3) Let \( F \) denote a field, and consider a vector space \( V \) over \( F \) with dimension \( 5 \). Let
\( T : V \rightarrow V \) denote a linear transformation whose Jordan canonical form consists of a single Jordan block
with eigenvalue \( 0. \) (i) Find the dimension of the subspace \( U = \{v \in V \; | \; T^3 v = 0 \}. \) (ii) Find all
subspaces \( W \) of \( V \) such that \( TW \subseteq W \) and the sum \( V = U + W \) is direct. Prove that your
answer is correct.
Proof. If \( J \) is the matrix in question, multiplying by \( J \) is the same as applying \( T \) for some
basis \( \{{v_i}\}_{i=1}^5 . \) Multiplying another \( 5 \times 5 \) matrix (or column vector!) by \( J \) on the left
has the effect of moving the rows of the matrix on the right up by one row, bringing in zeros along the bottom:
\[ J =
\begin{pmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}, \; \; J^2 =
\begin{pmatrix}
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}, \; \; J^3 =
\begin{pmatrix}
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}.
\]
By the definition of matrix representation of a linear transformation, the \( i^{th} \) column and \( j^{th} \) row of \( J^3 \)
is the weight of \( T^3(v_i) \) in terms of \( v_j \), so \( T^3(v_1) = T^3(v_2) = T^3(v_3) = 0, \; \) \(T^3(v_4) = v_1, \; T^3(v_5) = v_2. \)
It follows that \( \{v_1, v_2, v_3\} \) is a basis for \( U \), so \( \text{dim}(U) = \) \(\text{dim}(\{v \in V \; | \; T^3 v = 0 \}) = 3. \)
For (ii), we're seeking subspaces \( W \) such that \( TW \subseteq W \) and \( V = U \oplus W. \) Let \( W \) be such a subspace,
obviously of dimension \( 2. \) Write \( v_4 = u + w, u \in U, w \in W. \) Then
\( v_1 = T^3(v_4) = T^3(u + w) = \) \( T^3(u) + T^3(w) = T^3(w) \in W, \) but \( v_1 \in U \), which we can't have because \( v_1 \neq 0 \)
and \( U \) and \( W \) can have no nonzero common elements if they are to be direct summands. This contradiction shows that there
can be no such \( W. \) QED.