Let \(V\) be a finite-dimensional vector space over a field \(k\), let \(T:V\to V\) be linear, and let \(W\subseteq V\) be a \(T\)-invariant subspace. Show that the characteristic polynomial of \(T|_W\) divides the characteristic polynomial of \(T\).
Solution
Let \[ m\coloneqq\dim W, \qquad n\coloneqq\dim V. \]
1. There is a basis\[ v_1,\ldots,v_m,v_{m+1},\ldots,v_n \]of \(V\) such that \(v_1,\ldots,v_m\) is a basis of \(W\).
Choose any basis \(v_1,\ldots,v_m\) of the subspace \(W\) and extend it to a basis of the finite-dimensional space \(V\).
2. Relative to the basis in step 1, the matrix of \(T\) has block form\[ [T] = \begin{pmatrix} A&C\\ 0&D \end{pmatrix}, \]where \(A\) is the matrix of \(T|_W\) in the basis\(v_1,\ldots,v_m\).
Because \(W\) is \(T\)-invariant, \[ T(v_j)\in W \] for \(1\leq j\leq m\). Therefore the last \(n-m\) coordinates of each of the first \(m\) columns of \([T]\) vanish, giving the zero lower-left block. The upper-left block records exactly the coordinates of the restricted map \(T|_W\) in the chosen basis of \(W\).
3. The characteristic polynomial of \(T\) factors as\[ \chi_T(t) = \chi_{T|_W}(t)\det(tI_{n-m}-D). \]
By step 2, \[ tI_n-[T] = \begin{pmatrix} tI_m-A&-C\\ 0&tI_{n-m}-D \end{pmatrix}. \] The determinant of a block upper-triangular matrix is the product of the determinants of its diagonal blocks. Hence \[ \begin{aligned} \chi_T(t) &= \det(tI_n-[T])\\ &= \det(tI_m-A)\det(tI_{n-m}-D). \end{aligned} \] Since \(A\) represents \(T|_W\), \[ \det(tI_m-A)=\chi_{T|_W}(t). \]
4. The polynomial \(\chi_{T|_W}\) divides \(\chi_T\) in \(k[t]\).
Step 3 expresses \(\chi_T\) as \(\chi_{T|_W}\) times a polynomial in \(k[t]\).
5. Q.E.D.
Step 4 is the required divisibility statement.