Question:medium

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:

Show Hint

If \(Av=\lambda v\) for \(v\neq 0\), applying \(A\) again gives \(A^2v=\lambda^2v\); use \(A^2=0\) to pin down \(\lambda\).
Updated On: Jul 4, 2026
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Show Solution

The Correct Option is A

Solution and Explanation

Step 1: $A^2=0$ with $A \neq 0$ means the minimal polynomial $m(x)$ of $A$ divides $x^2$ but is not constant and is not $x$ alone, so $m(x) = x^2$. This makes $A$ a nilpotent matrix.
Step 2: For any nilpotent matrix, every eigenvalue must be 0, so the characteristic polynomial of the $3\times 3$ matrix $A$ is $x^3$.
Step 3: The roots of $x^3=0$ are $\lambda=0,0,0$, i.e. all three eigenvalues (with multiplicity) are zero. Consequently $\text{trace}(A)=0$ and $\det(A)=0$ too, consistent with $A$ being singular and nilpotent.
Step 4: So none of the eigenvalues of $A$ is non-zero.
\[\boxed{0}\]
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