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Show that the set $U \subset \mathbb{R}^{n^{2}}$ of matrices $A$ with $\det(A) \neq 0$ is open. Let $A^{-1}$ be the inverse of the matrix $A$. Show that the mapping $A \mapsto A^{-1}$ is continuous from $U$ to $U$.

My solution to the first part is that $\det(A)$ can be expressed as a polynomial in the entries of $A$, and since polynomial functions are continuous we have that the determinant function is continuous from which we can say the given set is indeed open. Any thoughts on the next part?

Hosein Rahnama
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AlexBowring
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3 Answers3

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First: $f=\det$ is continuous, and the set $\operatorname{GL}_n(\mathbb{R})$ of all invertible matrices can be written as $$ \operatorname{GL}_n(\mathbb{R}) = f^{-1}(\mathbb{R}^\ast) $$ Since $\mathbb{R}^\ast := \mathbb{R}-\{0\}$ is open and $f$ is continuous, $f^{-1}(\mathbb{R}^\ast)$ is open.

Now, for showing $g\colon A\in \operatorname{GL}_n(\mathbb{R}) \mapsto A^{-1}$ is continuous, use the fact that $\det$ is continuous, that $A\mapsto\operatorname{adj} A$ is continuous (not hard to show), and that $$ A^{-1} = \frac{1}{\det A}\operatorname{adj} A. $$

Hosein Rahnama
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Clement C.
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You're on the right track. You have been given a way to define a neighborhood. He's told you in the problem to think of $n\times n$ matrices as $\Bbb R^{n^2}$.

HINT: Do you have a concrete formula for the inverse of a nonsingular matrix? Say, something involving cofactors?

Ted Shifrin
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  • How would you define a neighborhood? I thought we couldn't as we haven't been given a metric... – AlexBowring May 27 '14 at 01:32
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    You have the standard metric on $\Bbb R^{n^2}$. Think of a matrix as a giant vector with $n^2$ entries. So, if you have two matrices $A$ and $B$, the distance between them will be $$\big(\sum_{i,j} (a_{ij}-b_{ij})^2\big)^{1/2}.$$ – Ted Shifrin May 27 '14 at 01:37
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    Also, it is useful to keep in mind that all possible norms will (induce metrics which) define the same topology (as the dimension of the space is finite). – Clement C. May 27 '14 at 02:36
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For funsies because this popped up on the front page...

Another way to see inversion is continuous (smooth, analytic) is to note that $$ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n $$ holds in any Banach algebra for $\|x\|<1$. In particular, if $A=I-B$ for small $B$, then $$ A^{-1}=\sum_{n=0}^{\infty}B^n. $$ You can translate any matrix to a neighborhood of the identity, invert, and translate it back, i.e. if $A=C(I-B)$, then $$ A^{-1}=C^{-1}\sum_{n=0}^{\infty}B^n. $$ This is clearly continuous (smooth, analytic) for fixed $C$ and varying $B$ in a neighborhood of the identity.

yoyo
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