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Let $A$ and $B$ be $n\times n$ matrices over the field $F$. Show that if $A$ is invertible, there are at most $n$ scalars $c$ in $F$ for which the matrix $cA+B$ is not invertible.

Here is what I am looking at so far. I've considered $cI+BA^{-1}$ and the determinant $\det(cI+BA^{-1})$. Now, this is not invertible provided that $\det(cI+BA^{-1})=0.$ I'm stuck trying to convince myself that this is an polynomial with $n$ roots.

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    $\begingroup$ If you are familiar with expansion by minors, you can use induction together with that to show that it is indeed a polynomial with at most $n$ roots. $\endgroup$ – Kurtland Chua Feb 23 '18 at 5:47
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$$\det(cI+BA^{-1})=0$$

Let $c=-\lambda$,

then we have $$\det(BA^{-1}-\lambda I)=0$$

This should relate it to a special polynomial known as charactheristic polynomial which is of degree $n$.

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Recall how the determinant is defined. It is $$ \sum_{\sigma \in S_n} \prod_{i=1}^n \varepsilon(\sigma) a_{i, \sigma(i)}. $$ Now each $a_{i, \sigma(i)}$ term will have at most instance of $c$ in it (i.e. it will have an instance of $c$ exactly when $i = \sigma(i)$), and so it is a sum of terms of degree at most $n$. So it is a polynomial of degree at most $n$ (and in fact equal to $n$), thus it has at most $n$ roots.

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