# Steps to find Kernel of $T - \lambda I$ where $T$ is a linear operator?

Let $$K(t,s)= st$$ for $$s, t \in [0,1]$$. Define the operator $$T$$ on $$C[0,1]$$ by

$$(Tf)(t) = \int_0^1 K(t,s)f(s)ds \,\,\,\,\,(\forall \,f \in C[0,1]).$$

a) For any $$\lambda$$, describe the Null space(= Kernel) of $$T - \lambda I$$.
b) Find all values of $$\lambda$$ such that the equation $$(T-\lambda I)f=g$$ has a unique solution for every $$g \in C[0,1]$$.

Also, can you explain what is the interpretation of $$g$$'s that satisfy the above equality?

• Note that $(Tf)(t) = \left(\int_0^1 sf(s) ds\right) t$ is just a linear function in $t$. – Arctic Char Dec 15 '19 at 22:42
• $T$ is a compact self adjoint operator and its only non-zero eigen value is $\frac 1 3$. Hence $\frac 1 3$ is the only possible value of $\lambda$. The kernel is $\{f: \int_0^{1} tf(t)dt=0\}$. – Kavi Rama Murthy Dec 15 '19 at 23:58
• @Kabo Murphy: Can you please show me how you would write $T- \lambda I$? – Saeed Dec 16 '19 at 3:19

If $$(T-\lambda I)f=0$$, you have $$\tag1 \lambda f(t)=\int_0^1 st f(s)\,ds=t\int_0^1 sf(s)\,ds.$$ So, if $$\lambda=0$$ then $$\ker T$$ consists of all those $$f$$ with $$\int_0^1 sf(s)\,ds=0$$. When $$\lambda\ne0$$: if $$(1)$$ holds for $$\lambda\ne0$$, we have $$\tag2 f(t)=\phi_\lambda(f)\,t,$$ where $$\tag3\phi_\lambda(f)=\frac1\lambda\,\int_0^1 sf(s)\,ds.$$ Then $$\tag4 \phi_\lambda(f)=\frac1\lambda\,\int_0^1 sf(s)\,ds=\frac1\lambda\int_0^1 s\,\phi_\lambda(f)\,s\,ds=\frac1{3\lambda}\,\phi_\lambda(f).$$ For $$f$$ to be an eigenvector we need $$f\ne0$$, which forces $$\phi_\lambda(f)\ne0$$. Then $$3\lambda=1$$, and so $$\lambda=\tfrac13$$ is the only possible nonzero eigenvalue. In summary, $$\ker(T-\lambda I)=\{0\}$$ when $$\lambda\not\in\{0,\tfrac13\}$$, and $$\ker(T-\tfrac13\,I)=\{\alpha g:\ \alpha\in\mathbb R\}$$ where $$g(t)=t$$.
For the second part, if the equation $$(T-\lambda I)f=g$$ has a unique solution for all $$g\in C[0,1]$$, this means that $$T-\lambda I$$ is invertible. Because $$T$$ is compact (or, instead of using a more general result, we could just notice that $$T$$ is rank-one), its spectrum consists only of eigenvalues, and so the equation $$(T-\lambda I)f=g$$ has a unique solution for all $$g$$ precisely when $$\lambda\not\in\{0,\tfrac13\}$$.