# Can we always for an invertible matrix $M$ find real number $\alpha \neq 0$ such that $M+\alpha$ is invertible?

I do not know enough about matrices, maybe only enough to be able to create question like this one, but I would like to see an answer.

Let $a_{ij}$ be some element of invertible $n\times n$ matrix $M$ which has real numbers as entries.

Let us now define the addition of the real number $\alpha$ with matrix $M$ in such a way that $\alpha+M$ gives matrix $N$ which has entries $a_{ij}+\alpha$.

The question is:

Can we for every real invertible matrix $M$ find real number $\alpha \neq 0$ such that $M+\alpha$ is invertible?

I created this question because it seems to me that the set of real invertible matrices is connected in such a way that if we have some real invertible matrix that we could alter coefficients at least a little bit by the same amount and still arrive at another real invertible matrix.

As I know almost nothing about this area of mathematics, I could be wrong, but because some of you surely know how to answer this question I will probably find out was my intuition right?

Also, I would like to have a proof or counterexample that is as elementary as possible.

• @arctictern But that only makes change to the numbers on the diagonal, right? – Farewell Dec 9 '15 at 16:05
• Yes but that is actually a common operation. $M+\alpha$ does not a priori actually mean anything. How do you add a Matrix to a number? Just adding $\alpha$ to all entries is not a very common operation. – Jorik Dec 9 '15 at 16:06
• @Jorik Maybe the operation is not common but it is unavoidable for the problem which is stated as it is stated. – Farewell Dec 9 '15 at 16:07
• You can define whatever you want as long as it have some sense. – Farewell Dec 9 '15 at 16:08

Yes. A matrix $A$ is invertible if and only if the determinant $\det A$ is non-zero. Since the determinant is a polynomial in the entries of $A$ (as can be seen, e.g. by cofactor expansion), the determinant is continuous in the entries of $A$.
Hence if $\det(M) \ne 0$, then for all sufficiently small $\alpha$, $\det(M + \alpha) \ne 0$.
• So it seems that we will have an uncountable number of choices for $\alpha$? – Farewell Dec 9 '15 at 16:14
• @GyroGearloose here $M+\alpha$ does not mean $M+\alpha I$. So that argument doesn't work. – Jorik Dec 9 '15 at 16:44
• @user by viewing $\det(M+\alpha)$ as a polynomial in $\alpha$ we see that there are just a finite number of values $\alpha$ can take so that $M+\alpha$ is not invertable. – Jorik Dec 9 '15 at 16:45