$T$-invariant subspaces, where the characteristic polynomial of $T$ is $x^4-3x^3$

Question

Let $T:V\to V$ be a linear transformation of a four-dimensional real vector spaces $V$. Assume that the characteristic polynomial of $T$ is $x^4-3x^3$.

1. Show that $V$ hash $T$-invariant subspaces of dimension 1,2,and 3.
2. What can you say about the rank of $T$?

Answer 1

The characteristic polynomial of $T$ splits as $p_T(x) = x^3(x - 3)$ and so the real eigenvalues of $T$ are $\lambda_1 = 0$ and $\lambda_2 = 3$. Consider the minimal polynomial of $T$:

1. If $m_T(x) = x(x-3)$ then $T$ is diagonalizable and in particular $\dim V_{\lambda_1} = \dim \ker(T - \lambda_1 I) = \dim \ker(T) = 3$. Any subspace of $V_{\lambda_1}$ is $T$-invariant and so $T$ has invariant subspaces of any dimension. Also, $T$ has rank $1$.
2. If $m_T(x) = x^2(x-3)$ then $V = \ker(T^2) \oplus \ker(T - \lambda_2 I)$. In this case, we have a basis of the form $(v, Tv, w)$ for $\ker(T^2)$ with $T^2v = Tw = 0$. The subspace $\ker(T^2)$ is a three-dimensional invariant subspace, the subspace $\operatorname{span} \{ v, Tv \}$ is a two-dimensional invariant subspace and the subspace $\operatorname{span} \{ w \}$ is a one-dimensional invariant subspace. In this case $\ker(T) = \operatorname{span} \{ v, w \}$ and so $T$ has rank $2$.
3. If $m_T(x) = x^3(x-3)$ then $V = \ker(T^3) \oplus \ker(T - \lambda_2 I)$. In this case, we have a basis of the form $(v, Tv, T^2v)$ for $\ker(T^3)$ with $T^3 v = 0$. The subspace $\ker(T^3)$ is a three-dimensional invariant subspace, the subspace $\operatorname{span} \{ Tv, T^2v \}$ is a two-dimensional invariant subspace and $\operatorname{span} \{ T^2 v \}$ is one-dimensional invariant subspace. In this case, $\ker(T) = \operatorname{span} \{ v \}$ and so $T$ has rank $3$.

Answer 2

The characteristic equation implies $T$ has eigenvalue $3$ of multiplicity $1$ and eigenvalue $0$ of multiplicity $3$ i.e. nullity of $T$ is $3$. So rank is $1$.

Construct an orthonormal basis as follows (Do you know how to construct such a basis from a given basis? like by the Grahm Schimdt procedure or other procedures)

Take the vector which has eigenvalue $3$ as $1$ basis element. Then with the help of the canonical basis (you know what it is right?) construct the other $3$ basis elements (say $b_2,b_3\ \&\ b_4$).

Clearly these $3$ vectors ($b_2,b_3\ \&\ b_4$) belong to Nullspace of $T$. So take any one of these $3$ vectors and its span will be $T$ invariant and have dimension $1$.

Similarly take any $2$ of these vectors and their span will be both $T$ invariant and have dimension $2$. Similar is the case for $3$.