For real matrices $A$ and $B$, with $B$ invertible, prove $A+\lambda B $ is invertible for some $\lambda$ I stumbled upon this question that I would like to ask about:

Let $A$ and $B$ be $n\times n$ matrices over $\mathbb R$ where $B$ is an invertible matrix. How do you show that there exists some $\lambda \in \mathbb R$ such that $A+\lambda B $ is invertible?

Do I have to split this into the two cases where (i) $A$ is invertible, and (ii) $A$ is not invertible and then make an argument about the sum above? I want to use some determinant rule, perhaps it would help, but since this is a sum, I can't simply apply it; And I think I remember that only the product of two invertible matrices is again an invertible matrix, but for sums it is not so clear.
 A: Consider $$\det(A + \lambda B)$$
This will be a polynomial of finite degree in $\lambda$. Unless the polynomial is uniformly zero, there will exist only a finite set of real roots. Then any $\lambda$ which is not a root of the polynomial will give you the required invertible matrix. This does not even require $B$ to be invertible.
I'm not sure under what conditions the above will be uniformly zero (if someone knows the answer, I'd be interested to know), but I can say that when $B$ is invertible that this never happens. For we have 
$$\det(AB^{-1} + \lambda I) = \det(A + \lambda B)\det(B^{-1})$$
so the expressions $\det(AB^{-1} + \lambda I)$ and $\det(A + \lambda B)$ are either together zero or together non-zero. The first expression is 
$$-\det(-AB^{-1} - \lambda I)$$
which is the (negative) characteristic polynomial of $-AB^{-1}$, and that's always of degree $n$.
A: See what you can do with $B^{-1}A + \lambda I.$ Or, if you prefer, $A B^{-1} + \lambda I.$ 
Given some square matrix $C$ and constant $t,$ what do we say when $C - t I$ is singular? And how many different values of such $t$ are possible?
A: Let's think about it in this way. Since we already know that $B$ is invertible (so is $\lambda B$ for any $\lambda\neq 0$) and we want to have $\lambda B+A$ invertible, so we want the behavior of $A+\lambda B$ to be dominated by $\lambda B$. So the guess is that we want to make $\lambda$ large.
To make it precise, in finite dimensional spaces, it suffices to show $A+\lambda B$ is injective. That is, if $(\lambda B+A)v=0$ then $v=0$. We can even assume $\|v\|\le 1$ by linearity. But $B$ is invertible, so $(\lambda B+A)v=0$ is the same as \begin{equation}
\lambda v=-B^{-1}Av=Cv
\end{equation}   if we choose $C=-B^{-1}A$.
The right hand side is a linear combination of columns of $C$, the left hand side is a linear combination of columns of $\lambda I$, now you can see when we choose $\lambda$ large enough, the equation above can be true only when $v=0$.
There is actually something more general then this. You may search for spectrum of bounded linear operators for more information, the key word is compactness.
