An operator A is called nilpotent if Ak = 0 for some k. Prove that if A isnilpotent, then σ(A) = {0

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

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 this equation to the power of k: (Ak)v = (0)v.

Step 4: Using the property of matrix multiplication, we have A^k v = 0v, which simplifies to A^k v = 0.

Step 5: Since Ak = 0, we can substitute this into the equation from step 4: 0v = 0.

Step 6: Since v is an eigenvector, it is nonzero. Therefore, we can divide both sides of the equation by v: 0 = 0/v.

Step 7: Dividing by a nonzero number is always defined, so we have 0/v = 0.

Step 8: This implies that λ = 0, since λ is the eigenvalue corresponding to the eigenvector v.

Step 9: Therefore, the only eigenvalue of A is 0, which means σ(A) = {0}.

Thus, we have proven that if A is nilpotent, then σ(A) = {0}.