Invariant subspaces with characteristic polynomial $\chi_T=x^4-x^3$

linear algebra

Let $V$ be a $4$-dimensional vector space, and let $T:V\to V$ be a linear map with characteristic polynomial $\chi_T=x^4-x^3$. Show $V$ has $T$-invariant subspaces of dimension $1$, $2$, and $3$.

I know that since $1$ is an eigenvalue of algebraic multiplicity $1$, the eigenspace corresponding to $1$ is going to be a dimension $1$ $T$-invariant subspace. However, I cannot figure out how to proceed. We have that $0$ is an eigenvalue of algebraic multiplicity $3$, and the bound
$$
0\leq \dim(\text{eigenspace corresponding to }0)\leq\text{algebraic multiplicity of }0=3.
$$
How can I go from here?

Best Answer

You know that the eigenspaces $ V_0 $ and $ V_1 $ (corresponding to eigenvalues $ 0 $ and $ 1 $ respectively) are nontrivial. By an algebraic multiplicity argument, you also know that $ \dim V_1 = 1 $ and $ \dim V_0 \geq 1 $. These are enough to give you an invariant subspace of dimension $ 1 $ (which is simply $ V_1 $) and of dimension $ 2 $ (take the direct sum $ V_1 \oplus \operatorname{span}(v) $ where $ v $ is any eigenvector with eigenvalue $ 0 $). For dimension $ 3 $, there are two cases:

  • $ \dim V_0 \geq 2 $. In this case, you can pick two linearly independent eigenvectors with eigenvalue $ 0 $ which form an invariant subspace of dimension $ 2 $, and take its direct sum with $ V_1 $ for an invariant subspace of dimension $ 3 $.
  • $ \dim V_0 = 1 $. In this case, $ T : V \to V $ has one-dimensional kernel and thus three-dimensional image by the rank-nullity theorem, so $ \operatorname{Im}(T) $ is the desired $ 3 $-dimensional invariant subspace of $ V $.
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