An operator A is called nilpotent if Ak = 0 for some k. Prove that if A isnilpotent, then σ(A) = {0} (i.e. that 0 is the only eigenvalue of A)

To prove that if A is nilpotent, then σ(A) = {0}, we need to show that 0 is the only eigenvalue of A.

Step 1: Assume A is nilpotent, which means there exists a positive integer k such that Ak = 0.

Step 2: Let λ be an eigenvalue of A, and let v be the corresponding eigenvector. This means that Av = λv.

Step 3: We can raise both sides of the equation Av = λv to the power of k to get Akv = λ^kv.

Step 4: Since Ak = 0, the equation simplifies to 0v = λ^kv.

Step 5: Since v is an eigenvector, it is nonzero. Therefore, λ^k must be zero for the equation to hold.

Step 6: The only way for λ^k to be zero is if λ is zero.

Step 7: Therefore, if A is nilpotent, the only eigenvalue of A is 0.

Step 8: Hence, we can conclude that if A is nilpotent, then σ(A) = {0}, where σ(A) represents the set of eigenvalues of A.