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Why is the determinant as a function from $M_n(\mathbb{R})$ to $\mathbb{R}$ continuous, please can anyone explain precisely and rigorously? So far I know the explanation which comes from the facts: polynomials are continuous, sum and product of continuous functions are continuous. Also I have the confusion regarding the metric on $M_n(\mathbb{R})$

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3  
Isn't it clear that polynomials are continuous functions? $M_n(\mathbb R)$ is the same as $\mathbb R^{n^2}$ under a different disguise. – azarel Mar 18 '12 at 20:50
Non-rigorous, but perhaps helpful.... Determinants can be show to be equivalent to the hyper-volume of a hyper-parallelepiped with all the edges extending from one vertex defined by the columns (or rows) of a matrix. Given that this volume changes continuously with a change in any component vector, the determinant may be seen to be continuous. – Tpofofn May 17 '12 at 2:08

5 Answers

up vote 12 down vote accepted

$M_n(\mathbb R)$ is just $\mathbb R^{n^2}$ with the euclidian metric.

det is countinous, because it is a polynomial in the coordinates $$ \text{det}(x_{i,j})= \sum_\sigma \text{sgn}(\sigma) \prod_{i=1}^{n} x_{\sigma(i),i} $$

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It should be noted that $M_n(\mathbb{R})$ is often also given the metric induced by the operator norm. Of course the point is, it doesn't matter because all norms on a finite dimensional space induce the same topology. – user12014 May 17 '12 at 1:04
up vote 4 down vote

The coefficient maps $A\longmapsto a_{i,j}$ are continuous because they are linear on the finite-dimensional vector space $M_n(\mathbb{R})$. Here you want to refer to the topology of the latter as a normed space, which does not depend on the norm since they are all equivalent in finite dimension. Then the determinant is a polynomial in the coefficients, so it is continuous by composition of continuous maps.

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up vote 2 down vote

Recall that the determinant can be computed by a sum of determinants of minors, that is "sub"-matrices of smaller dimension.

Now we can prove by induction that $\det$ is continuous:

  • For $n=1$, $A\in M_1(\mathbb R)$ is simply a scalar we have that $\det A=A$, and surely the identity function is continuous.

  • Suppose that for $n$ we have that $\det$ is continuous on $M_n(\mathbb R)$, let $A\in M_{n+1}(\mathbb R)$. We know that $\det A$ can be calculated as the alternating sum over one of first row, when calculating the $\det$ of the appropriate minor.

    So $\det A$ is written as a sum and scalar multiplication of $\det$ on a smaller dimension. From the induction hypothesis these are continuous and therefore $\det$ is continuous on $n+1\times n+1$ matrices.

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up vote 1 down vote

It's continuous because it's computable as a function from $\mathbb{R}^{n^2}$ to $\mathbb{R}$.

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Would whoever downvoted this please speak up; especially if you think what I've said is incorrect. – Quinn Culver May 17 '12 at 17:40
I didn't downvote you, but I'm curious if you can elaborate on why computable functions (in the relevant sense) are continuous? – Isaac Solomon May 26 '12 at 0:10
@IsaacSolomon It's essentially the finite use principal. See here for example. – Quinn Culver May 26 '12 at 18:23
Thanks for the link, this is very interesting! :) – Isaac Solomon May 26 '12 at 18:51
@QuinnCulver If someone downvoted, it was probably because they thought the terminology was a little vague. I can see your idea (and this is how I think of it, too) is that the determinant is a composition of finitely many addition and multiplication operations, all of which are jointly continuous, and so the composition is continuous. – rschwieb Jun 8 '12 at 14:12
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up vote -5 down vote

The function $$\det:\mathcal{M}_n(\mathbb{R})\rightarrow\mathbb{R}$$ is continuous because is a escalar function and bounded. (Theory of operators)

And not all polynomials are continuous…

P.D.: Excuse my English, please.

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5  
Could you say more about what you mean? In our context, I think the claim is simply that a polynomial $f(x_1, \ldots, x_m)$ in $m$ variables with coefficients in $\mathbf R$ induces a continuous function on $\mathbf R^m$. This is definitely true. And when you mention bounded operators it seems like you're implying that $\det$ is linear, which doesn't seem to be true in any obvious sense. Cheers, – Dylan Moreland Mar 18 '12 at 23:37
@DylanMoreland You mean like my answer right? – Gastón Burrull May 16 '12 at 23:44

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