# Linear algebra: Find the matrix of linear transformation

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1. Feb 13, 2016

### gruba

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

2. Relevant equations
-Linear transformations

3. The attempt at a solution

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

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

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

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

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

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

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

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

Also, how to check if $L(p)(x)$ is bijective?

2. Feb 13, 2016

### geoffrey159

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: Feb 13, 2016
3. Feb 13, 2016

### gruba

Are matrices of $L$ and $L^{-1}$ correct?

4. Feb 13, 2016

### geoffrey159

$L$ is correct, and $L^{-1}$ is correct if $LL^{-1} = I_3$

5. Feb 13, 2016

### gruba

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

6. Feb 13, 2016

### geoffrey159

First of all, your results are correct, I didn't tell you anything because you could check the answer by yourself.

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. Feb 13, 2016

### geoffrey159

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. Feb 13, 2016

### gruba

I don't understand what are the polynomials of $L(1),L(x),L(x^2)$ if $L(p)(x)=a+x(4a+2b)+(3b+c)x^2$.
Could you explain how to evaluate $L(1),L(x),L(x^2)$?

9. Feb 13, 2016

### geoffrey159

Is your transformation linear with respect to polynomial $p$ ?

10. Feb 13, 2016

### gruba

Yes, and the basis is standard for $\mathbb{R_2}[x]$.

11. Feb 13, 2016

### geoffrey159

Really ? Do we have $L(\lambda p + q) = \lambda L(p) + L(q)$ ?

12. Feb 13, 2016

### geoffrey159

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. Feb 13, 2016

### gruba

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

But for this new basis I get that $\det \neq 0$ and in standard basis (correction from first post), $\det = 0$.
What is wrong here?

14. Feb 13, 2016

### geoffrey159

No this is wrong, you didn't decompose on $(3,X-1,X^2 + 1)$ but on $(1,X,X^2)$

15. Feb 13, 2016

### gruba

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

What is wrong here?

16. Feb 13, 2016

### geoffrey159

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. Feb 13, 2016

### gruba

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

Last edited: Feb 13, 2016