Let \(B\) be a \(3\times3\) matrix whose null space is two-dimensional, and let \(\chi(\lambda)\) be its characteristic polynomial. For each assertion, give a proof or a counterexample.
- \(\lambda^2\) divides \(\chi(\lambda)\).
- The trace of \(B\) is an eigenvalue of \(B\).
- \(B\) is diagonalizable.
Solution
1. Assertion (1) is true: \(\lambda^2\) divides \(\chi(\lambda)\).
The null space of \(B\) is the eigenspace for the eigenvalue \(0\), so the geometric multiplicity of \(0\) is \(2\). The geometric multiplicity of an eigenvalue is at most its algebraic multiplicity. Hence \(0\) has algebraic multiplicity at least \(2\), which says exactly that \(\lambda^2\) divides \(\chi(\lambda)\).
2. The characteristic polynomial has the form\[ \chi(\lambda)=\lambda^2(\lambda-\operatorname{tr} B). \]
By step 1 and the fact that \(\chi\) is monic of degree \(3\), there is a scalar \(c\) such that \[ \chi(\lambda)=\lambda^2(\lambda-c) =\lambda^3-c\lambda^2. \] For a \(3\times3\) matrix, the coefficient of \(\lambda^2\) in \(\det(\lambda I-B)\) is \(-\operatorname{tr} B\). Comparing coefficients gives \(c=\operatorname{tr} B\).
3. Assertion (2) is true: \(\operatorname{tr} B\) is an eigenvalue of \(B\).
By step 2, \[ \chi(\operatorname{tr} B)=0. \] Thus \(\operatorname{tr} B\) is a root of the characteristic polynomial and hence an eigenvalue of \(B\).
4. Assertion (3) is false.
Consider \[ B= \begin{pmatrix} 0&1&0\\ 0&0&0\\ 0&0&0 \end{pmatrix}. \] Its kernel is \[ \ker B = \operatorname{span} \left\{ \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix} \right\}, \] which is two-dimensional. Also \(B^2=0\) but \(B\neq0\). If \(B\) were diagonalizable, then \(B=SDS^{-1}\) for a diagonal matrix \(D\), and \(B^2=0\) would imply \(D^2=0\). Every diagonal entry of \(D\) would therefore be zero, so \(D=0\) and hence \(B=0\), a contradiction. Thus this \(B\) is not diagonalizable.
5. Q.E.D.
Steps 1 and 3 prove assertions (1) and (2), respectively, while step 4 supplies a counterexample to assertion (3).