Eigenvalues & Eigenvectors

Algebra Level 3

True or false :

Let A : V → V A:V\rightarrow V and B : V → V B:V\rightarrow V be linear operators on a vector space V V over a field F F .

If v ∈ V \mathbf{v}\in V is an eigenvector of A A corresponding to λ ∈ F \lambda\in F , then v \mathbf{v} is an eigenvector of B A − A − λ B BA-A-\lambda B corresponding to − λ ∈ F -\lambda\in F , for any B B .

False True

Stephen Pablo by Stephen Pablo , 27, USA

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1 solution

Stephen Pablo
May 22, 2016

This is a true statement.

Since v \mathbf{v} is an eigenvector of A A corresponding to λ \lambda , then we know that A v = λ v A\mathbf{v}=\lambda\mathbf{v} .

We want to show that ( B A − A − λ B ) v = − λ v (BA-A-\lambda B)\mathbf{v}=-\lambda\mathbf{v} .

Observe:

( B A − A − λ B ) v = B A v − A v − λ B v = B ( A v ) − λ v − λ B v (BA-A-\lambda B)\mathbf{v}=BA\mathbf{v}-A\mathbf{v}-\lambda B\mathbf{v}\ = B(A\mathbf{v})-\lambda\mathbf{v}-\lambda B\mathbf{v}

= B ( λ v ) − λ v − λ B v = λ B v − λ v − λ B v = − λ v . = B(\lambda\mathbf{v})-\lambda\mathbf{v}-\lambda B\mathbf{v} = \lambda B\mathbf{v}-\lambda\mathbf{v}-\lambda B\mathbf{v} = -\lambda\mathbf{v}.

Thus, ( B A − A − λ B ) v = − λ v (BA-A-\lambda B)\mathbf{v}=-\lambda\mathbf{v} , so v \mathbf{v} is an eigenvector of B A − A − λ B BA-A-\lambda B corresponding to − λ -\lambda .

Therefore, if v \mathbf{v} is an eigenvector of A A corresponding to λ \lambda , then v \mathbf{v} is an eigenvector of B A − A − λ B BA-A-\lambda B corresponding to − λ -\lambda , for any B : V → V B: V\rightarrow V .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Awesome problem and solution, Stephen.

Pranshu Gaba - 5 years ago

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