Let \(W\subset V\) be finite-dimensional vector spaces over a field, and let \(L:V\to V\) satisfy \[ L(V)\subset W. \] Let \(L_W:W\to W\) be the restriction of \(L\). Prove that \[ \det(I_V-tL)=\det(I_W-tL_W). \]
Solution
Choose a basis \(w_1,\ldots,w_r\) of \(W\) and extend it to a basis \[ w_1,\ldots,w_r,v_1,\ldots,v_s \] of \(V\).
1. In this basis, the matrix of \(L\) has block form\[ [L]= \begin{pmatrix} [L_W]&B\\ 0&0 \end{pmatrix} \]for some matrix \(B\).
Because \(L(W)\subseteq L(V)\subseteq W\), the restriction \(L_W\) is well-defined and gives the upper-left block. Since every vector \(L(v_j)\) lies in \(W\), all coordinates of \(L(v_j)\) in the complementary basis vectors \(v_1,\ldots,v_s\) vanish. Hence the entire lower row of blocks is zero.
2. In the same basis,\[ [I_V-tL] = \begin{pmatrix} I_W-t[L_W]&-tB\\ 0&I_s \end{pmatrix}. \]
Subtract \(t[L]\) from the identity matrix, using the block form from step 1.
3.\[ \det(I_V-tL)=\det(I_W-tL_W). \]
The matrix in step 2 is block upper triangular. Therefore \[ \det(I_V-tL) = \det(I_W-tL_W)\det(I_s) = \det(I_W-tL_W). \] This calculation holds in the polynomial ring over the ground field, so it proves the asserted identity in \(t\).
4. Q.E.D.
Step 3 is the required determinant identity.