# The Unapologetic Mathematician

## The Determinant

Let’s look at the dimensions of antisymmetric tensor spaces. We worked out that if $V$ has dimension $d$, then the space of antisymmetric tensors with $n$ tensorands has dimension

$\displaystyle\dim\left(A^n(V)\right)=\binom{d}{n}=\frac{d!}{n!(d-n)!}$

One thing should leap out about this: if $n$ is greater than $d$, then the dimension formula breaks down. This is connected with the fact that at that point we can’t pick any $n$-tuples without repetition from $d$ basis vectors.

So what happens right before everything breaks down? If $n=d$, then we find

$\displaystyle\dim\left(A^d(V)\right)=\binom{d}{d}=\frac{d!}{d!(d-d)!}=\frac{d!}{d!}=1$

There’s only one independent antisymmetric tensor of this type, and so we have a one-dimensional vector space. But remember that this isn’t just a vector space. The tensor power $V^{\otimes d}$ is both a representation of $\mathrm{GL}(V)$ and a representation of $S_d$, which actions commute with each other. Our antisymmetric tensors are the image of a certain action from the symmetric group, which is an intertwiner of the $\mathrm{GL}(V)$ action. Thus we have a one-dimensional representation of $\mathrm{GL}(V)$, which we call the determinant representation.

I want to pause here and point out something that’s extremely important. We’ve mentioned a basis for $V$ in the process of calculating the dimension of this space, but the space itself was defined without reference to such a basis. Similarly, the representation of any element of $\mathrm{GL}(V)$ is defined completely without reference to any basis of $V$. It needs only the abstract vector space itself to be defined. Calculating the determinant of a linear transformation, though, is a different story. We’ll use a basis to calculate it, but as we’ve just said the particular choice of a basis won’t matter in the slightest to the answer we get. We’d get the same answer no matter what basis we chose.

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## 7 Comments	»

1. [...] the Determinant Today we’ll actually calculate the determinant representation of an element of for a finite-dimensional vector space . That is, what [...]

Pingback by | January 2, 2009 | Reply

2. [...] Determinant of a Noninvertible Transformation We’ve defined and calculated the determinant representation of for a finite-dimensional vector space . But we [...]

Pingback by | January 14, 2009 | Reply

3. [...] polynomial of an endomorphism on a vector space of finite dimension , then we can get its determinant from the constant [...]

Pingback by | January 29, 2009 | Reply

4. [...] always unapologetic John Armstrong is currently on an exposition of the determinant, starting here and now here . This must be about the fifth time I’m relearning the determinant, each time [...]

Pingback by | January 30, 2009 | Reply

5. [...] of some appropriate — we have a homomorphism to the multiplicative group of given by the determinant. We originally defined the determinant on itself, but we can easily restrict it to any subgroup. [...]

Pingback by | September 8, 2009 | Reply

6. [...] we can use this to get back to our original definition of the determinant of a linear transformation . Pick a orthonormal basis for and wedge them all [...]

Pingback by | October 30, 2009 | Reply

7. [...] these matrices are exactly the Jacobian matrices of the functions! And since the by definition, the determinant of the product of two matrices is the product of their determinants. That is, we [...]

Pingback by | November 12, 2009 | Reply

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## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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