2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $n$.

$\endgroup$
2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.