MAT4858 Assignment 1 (197343) |ANSWERS| - DUE 30 JUNE 2026

MAT4858
Assignment 1

(197343)
DUE: 30 JUNE 2026

, MAT4858 – Linear Algebra

Assignment 01 | Unique Number: 197343

Closing Date: 30 June 2026 | Total Marks: 50


Question 1 [10 Marks]

1.1 Eigenvectors and the Same Eigenvalue (Chapter 12, Exercise 765)

Statement: Let α be an endomorphism of a vector space V over a field F, and let v, w
be eigenvectors of α with v + w ≠ 0V. Show that v + w is an eigenvector of α if and only
if v and w correspond to the same eigenvalue.



Proof.

(⇒) Suppose v + w is an eigenvector of α.

Since v and w are eigenvectors of α, there exist scalars λ, μ ∈ F such that

α(v) = λv and α(w) = μw.

Since v + w ≠ 0V and v + w is assumed to be an eigenvector, there exists a scalar θ ∈ F
with

α(v + w) = θ(v + w).

By linearity of α:

α(v + w) = α(v) + α(w) = λv + μw.

Hence λv + μw = θ(v + w) = θv + θw, which gives

(λ − θ)v + (μ − θ)w = 0.

Now we show v and w are linearly independent. Suppose av + bw = 0 for a, b ∈ F.
Since v ≠ 0V, if a ≠ 0 then v = −(b/a)w, so α(v) = −(b/a)α(w) = −(b/a)μw = μv (scaling),
meaning λ = μ and v, w are scalar multiples of each other. In that case (λ−θ)v + (μ−θ)w
= (λ−θ)v + (λ−θ)w = (λ−θ)(v+w) = 0, and since v + w ≠ 0 we get λ = θ = μ, so λ = μ as
required.

If instead v and w are linearly independent, then from (λ − θ)v + (μ − θ)w = 0 and linear
independence, we deduce λ − θ = 0 and μ − θ = 0, giving λ = μ.