Question

If \(A\) is a \(3\times 3\) non-zero matrix such that \(A^2=0\), then the number of non-zero eigen values of \(A\) is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

If \(Av=\lambda v\) for \(v\neq 0\), applying \(A\) again gives \(A^2v=\lambda^2v\); use \(A^2=0\) to pin down \(\lambda\).

Answer 1

The Correct Option is A

Step 1: Let \(\lambda\) be any eigenvalue of \(A\) with eigenvector \(v \neq 0\), so \(Av=\lambda v\).
Step 2: Apply \(A\) again: \(A^2v = A(Av) = A(\lambda v) = \lambda(Av) = \lambda^2 v\).
Step 3: Since \(A^2=0\), we get \(A^2v=0\), so \(\lambda^2v=0\). As \(v\neq 0\), this forces \(\lambda^2=0\), i.e. \(\lambda=0\).
Step 4: This argument applies to every eigenvalue of \(A\) (all 3, counted with multiplicity), so all eigenvalues of \(A\) equal 0.
Step 5: Hence the number of non-zero eigenvalues of \(A\) is 0.
\(\boxed{0}\)