Find eigenvalues & eigenvectors for an integral. Can anyone please explain me how to solve it?
Find the nonzero eigenvalues and the corresponding eigenvectors:
$T:[-1,1]\rightarrow[-1,1]$
$$T((f(x))=\int_{-1}^1(x^2 y + y^2 x) f(y) \, dy$$
 A: 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. 
A: 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. 
A: 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. $$f(x) = x^3-\frac53 x^2 - \frac35 x + 1$$
A: $$Tf(x)=x^2\int_{-1}^{1}yf(y)dy+x\int_{-1}^{1}y^2f(y)dy$$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
