Question

If T is an n-thorder linear differential operator and$\mathit{\lambda }$ is an arbitrary scalar, is$\mathit{\lambda }$ necessarily an eigenvalue of T? If so, what is the dimension of the eigenspace associated with $\mathit{\lambda }$?

Short answer

The dimension of eigenspace associated with$\lambda$ is n .

Step-by-step solution

Determine the function.

Consider the linear differential equation $Tx=\lambda x$where T is an n-th order linear differential equation.

Simplify the equation$Tx=\lambda x$ as follows.

$$\begin{gathered} Tx=\lambda x \\ Tx-\lambda x=0 \\ \left\{T-\lambda l\right\}x=0 \end{gathered}$$

By the definition of eigenvalue,$\lambda$ is an eigenvalue of linear differential equation T .

Determine the dimension of Eλ .

Consider the liner transformation T then nullity T of is defined as$ker\left(T\right)=\left\{x\in \mathrm{ℝ}:T\left(x\right)=0\right\}$ .

By the definition of kernel, the dimension of$E_{\lambda }$ is n because$E_{\lambda }$ is a kernel of the function $\left\{T-\lambda l\right\}$ and T is an n-th order linear differential equation.

Hence, the dimension of$\left\{T-\lambda l\right\}$ is n and$\lambda$ is an eigenvalue of T .