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Consider the space of functions $F=\operatorname{span}\lbrace \sin(x),\cos(x),x\cdot \sin(x), x\cdot \cos(x)\rbrace$ and the image $L:F\rightarrow F:f\rightarrow -f''-2f$.

  1. Show that $L$ is a linear image.
  2. Find the matrix $M$ of the linear images from $F$ to $F$ with respect to basis $B=\lbrace \sin(x),\cos(x),x\cdot \sin(x), x\cdot \cos(x)\rbrace$.
  3. What are the eigenvalues of $L$ (and also $M$).
  4. Give the eigenvectors of $M$ and $L$.
  5. What is the dimension of the eigenspace?

my attempt:

  1. The image is linear because the image of $0$ is $0$?

  2. $$\left(\begin{matrix}-\sin(x)\\ -\cos(x)\\ -x\cdot \sin(x)-2\cdot \cos(x)\\ 2\cdot \sin(x)-x\cdot \cos(x)\end{matrix}\right)=\sin(x)\cdot\left(\begin{matrix}-1\\ 0\\0\\2\end{matrix}\right)+\cos(x)\cdot\left(\begin{matrix}0\\ -1\\-2\\0\end{matrix}\right)+x\cdot \sin(x)\cdot\left(\begin{matrix}0\\ 0\\-1\\0\end{matrix}\right) +x \cdot \cos(x)\cdot\left(\begin{matrix}0\\ 0\\0\\-1\end{matrix}\right)$$ So $M=\left(\begin{matrix}-1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & -2 & -1 & 0\\ 2 & 0 & 0 & -1\end{matrix}\right)$

  3. eigenvalues: $-1, -1, -1, -1$

  4. eigenvectors: $\left(\begin{matrix}0\\ 0\\0\\1\end{matrix}\right)$,$\left(\begin{matrix}0\\ 0\\0\\-1\end{matrix}\right)$,$\left(\begin{matrix}0\\ 0\\1\\0\end{matrix}\right)$,$\left(\begin{matrix}0\\ 0\\1\\0\end{matrix}\right)$

  5. dimension eigenspace: $4$?

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  • 1
am87gu
  • 101
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    Some of what you have done is clearly wrong. A list of eigenvectors ought to be a linearly independent set, but your first and second eigenvectors are not linearly independent, neither are your third and fourth. Also, if those are the only eigenvectors, then (because of the linear dependence) the dimension of the eigenspace is 2, not 4. – Gerry Myerson Aug 22 '13 at 12:44
  • matlab gave me this: A = 
    -1     0     0     0
 0    -1     0     0
 0    -2    -1     0
 2     0     0    -1

    [V,D] = eig(A)

    V =

         0    0.0000         0         0
     0         0         0    0.0000
     0         0    1.0000    1.0000
1.0000   -1.0000         0         0

    D =

    -1     0     0     0
 0    -1     0     0
 0     0    -1     0
 0     0     0    -1
    – am87gu Aug 22 '13 at 12:48
  • Matlab? I thought you said this was your attempt. So, what is your real question? Do you understand my comments? – Gerry Myerson Aug 22 '13 at 12:51
  • Yes, just entered the M matrix in matlab to see the eigenvalues and eigenvectors. I understand the eigenvectors i wrote are not linearly independent. – am87gu Aug 22 '13 at 12:58
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    $L(0) = 0$ would tell you that $L$ is injective IF you already knew that $L$ was linear. To show that $L$ is linear, you want to show that $L(f+g) = L(f) + L(g)$. – Greebo Aug 22 '13 at 13:01
  • As you understand they are not linearly independent, do you understand the dimension is not 4? – Gerry Myerson Aug 22 '13 at 13:04
  • The eigenvalues are -1,-1,-1,-1 because they are the diagonal entries. I find two eigenvectors (0,0,0,1)^T and (0,0,1,0)^T. Is that correct? – am87gu Aug 22 '13 at 13:05
  • So the eigenspace consists of all the eigenvectors corresponding to the eigenvectors, which is 2? – am87gu Aug 22 '13 at 13:12
  • I haven't checked to see whether $M$ is right, but, if it is, then your eigenvectors are now correct. The eigenspace is not 2; the eigenspace is a vector space, whereas 2 is a number. But the dimension of the eigenspace is 2. – Gerry Myerson Aug 23 '13 at 10:41

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