Vector space $C_0$ consists of every continuous function in [0,1].
Tensor $A$ on the space $C_0$ is defined as $Af=\int_0^1$$(1-3t\varphi)f(\varphi)d\varphi$
where $f(t)$ is a vector on space $C_0$.
Tensor $A$ has three eigenvalues $\lambda_1 =0$, $\lambda_2 = 1/2$, $\lambda_3 =-1/2$
and the corresponding eigenvector $z_1 (t)$ satisfies $\int_0^1 f(\varphi)d\varphi=\int_0^1 \varphi f(\varphi)d\varphi=0$
How can you show that $z_2 (t)=1-t$,$z_3 (t)=1-3t$ ?
[It says to get integral equation from $Af=\lambda f$ and to divide into cases-when $\lambda=0 $ and when $\lambda\neq 0$. And when $\lambda\neq0$,suppose that $z(t)$ is a linear function of $t$ to get the eigenvector]
Any help would be much appreciated!!
For $z_2$: $$Az_2(t) = \int_0^1 (1-3t\varphi)z_2(\varphi) \, d\varphi = \int_0^1 (1-\varphi-3t\varphi+ 3 t \varphi^2)\, d \varphi \\ =1 -\frac{1}{2} - \frac{3}{2}t + t = \frac{1}{2}(1 - t) = \frac{1}{2}z_2(t)$$ so $Az_2 = \lambda_2 z_2$.
Now, can you finish for the $z_3$ case?