Question

2.5.8 Let U be a vector space over a field F and T ∈ L(U). If λ1, λ2 ∈ F are distinct eigenvalues and u1, u2 are the respectively associated eigenvectors of T , show that, for any nonzero a1, a2 ∈ F, the vector u = a1u1 + a2u2 cannot be an eigenvector of T .

Answer 1

If λ₂, λ₂ ∈ F are distinct eigenvalues and u₁, u₂ are the respectively associated eigenvectors of T, then u₁, u₂ are linearly independent vectors.

Now, let us assume that u = a₁u₁ + a₂u₂ is an eigenvector of T corresponding to the eigenvalue ʎ. Then Tu = ʎu or, T(a₁u₁ + a₂u₂) = ʎ(a₁u₁ + a₂u₂) or, a₁T(u₁)+a₂T(u) = ʎ(a₁u₁ + a₂u₂) or, a₁ʎ₁u₁+    a₂ ʎ₂u₂= ʎ(a₁u₁ + a₂u₂) or, a₁u₁(ʎ₁- ʎ) = a₂u₂(ʎ -ʎ₂).Further, since u₁, u₂ are linearly independent vectors, neither of these can be a scalar multiple of the other. This implies that either ʎ₁- ʎ = 0 = ʎ -ʎ₂ or, a₁= a₂ = 0.

Now, ʎ₁- ʎ = 0 = ʎ -ʎ₂   means that ʎ₁ =ʎ = ʎ₂ which is a contradiction as λ₂, λ₂ are distinct eigenvalues of T.

Also, a₁, a₂ are given to be non-zero.

This implies that we cannot have T(a₁u₁ + a₂u₂) = ʎ(a₁u₁ + a₂u₂) for any ʎ, i.e. a₁u₁ + a₂u₂ cannot be an  eigenvector of T.