# $A+\lambda B$ is invertible

I stumbled upon this question that I would like to ask you about:

Let $A$ be a $n\times n$ matrix $(\mathbb R)$ and $B$ an invertible Matrix of size $n$ with real coefficients.

I need to show that there exists some $\lambda \in \mathbb R$ such that $$A+\lambda B$$ is invertible.

So do I have to split this into two cases?

i) $A$ is invertible ii) $A$ is not invertible

and make an argument for 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 to remember that only the product of two invertible matrices is again an invertible matrix, but for sums it is not so clear.

Any help would be greatly appreciated!

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Please avoid $$in titles. – EuYu Oct 31 '12 at 2:48 Thanks for letting me know! – user41725 Oct 31 '12 at 2:50 ## 3 Answers Consider$$\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 $n$.

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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?

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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 $$\lambda v=-B^{-1}Av=Cv$$ 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.

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