- Prove that if \(A\) and \(B\) are invertible matrices over a field \(\boldsymbol{k}\), then \(A+\lambda B\) is invertible for all but finitely many \(\lambda \in \boldsymbol{k}\).
Solution
1. Suppose \(A,B\in M_n(k)\) are invertible.For \(\lambda\in k\),\[ A+\lambda B=B\bigl(B^{-1}A+\lambda I_n\bigr). \]
Expanding the right-hand side gives \(BB^{-1}A+\lambda BI_n=A+\lambda B\).
2. Hence\[ \det(A+\lambda B)=\det(B)\,p(\lambda), \qquad p(t):=\det(B^{-1}A+tI_n)\in k[t]. \]
Apply multiplicativity of the determinant to 1.
3. The polynomial \(p(t)\) is nonzero and has degree \(n\).
In the determinant expansion of \(B^{-1}A+tI_n\), the product of the \(t\)-terms on the diagonal contributes \(t^n\), and no other term has degree \(n\). Thus \(p(t)\) is monic of degree \(n\).
4. Therefore \(A+\lambda B\) is singular for at most \(n\) values of \(\lambda\in k\).
Because \(B\) is invertible, \(\det(B)\neq0\). Thus \[ \det(A+\lambda B)=0 \quad\Longleftrightarrow\quad p(\lambda)=0. \] A nonzero polynomial of degree \(n\) over a field has at most \(n\) roots.
5. Hence \(A+\lambda B\) is invertible for all but finitely many \(\lambda\in k\).
A square matrix over a field is invertible exactly when its determinant is nonzero, so the conclusion follows from 4.