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 $M_n(\mathbb{R})$
Why is the determinant $M_n(\mathbb R) \to \mathbb R$ continuous?
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0Non-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. – 2012-05-17
4 Answers
$M_n(\mathbb R)$ 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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1I do not understand your expression at the right side of the inequality.also what do you mean by $\det (x_{i,j})$? – 2013-12-16
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 $\mathbb{R}$.
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3@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. – 2012-06-09