Matrix of linear transformation

[gruba]

Homework Statement

Let A:\mathbb R_2[x]\rightarrow \mathbb R_2[x] is a linear transformation defined as (A(p))(x)=p&#039;(x+1) where \mathbb R_2[x] is the space of polynomials of the second order. Find all a,b,c\in\mathbb R such that the matrix \begin{bmatrix}<br /> a &amp; 1 &amp; 0 \\<br /> b &amp; 0 &amp; 1 \\<br /> c &amp; 0 &amp; 0 \\<br /> \end{bmatrix} is the matrix of linear transformation A with respect to some arbitrary basis of \mathbb R_2[x].

Homework Equations

-Polynomial vector space
-Basis

The Attempt at a Solution

If we choose the standard basis, \mathcal B=\{1,x,x^2\}\Rightarrow p(x)=\alpha+\beta x+\gamma x^2,\alpha,\beta,\gamma\in\mathbb R\Rightarrow (A(p))(x)=\beta+(\beta+2\gamma)x+2\gamma x^2\Rightarrow

A(1)=0x^2+0x+1,A(x)=0x^2+1x+1,A(x^2)=2x^2+0x+0

Setting A(1),A(x),A(x^2) as column vectors gives the matrix \begin{bmatrix}<br /> 0 &amp; 0 &amp; 2 \\<br /> 0 &amp; 1 &amp; 0 \\<br /> 1 &amp; 1 &amp; 0 \\<br /> \end{bmatrix} that is not of the form of given matrix \begin{bmatrix}<br /> a &amp; 1 &amp; 0 \\<br /> b &amp; 0 &amp; 1 \\<br /> c &amp; 0 &amp; 0 \\<br /> \end{bmatrix}.

This means that we can't choose the standard basis to get matrix of A that will be of the form \begin{bmatrix}<br /> a &amp; 1 &amp; 0 \\<br /> b &amp; 0 &amp; 1 \\<br /> c &amp; 0 &amp; 0 \\<br /> \end{bmatrix}.

Question: Do we have to guess a proper basis? If not, then how to find one?

 

[micromass]

What is the relation between the matrix of a linear map in one basis and the matrix of the same linear map in another basis?

 

[micromass]

Alternatively, do you know a property of linear maps which is independent of the basis?

 

[gruba]

What is the relation between the matrix of a linear map in one basis and the matrix of the same linear map in another basis?

Let B=\{b_1,b_2,b_3\} is a standard basis, and B&#039;=\{{b&#039;}_1,{b&#039;}_2,{b&#039;}_3\}.
Matrix of changing basis from B to B&#039; is defined as S=[[b_1]_{B&#039;},...,[{b&#039;}_n]_{B&#039;}].

Finding [b_1]_{B&#039;},...,[{b&#039;}_n]_{B&#039;} gives matrix S.

This should be the reversed process, right? We know vectors [b_1]_{B&#039;},...,[{b&#039;}_n]_{B&#039;} in some basis,
and we need to find b_1,...,b_n from basis B.

 

[micromass]

Right, but that is not my point. Let's say I give you two matrices $A$ and $B$ which are matrices of the same linear map but with different bases. Do you know anything about how $A$ and $B$ are related? Does similarity tell you something?

 

[gruba]

Right, but that is not my point. Let's say I give you two matrices $A$ and $B$ which are matrices of the same linear map but with different bases. Do you know anything about how $A$ and $B$ are related? Does similarity tell you something?

If B is the matrix after transition to new basis, then B=A^{-1}?

 

[gruba]

@micromass The only condition is c\neq 0 since from b_1=1,b_2=x,b_3=x^2 follows {b&#039;}_1=x,{b&#039;}_2=x^2,{b&#039;}_3=\frac{1-ax-bx^2}{c}.
Is this correct?

 

[gruba]

ATTEMPT EDITED:

Let p(x)=a+bx+cx^2 be a polynomial in standard basis \mathcal B=\{1,x,x^2\} of \mathbb R^2[x].
Then, linear transformation A is defined as

(A(p))(x)=p&#039;(x+1)=(2+b)+2cx+0x^2

From the given matrix, \begin{bmatrix}<br /> a &amp; 1 &amp; 0 \\<br /> b &amp; 0 &amp; 1 \\<br /> c &amp; 0 &amp; 0 \\<br /> \end{bmatrix}

we have

[b_1]_{B&#039;}= \begin{bmatrix}<br /> a \\<br /> b \\<br /> c \\<br /> \end{bmatrix},[b_2]_{B&#039;}= \begin{bmatrix}<br /> 1 \\<br /> 0 \\<br /> 0 \\<br /> \end{bmatrix},[b_3]_{B&#039;}= \begin{bmatrix}<br /> 0 \\<br /> 1 \\<br /> 0 \\<br /> \end{bmatrix}

where \mathcal {B&#039;} is some basis different from standard basis \mathcal B.

Now, we have

b_1=a\cdot {b&#039;}_1+b\cdot {b&#039;}_2+c\cdot {b&#039;}_3

b_2=1\cdot {b&#039;}_1+0\cdot {b&#039;}_2+0\cdot {b&#039;}_3

b_3=0\cdot {b&#039;}_1+1\cdot {b&#039;}_2+0\cdot {b&#039;}_3

where b_1,b_2,b_3 form standard basis \mathcal B, and
{b&#039;}_1,{b&#039;}_2,{b&#039;}_3 form new basis \mathcal{B&#039;}.

From above equations,

{b&#039;}_1=x

{b&#039;}_2=x^2

{b&#039;}_3=\frac{1-ax-bx^2}{c},c\neq 0

and we have \mathcal{B&#039;}=\left\{x,x^2,\frac{1-ax-bx^2}{c}\right\}.

Now we need to find the matrix of linear transformation A with respect to basis \mathcal{B&#039;}.

\begin{bmatrix}<br /> a &amp; 1 &amp; 0\\<br /> b &amp; 0 &amp; 1\\<br /> c &amp; 0 &amp; 0\\<br /> \end{bmatrix} \begin{bmatrix}<br /> 2+b \\<br /> 2c \\<br /> 0 \\<br /> \end{bmatrix} =\begin{bmatrix}<br /> 2a+ab+2c \\<br /> 2b+b^2 \\<br /> 2c+bc \\<br /> \end{bmatrix}

Linear transformation A in basis \mathcal B&#039; is given by

((A(p))(x))&#039;=(2a+ab+2c)x+b(2+b)x^2+c(2+b)\frac{1-ax-bx^2}{c}

=(2a+ab+2c)x+b(2+b)x^2+(2+b)(1-ax-bx^2)

Conclusion: For all a,b and for all c\neq 0, the matrix given by \begin{bmatrix}<br /> a &amp; 1 &amp; 0 \\<br /> b &amp; 0 &amp; 1 \\<br /> c &amp; 0 &amp; 0 \\<br /> \end{bmatrix} is the matrix of linear transformation A in some basis \mathcal B&#039;.

Is this correct?

 

[Mark44]

@gruba, this is a pretty strong hint:

Do you know anything about how AAA and BBB are related? Does similarity tell you something?

 

[gruba]

@gruba, this is a pretty strong hint:

So, is it correct that one is the inverse of the other?

 

[Mark44]

So, is it correct that one is the inverse of the other?

No.