Let TVV be a linear operator that has only two distinct eige

Let T:VV be a linear operator that has only two distinct eigenvalues ₁ and ₂ . Prove that T is diagonalizable if and only if V=E₁ E2

Solution

. Assume that T is diagonalizable.

Then we can find a basis B for V consisting of eigenvectors for T. Each of these vectors is associated with a particular eigenvalue, so write 1, . . . , k for the distinct ones.

We can then group together the elements of B associated with i , span them, and call the resulting subspace Ei . It follows then that E1 · · · Ek = E1 + · · · + Ek = Span B = V

Here we have 2 distinct eigen values.This theorem hlds for n distinct eigen values.