An operator A is called nilpotent if Ak = 0 for some k. Prove that if A isnilpotent, then σ(A) = {0
Question
An operator A is called nilpotent if for some . Prove that if A is nilpotent, then .
Solution
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}.
Similar Questions
An operator A is called nilpotent if Ak = 0 for some k. Prove that if A isnilpotent, then σ(A) = {0
Let v be a vector and let k be a scalar. From the axioms of a vector space, prove that if kv = 0 theneither k = 0 or v = 0.
Let 𝐴 𝑏𝑒 𝑎𝑛 𝑛 × 𝑛 matrix. The characteristic polynomial of A is a polynomial of degree
An n×n matrix A is diagonalizable if A has n linearly independent eigenvectors.A. TrueB. False
Every linear operator in an n-dimensional vector space has n distinct eigen-values
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.