3
$\begingroup$

Can anyone please explain me how to solve it?

Find the nonzero eigenvalues and the corresponding eigenvectors:

$T:[-1,1]\rightarrow[-1,1]$

$$T((f(x))=\int_{-1}^1(x^2 y + y^2 x) f(y) \, dy$$

$\endgroup$
  • 5
    $\begingroup$ Why did you mess up the MathJax I just fixed? $\endgroup$ – Moo Jul 23 '18 at 17:04
  • $\begingroup$ Sorry, i'm new on this website and i don't know how to properly use it. I wanted to add a term which i forgot to put the first time between the parantheses. $\endgroup$ – A.Sheriff Jul 23 '18 at 17:08
  • $\begingroup$ Is it now right like this? $\endgroup$ – mrtaurho Jul 23 '18 at 17:09
  • $\begingroup$ yes, but the MathJax probably isn't right $\endgroup$ – A.Sheriff Jul 23 '18 at 17:12
  • $\begingroup$ Do you know how MathJax works or respectively LaTeX or could you at least explain what is not right now? $\endgroup$ – mrtaurho Jul 23 '18 at 17:16
1
$\begingroup$

The eigenvalues are $$\pm\sqrt{\frac{4}{15}}$$ with corresponding eigenvectors $$f\left(x\right)=\sqrt{5}x^{2}\pm\sqrt{3}x$$ We get this by noting that $Tf\left(x\right)$ is always a quadratic, so eigenvectors must be quadratics. We insert $f\left(x\right)$ in this form and solve the algebraic equations.

$\endgroup$
  • 1
    $\begingroup$ Can you explain me how to solve it step by step? I have no idea how to solve this type of problems since we didn't use integrals in the linear algebra classes, but the teacher got mad at us and asked this question at the exam. $\endgroup$ – A.Sheriff Jul 23 '18 at 17:30
0
$\begingroup$

Since $$Tf\left(x\right)=x^2\int_{-1}^{1} yf\left(y\right)dy+x\int_{-1}^{1} y^2f\left(y\right)dy$$ we have that $Tf\left(x\right)=ax^2+bx$ for some $a,b$. Hence, if $f\left(x\right)$ is an eigenvector it must take this form. Plugging this in we get $$Tf\left(x\right)=x^2\int_{-1}^{1} y\left(ay^2+by\right)dy+x\int_{-1}^{1} y^2\left(ay^2+by\right)dy$$ $$=x^2\int_{-1}^{1} \left(ay^3+by^2\right)dy+x\int_{-1}^{1} \left(ay^4+by^3\right)dy$$ $$=bx^2\int_{-1}^{1} \left(y^2\right)dy+ax\int_{-1}^{1} \left(y^4\right)dy$$ $$=\frac {2} {3} bx^2+\frac {2} {5} ax$$ and if this is an eigenvector with eigenvalue $\lambda$ then we find $\lambda a=\frac {2} {3} b$ and $\lambda b=\frac {2} {5} a$. Solving this for $\lambda$ and $a,b$ we find the eigenvalues and the eigenvectors in my other post.

$\endgroup$
0
$\begingroup$

I'm assuming that $T$ is acting on the space of continuous functions on $[-1,1]$, i.e. $T : C[-1,1] \to C[-1,1]$.

The eigenvalues of $T$ are $0$ and $\pm\sqrt{\frac4{15}}$.

Assume that $Tf = \lambda f$ for some $\lambda \in \mathbb{R}$.

We have

$$\lambda f(x) = (Tf)(x) = \int_{-1}^1 (x^2y + y^2x)f(y)\,dy$$

so $f$ is differentiable and we have

$$\lambda f'(x) = (Tf)(x) = \int_{-1}^1 (2xy + y^2)f(y)\,dy$$

so $f'$ is differentiable and we have

$$\lambda f''(x) = (Tf)(x) = \int_{-1}^1 2yf(y)\,dy = \text{const.}$$

If $\lambda \ne 0$, then $f'' = \text{const.}$ so $f$ is a quadratic polynomial $f(x) = Ax^2 + Bx + C$.

We get $$A = \frac12f''(x) = \frac1\lambda \int_{-1}^1 yf(y)\,dy = \frac1\lambda \frac{2B}3$$ $$B = f'(0) = \frac1\lambda \int_{-1}^1 y^2f(y)\,dy = \frac1\lambda\left(\frac{2A}5 + \frac{2C}3\right)$$ $$C = f(0) = 0$$

so if we fix $A = 1$ we get $\lambda = \frac43 B$ and $\lambda B = \frac25$ so $$\lambda = \pm \sqrt{\frac{4}{15}}$$

with eigenvectors $f(x) = x^2 \pm \frac12\sqrt{\frac35}$.

If $\lambda = 0$ then we get the condition

$$0 = x^2 \int_{-1}^1 yf(y) \,dy + x \int_{-1}^1 y^2f(y)\,dy$$

so $\int_{-1}^1 yf(y) \,dy = \int_{-1}^1 y^2f(y)\,dy = 0$.

This is indeed an eigenvalue with an eigenvector e.g. $$f(x) = x^3-\frac53 x^2 - \frac35 x + 1$$

$\endgroup$
0
$\begingroup$

$$Tf(x)=x^2\int_{-1}^{1}yf(y)dy+x\int_{-1}^{1}y^2f(y)dy$$then any eigenfunction is of form$$f(x)=ax^2+bx$$by substitution we obtain$$Tf(x)=\dfrac{2}{3}bx^2+\dfrac{2}{5}ax=\lambda ax^2+\lambda bx$$which means that $$\dfrac{2}{3}b=\lambda a\\\dfrac{2}{5}a=\lambda b$$one trivial eigenfunction is $f(x)=0$ the other ones exist when $$\dfrac{4}{15}=\lambda^2\\\lambda=\pm\dfrac{2}{\sqrt 15}$$therefore $$f(x)=ax^2\pm\sqrt{\dfrac{3}{5}}ax$$ corresponding to $\lambda=\pm\dfrac{2}{\sqrt {15}}$ respectively

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.