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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    $\begingroup$ 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. $\endgroup$
    – azarel
    Mar 18, 2012 at 20:50
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    $\begingroup$ 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. $\endgroup$
    – Tpofofn
    May 17, 2012 at 2:08

4 Answers 4


$M_n(\mathbb R)$ 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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    $\begingroup$ 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. $\endgroup$
    – user12014
    May 17, 2012 at 1:04
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    $\begingroup$ I do not understand your expression at the right side of the inequality.also what do you mean by $\det (x_{i,j})$? $\endgroup$
    – Marso
    Dec 16, 2013 at 9:37
  • $\begingroup$ Is this argument independent of the metric you define on $M_n(\mathbb{R})$? (haven't taken topology yet) $\endgroup$
    – Chris Jing
    Apr 12, 2020 at 2:50
  • $\begingroup$ @ChrisJing all norms induce the same topology $\endgroup$
    – zaira
    Dec 7, 2020 at 8:10

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.


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.


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

  • $\begingroup$ I didn't downvote you, but I'm curious if you can elaborate on why computable functions (in the relevant sense) are continuous? $\endgroup$ May 26, 2012 at 0:10
  • $\begingroup$ @IsaacSolomon It's essentially the finite use principal. See here for example. $\endgroup$ May 26, 2012 at 18:23
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    $\begingroup$ @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. $\endgroup$
    – rschwieb
    Jun 8, 2012 at 14:12
  • $\begingroup$ @rschwieb Actually the notion of a computable function on $\mathbb{R}^{n}$ is a precisely defined one. See for example, Computable Analysis by Weihrauch. $\endgroup$ Jun 9, 2012 at 15:12
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    $\begingroup$ @QuinnCulver No objection here, since the veracity of your claim has never been in doubt: only the accessibility of it. It's good policy to assume others have not read everything you have. $\endgroup$
    – rschwieb
    Jun 9, 2012 at 17:49

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