Sum of eigenvectors of linear transformation

[gruba]

Homework Statement

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

Homework Equations

-Eigenvalues and eigenvectors

The Attempt at a Solution

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

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

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

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

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

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

Corresponding eigenvectors are $v_1=\begin{bmatrix}<br /> 1 \\<br /> 1 \\<br /> a \\<br /> \end{bmatrix},v_2=\begin{bmatrix}<br /> 2 \\<br /> 1 \\<br /> 0 \\<br /> \end{bmatrix},v_3=\begin{bmatrix}<br /> 1 \\<br /> 1 \\<br /> 0 \\<br /> \end{bmatrix}$.

What does it mean that the space $V$ is the sum of eigenvectors of $L$?

 

[fresh_42]

Homework Statement

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

Homework Equations

-Eigenvalues and eigenvectors

The Attempt at a Solution

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

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

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

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

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

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

Corresponding eigenvectors are $v_1=\begin{bmatrix}<br /> 1 \\<br /> 1 \\<br /> a \\<br /> \end{bmatrix},v_2=\begin{bmatrix}<br /> 2 \\<br /> 1 \\<br /> 0 \\<br /> \end{bmatrix},v_3=\begin{bmatrix}<br /> 1 \\<br /> 1 \\<br /> 0 \\<br /> \end{bmatrix}$.

What does it mean that the space $V$ is the sum of eigenvectors of $L$?

$V$ 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.)