Sum of eigenvectors of linear transformation

[gruba]

Homework Statement

Find all values [itex]a\in\mathbb{R}[/itex] such that vector space [itex]V=P_2(x)[/itex] is the sum of eigenvectors of linear transformation [itex]L: V\rightarrow V[/itex] defined as [itex]L(u)(x)=(4+x)u(0)+(x-2)u'(x)+(1+3x+ax^2)u''(x)[/itex]. [itex]P_2(x)[/itex] is the space of polynomials of order [itex]2[/itex].

Homework Equations

-Eigenvalues and eigenvectors

The Attempt at a Solution

First, we find the matrix of [itex]L[/itex] (choose a standard basis [itex]\{1,x,x^2\}[/itex]).

[tex]L(1)\Rightarrow L(u)(x)=0x^2+1x+4\Rightarrow [L(1)]= \begin{bmatrix}
4 \\
1 \\
0 \\
\end{bmatrix}[/tex]

[tex]L(x)=0x^2+1x-2\Rightarrow [L(x)]= \begin{bmatrix}
-2 \\
1 \\
0 \\
\end{bmatrix}[/tex]

[tex]L(x^2)=(2+2a)x^2+2x+2\Rightarrow [L(1)]= \begin{bmatrix}
2 \\
2 \\
2+2a \\
\end{bmatrix}\Rightarrow [L]_{\mathcal{B}}= \begin{bmatrix}
4 & -2 & 2 \\
1 & 1 & 2 \\
0 & 0 & 2+2a \\
\end{bmatrix}[/tex]

Next, we find eigenvalues and eigenvectors of [itex][L]_{\mathcal{B}}[/itex]:

[tex]\det([L]_{\mathcal{B}}-\lambda I)=(2+2a-\lambda)(\lambda-3)(\lambda-2)\Rightarrow[/tex] eigenvalues are [itex]\lambda_1=2+2a,\lambda_2=3,\lambda_3=2,a\neq 0,a\neq \frac{1}{2}[/itex].

Corresponding eigenvectors are [itex]v_1=\begin{bmatrix}
1 \\
1 \\
a \\
\end{bmatrix},v_2=\begin{bmatrix}
2 \\
1 \\
0 \\
\end{bmatrix},v_3=\begin{bmatrix}
1 \\
1 \\
0 \\
\end{bmatrix}[/itex].

What does it mean that the space [itex]V[/itex] is the sum of eigenvectors of [itex]L[/itex]?

 

[fresh_42]

Homework Statement

Find all values [itex]a\in\mathbb{R}[/itex] such that vector space [itex]V=P_2(x)[/itex] is the sum of eigenvectors of linear transformation [itex]L: V\rightarrow V[/itex] defined as [itex]L(u)(x)=(4+x)u(0)+(x-2)u'(x)+(1+3x+ax^2)u''(x)[/itex]. [itex]P_2(x)[/itex] is the space of polynomials of order [itex]2[/itex].

Homework Equations

-Eigenvalues and eigenvectors

The Attempt at a Solution

First, we find the matrix of [itex]L[/itex] (choose a standard basis [itex]\{1,x,x^2\}[/itex]).

[tex]L(1)\Rightarrow L(u)(x)=0x^2+1x+4\Rightarrow [L(1)]= \begin{bmatrix}
4 \\
1 \\
0 \\
\end{bmatrix}[/tex]

[tex]L(x)=0x^2+1x-2\Rightarrow [L(x)]= \begin{bmatrix}
-2 \\
1 \\
0 \\
\end{bmatrix}[/tex]

[tex]L(x^2)=(2+2a)x^2+2x+2\Rightarrow [L(1)]= \begin{bmatrix}
2 \\
2 \\
2+2a \\
\end{bmatrix}\Rightarrow [L]_{\mathcal{B}}= \begin{bmatrix}
4 & -2 & 2 \\
1 & 1 & 2 \\
0 & 0 & 2+2a \\
\end{bmatrix}[/tex]

Next, we find eigenvalues and eigenvectors of [itex][L]_{\mathcal{B}}[/itex]:

[tex]\det([L]_{\mathcal{B}}-\lambda I)=(2+2a-\lambda)(\lambda-3)(\lambda-2)\Rightarrow[/tex] eigenvalues are [itex]\lambda_1=2+2a,\lambda_2=3,\lambda_3=2,a\neq 0,a\neq \frac{1}{2}[/itex].

Corresponding eigenvectors are [itex]v_1=\begin{bmatrix}
1 \\
1 \\
a \\
\end{bmatrix},v_2=\begin{bmatrix}
2 \\
1 \\
0 \\
\end{bmatrix},v_3=\begin{bmatrix}
1 \\
1 \\
0 \\
\end{bmatrix}[/itex].

What does it mean that the space [itex]V[/itex] is the sum of eigenvectors of [itex]L[/itex]?

##V## is the span or linear hull of eigenvectors of ##L##. It means the eigenvectors of ##L## build a basis of the three dimensional vector space ##V=P_2(x)##. (Assuming your calculations are correct.)