# Why is the determinant continuous?

Why is the determinant as a function from $$M_n(\mathbb{R})$$ to $$\mathbb{R}$$ continuous? Can anyone explain precisely and rigorously? So far, I know the explanation which comes from the facts that

• polynomials are continuous,
• sum and product of continuous functions are continuous.

Also I have the confusion regarding the metric on $$M_n(\mathbb{R})$$.

• 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. Commented Mar 18, 2012 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. Commented May 17, 2012 at 2:08

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

determinant is continuous, because it is a polynomial in the coordinates $$\det(X) = \det ([x_{i,j}])= \sum_\sigma \text{sgn}(\sigma) \prod_{i=1}^{n} x_{\sigma(i),i},$$ where $$\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$$ is a permutation.

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

This answer is for $$\epsilon-\delta$$ fans!

Consider determinant as a function of the form $$\det: \mathcal{M}_n(\mathbb{R})\to\mathbb{R}$$. What are the elements of its domain? Indeed, they are functions! More specifically, if $$A\in\mathcal{M}_n(\mathbb{R})$$ then $$A$$ is a function of the form $$A:\{1,\dots,n\}\times\{1,\dots,n\} \to \mathbb{R}$$ which we usually call a matrix. Furthermore, the domain and target space of $$\det$$, i.e. $$\mathcal{M}_n(\mathbb{R})$$ and $$\mathbb{R}$$, are finite dimensional normed vector spaces. To investigate the continuity of $$\det$$, we have to choose norms on $$\mathcal{M}_n(\mathbb{R})$$ and $$\mathbb{R}$$. However, these normed spaces are finite dimensional and it does not really matter what type of norm we choose as they are all equivalent in such spaces. Consequently, we shall choose the Frobenius norm $$\lVert\cdot\rVert_F$$ on $$\mathcal{M}_n(\mathbb{R})$$ and absolute value $$|\cdot|$$ on $$\mathbb{R}$$ for further investigation. Then, continuity of determinant at $$A \in \mathcal{M}_n(\mathbb{R})$$ reads as

$$$$\forall\epsilon>0, \quad \exists\delta>0, \quad \lVert B-A \rVert_F < \delta \implies |\det B - \det A| <\epsilon.$$$$

The hard work is to obtain an upper bound for $$|\det B - \det A|$$ in terms of $$\lVert B-A \rVert_F$$. Fortunately, this can be done.

Lemma. Let $$A, B \in \mathcal{M}_n(\mathbb{R})$$. If $$\lVert B-A \rVert_F \leq 1$$ then we have $$$$|\det B - \det A|\leq M \,\lVert B-A \rVert_F,$$$$ where $$M = n!\sum_{m=0}^{n-1}\,C_m^n \lVert A\rVert^{m}_F$$ and $$C_m^n$$ is $$n$$ choose $$m$$.

Continuity of determinant follows imidiately from this lemma. More specifically, given any $$\epsilon > 0$$, some $$0 < \delta \leq \min\{1, \frac{\epsilon}{M}\}$$ would work.

Remark 1. There is an extension of the above lemma for multi-linear forms on normed vector spaces. Determinant can be considered as an n-linear form on the normed vector space $$\mathbb{R}^n\times\dots\times\mathbb{R}^n$$.

Remark 2. Note that the constant $$M$$ depends on $$A$$ so that determinant is not uniformly continuous. Indeed, this is the case for multi-linear forms on normed vector spaces.

Proof. Firstly, pay attention to some definitions and notation. We know that

$$$$\det A = \sum_{\sigma\in \mathcal{S}_n} \text{sgn}(\sigma) A_{\sigma(1),1}\dots A_{\sigma(n),n},$$$$ where $$\sigma:\{1,\dots,n\}\to\{1,\dots,n\}$$ is a permutation of $$(1,\dots,n)$$ and $$\mathcal{S}_n$$ is the set of all permutations. Also, let $$H:=B-A$$ be the change in $$A$$ and define the following index set $$$$I_m:=\{ (i_1,\dots,i_n)\,|\,i_j\in\{0, 1\},\,i_1+\dots+i_n=m \}.$$$$ This is no magic and will show up in finding the upper bound naturally but I just wanted to make the definition before getting into that. Now, let us dive into it.

\begin{align} |\det B - \det A| &= \Big|\sum_{\sigma\in\mathcal{S}_n}\text{sgn}(\sigma)\big(B_{\sigma(1),1} \dots B_{\sigma(n),n} - A_{\sigma(1),1} \dots A_{\sigma(n),n}\big)\Big| \\ &\leq \sum_{\sigma\in\mathcal{S}_n}\big|\text{sgn}(\sigma)\big|\big|B_{\sigma(1),1} \dots B_{\sigma(n),n} - A_{\sigma(1),1} \dots A_{\sigma(n),n}\big| \\ &= \sum_{\sigma\in\mathcal{S}_n}\big|(A_{\sigma(1),1}+H_{\sigma(1),1}) \dots (A_{\sigma(n),n}+H_{\sigma(n),n}) - A_{\sigma(1),1} \dots A_{\sigma(n),n}\big| \\ &= \sum_{\sigma\in\mathcal{S}_n}\Big|\sum_{m=0}^{n-1}\sum_{i\in I_m}\big(H_{\sigma(1), 1}^{1-i_1}A_{\sigma(1),1}^{i_1}\big)\dots\big(H_{\sigma(n), n}^{1-i_n}A_{\sigma(n),n}^{i_n}\big)\Big| \\ &\leq \sum_{\sigma\in\mathcal{S}_n}\sum_{m=0}^{n-1}\sum_{i\in I_m}\big|H_{\sigma(1), 1}^{1-i_1}\dots H_{\sigma(n),n}^{1-i_n}\big|\big|A_{\sigma(1), 1}^{i_1}\dots A_{\sigma(n),n}^{i_n}\big| \\ &\leq \sum_{\sigma\in\mathcal{S}_n}\sum_{m=0}^{n-1}\sum_{i\in I_m} \lVert H\rVert^{n-(i_1+\dots+i_n)}_F \lVert A\rVert^{i_1+\dots+i_n}_F \\ &= \sum_{\sigma\in\mathcal{S}_n}\sum_{m=0}^{n-1}|I_m|\lVert H\rVert^{n-m}_F \lVert A\rVert^{m}_F \\ &= \sum_{m=0}^{n-1} |\mathcal{S}_n| \,C_m^n \lVert H\rVert^{n-m}_F \lVert A\rVert^{m}_F \\ &=\lVert H \rVert_F \sum_{m=0}^{n-1} n!\,C_m^n \lVert H\rVert^{n-m-1}_F \lVert A\rVert^{m}_F \\ &\leq \lVert H \rVert_F \, \Big(n! \sum_{m=0}^{n-1}C_m^n \lVert A\rVert^{m}_F\Big), \end{align} where the last inequality holds provided that $$\lVert H \rVert_F \leq 1$$. This completes the proof.

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

• I didn't downvote you, but I'm curious if you can elaborate on why computable functions (in the relevant sense) are continuous? Commented May 26, 2012 at 0:10
• @IsaacSolomon It's essentially the finite use principal. See here for example. Commented May 26, 2012 at 18:23
• @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. Commented Jun 8, 2012 at 14:12
• @rschwieb Actually the notion of a computable function on $\mathbb{R}^{n}$ is a precisely defined one. See for example, Computable Analysis by Weihrauch. Commented Jun 9, 2012 at 15:12
• @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. Commented Jun 9, 2012 at 17:49