1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sum of eigenvectors of linear transformation

  1. Feb 22, 2016 #1
    1. The problem statement, all variables and given/known data
    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].

    2. Relevant equations
    -Eigenvalues and eigenvectors

    3. 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]?
     
  2. jcsd
  3. Feb 22, 2016 #2

    fresh_42

    Staff: Mentor

    ##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.)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted