column-wise. This is the matrix of your transformation
 
  • #7
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72
If your decomposition gives
## L(1) = a + b X + c X^2##, ## L(X) = d + e X + f X^2##, ## L(X^2) = g + h X + i X^2##,

##[L]_{\cal B} = \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \end{pmatrix}##
 
  • #8
208
1
If your decomposition gives
## L(1) = a + b X + c X^2##, ## L(X) = d + e X + f X^2##, ## L(X^2) = g + h X + i X^2##,

##[L]_{\cal B} = \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \end{pmatrix}##

I don't understand what are the polynomials of [itex]L(1),L(x),L(x^2)[/itex] if [itex]L(p)(x)=a+x(4a+2b)+(3b+c)x^2[/itex].
Could you explain how to evaluate [itex]L(1),L(x),L(x^2)[/itex]?
 
  • #9
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72
Is your transformation linear with respect to polynomial ##p## ?
 
  • #10
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Is your transformation linear with respect to polynomial ##p## ?

Yes, and the basis is standard for [itex]\mathbb{R_2}[x][/itex].
 
  • #11
535
72
Yes, and the basis is standard for [itex]\mathbb{R_2}[x][/itex].

Really ? Do we have ##L(\lambda p + q) = \lambda L(p) + L(q) ## ?
 
  • #12
535
72
You are going nowhere with the last transformation because it is not even linear. It doesn't make any sense to look for its matrix representation in a given basis.
If you want to exercise, take the original transformation in post #1, and tell us what is the matrix of ##L## in basis ##{\cal B} = ( 3, X-1, X^2 + 1) ##
 
  • #13
208
1
Really ? Do we have ##L(\lambda p + q) = \lambda L(p) + L(q) ## ?

You are going nowhere with the last transformation because it is not even linear. It doesn't make any sense to look for its matrix representation in a given basis.
If you want to exercise, take the original transformation in post #1, and tell us what is the matrix of ##L## in basis ##{\cal B} = ( 3, X-1, X^2 + 1) ##

In basis [itex]\mathcal{B}=\{3,x-1,x^2+1\}\Rightarrow L(3)=12x+3,L(x-1)=3x^2-2x-1,L(x^2+1)=x^2+4x+1[/itex].
[itex][L]_{\mathcal{B}}=
\begin{bmatrix}
0 & 3 & 1 \\
12 & -2 & 4 \\
3 & -1 & 1 \\
\end{bmatrix}
[/itex].

But for this new basis I get that [itex]\det \neq 0[/itex] and in standard basis (correction from first post), [itex]\det = 0[/itex].
What is wrong here?
 
  • #14
535
72
No this is wrong, you didn't decompose on ##(3,X-1,X^2 + 1) ## but on ##(1,X,X^2)##
 
  • #15
208
1
No this is wrong, you didn't decompose on ##(3,X-1,X^2 + 1) ## but on ##(1,X,X^2)##

Matrix in standard basis [itex]\mathcal{B}=\{1,x,x^2\}[/itex] is [itex]
\begin{bmatrix}
4 & 3 & 1 \\
1 & 2 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}
[/itex] and in [itex]\mathcal{B}=\{3,x-1,x^2+1\}[/itex] is
[itex]
\begin{bmatrix}
0 & 3 & 1 \\
12 & -2 & 4 \\
3 & -1 & 1 \\
\end{bmatrix}
[/itex].

What is wrong here?
 
  • #16
535
72
The matrix ##[L]_{\cal B} ## is the column wise assembly of the coordinates of ##L({\cal B}) = ( L(e_1), L(e_2), L(e_3) ) ## expressed in ##{ \cal B} ##

You should find :

##L(e_1) = a e_1 + b e_2 + c e_3 ##
##L(e_2 ) = d e_1 + e e_2 + f e_3##
##L(e_3) = g e_1 + h e_2 + i e_3 ##
 
  • #17
208
1
You should find :

##L(e_1) = a e_1 + b e_2 + c e_3 ##
##L(e_2 ) = d e_1 + e e_2 + f e_3##
##L(e_3) = g e_1 + h e_2 + i e_3 ##

So in standard basis the matrix is [itex]
\begin{bmatrix}
1 & 0 & 0 \\
4 & 2 & 0 \\
0 & 3 & 1 \\
\end{bmatrix}
[/itex] and in the new basis is [itex]
\begin{bmatrix}
5 & -2 & 4/3 \\
12 & -2 & 4 \\
0 & 3 & 1 \\
\end{bmatrix}
[/itex].
 
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