Linear algebra: Find the matrix of linear transformation

In summary: First, you use the standard basis, so the matrix is the one you found. Then you are told to use a new basis, so you should start from the beginning, that is, find the decomposition of ##L(1),L(X),L(X^2)## on this new basis. You don't do that, and you are not even close to be rightFirst, you use the standard basis, so the matrix is the one you found. Then you are told to use a new basis, so you should start from the beginning, that is, find the decomposition of ##L(1
  • #1
gruba
206
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?
 
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  • #2
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
geoffrey159 said:
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
##L## is correct, and ##L^{-1}## is correct if ##LL^{-1} = I_3##
 
  • #5
geoffrey159 said:
##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
First of all, your results are correct, I didn't tell you anything because you could check the answer by yourself.

gruba said:
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
 
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  • #7
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}##
 
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  • #8
geoffrey159 said:
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
Is your transformation linear with respect to polynomial ##p## ?
 
  • #10
geoffrey159 said:
Is your transformation linear with respect to polynomial ##p## ?

Yes, and the basis is standard for [itex]\mathbb{R_2}[x][/itex].
 
  • #11
gruba said:
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
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
geoffrey159 said:
Really ? Do we have ##L(\lambda p + q) = \lambda L(p) + L(q) ## ?

geoffrey159 said:
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
No this is wrong, you didn't decompose on ##(3,X-1,X^2 + 1) ## but on ##(1,X,X^2)##
 
  • #15
geoffrey159 said:
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
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 ##
 
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  • #17
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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What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations and their representations in vector spaces. It involves the use of matrices and vectors to solve problems related to geometry, physics, engineering, and many other fields.

What is a matrix?

A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. It is used to represent linear transformations, such as rotations, reflections, and translations, and to solve systems of linear equations.

What is a linear transformation?

A linear transformation is a function that maps one vector space to another in such a way that it preserves the algebraic structure of the space. It is a fundamental concept in linear algebra and is used to describe geometric transformations, such as rotations and reflections.

How do you find the matrix of a linear transformation?

To find the matrix of a linear transformation, you need to first choose a basis for the domain and the codomain of the transformation. Then, you can represent the transformation as a matrix by writing the images of the basis vectors in terms of the basis vectors of the codomain.

What are the applications of linear algebra?

Linear algebra has many practical applications in fields such as computer graphics, data analysis, machine learning, and quantum mechanics. It is used to solve problems involving systems of linear equations, eigenvalues and eigenvectors, and matrix operations.

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