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Chapter 9: Q41E (page 442)

If T is an n-thorder linear differential operator and位 is an arbitrary scalar, is位 necessarily an eigenvalue of T? If so, what is the dimension of the eigenspace associated with 位?

Short Answer

Expert verified

The dimension of eigenspace associated with位 is n .

Step by step solution

01

Determine the function.

Consider the linear differential equation Tx=位xwhere T is an n-th order linear differential equation.

Simplify the equationTx=位x as follows.

Tx=位xTx-位x=0T-位lx=0

By the definition of eigenvalue,位 is an eigenvalue of linear differential equation T .

02

Determine the dimension of Eλ .

Consider the liner transformation T then nullity T of is defined askerT=x鈭�鈩�:Tx=0 .

By the definition of kernel, the dimension ofE位 is n becauseE位 is a kernel of the function T-位l and T is an n-th order linear differential equation.

Hence, the dimension ofT-位l is n and位 is an eigenvalue of T .

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