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


*

*Show that $V$ hash $T$-invariant subspaces of dimension 1,2,and 3.

*What can you say about the rank of $T$?

 A: 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 $T$:


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*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 $1$.

*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 $2$.

*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 $3$.

A: 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 $3$.
A: Write the characteristic polynomial as $x^3(x-3)$. Recall that $V\cong \mathbb{R}^4$ can be considered as a finitely generated torsion $\mathbb{R}[x]$-module via the multiplication $f(x)\cdot v:=f(T)(v)$. Since the invariant factors of $T$ multiply to give the characteristic polynomial, and the invariant factor of largest degree is the minimal polynomial, the only possibilities for the minimal polynomial are 
(1) $x^3(x-3)$, or
(2) $x^2(x-3)$.
Using the Fundamental Theorem for finitely generated modules over PIDs, we get
$$
(1)\implies V\cong \frac{\mathbb{R}[x]}{(x^3)}\oplus\frac{\mathbb{R}[x]}{(x-3)},
$$
and 
$$
(2)\implies V\cong \frac{\mathbb{R}[x]}{(x)}\oplus\frac{\mathbb{R}[x]}{(x^2)}\oplus\frac{\mathbb{R}[x]}{(x-3)}.
$$
In the first case, the first summand has $\mathbb{R}$-basis $\{1,\bar{x},\bar{x}^2\}$, therefore dimension 3, and the second summand has dimension $1$. Note that the subspace $\langle \bar{x},\bar{x}^2\rangle$ of the first summand has dimension 2. In the second case, there are clearly subspaces of dimensions 1,2, and 3. Regarding the rank of $T$, recall that $T$ acts on the right hand side of the above isomorphisms by multiplication by $\bar x$, and so we need to determine the dimension of the image each summand under this multiplication. In the first case, the new bases  (after applying the transformation) for the first and second summands become $\{\bar{x},\bar{x}^2\}$ (since $\bar x^3=0$) and $\{\bar x\}$, respectively, so the rank of $T$ in this case is 3. In the second case, the first summand vanishes and the $\mathbb{R}$-bases for the second two are both $\{\bar x\}$, so the rank of $T$ in this case is 2.
