Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $M_n(\mathbb{R})$

share|improve this question
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 5

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.

share|improve this answer

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

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

share|improve this answer
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
I do not understand your expression at the right side of the inequality.also what do you mean by $\det (x_{i,j})$? –  Une Femme Douce Dec 16 '13 at 9:37

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.

share|improve this answer
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

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.

share|improve this answer

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

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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