# Interpretation of the determinant of an operator in complex vector space

1. Aug 8, 2012

### dEdt

There's a geometric interpretation of the determinant of an operator in a real vector space that I've always found intuitive. Suppose we have a n-dimensional real-valued vector space. We can plot n vectors in an n-dimensional Cartesian coordinate system, and in general we'll have an n-dimensional parallelepiped. If we apply an operator to these n vectors, the result is a new parallelepiped whose "n-dimensional volume" is equal to the volume of the original parallelepiped time the (absolute value of the) determinant of the operator.

Is there an extension of this interpretation for an operator in a complex vector space?

2. Aug 8, 2012

### DonAntonio

Yes: exactly the same, as you can think of an n-dimensional complex space as an 2n-dimensional real space...:)

DonAntonio

3. Aug 8, 2012

### christoff

DonAntonio's argument seems correct. However, personally, I have a bit of trouble getting a handle on the idea that C^n and R^2n are the same in the sense of vector spaces (The same in what sense? Different field, so they aren't isomorphic as vector spaces. Are the norms equivalent?). Although I'm sure it is correct, I still have a bit of trouble believing it since I haven't seen it in the two linear algebra courses I've taken.

If you feel as I do, then try this out... draw a picture without even considering it as a 2n-dimensional real space. Start by looking at the complex numbers as a vector space over themselves.

A linear operator in this space corresponds simply to scaling by a complex number. But multiplication of complex numbers is associative under taking absolute values (ie. $| cx | = | c || x |$ ). The determinant of such an operator will simply be the scaling constant, and since C over itself is 1-dimensional, it follows that the 1-dimensional paralellepiped (ie. the line) has its 1-dimensional volume (length) scaled by the absolute value of the determinant.

Higher dimensions get harder to visualize, but the principle is the same.

Last edited: Aug 8, 2012
4. Aug 9, 2012

### DonAntonio

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They are the same (i.e., isomorphic) as real vector spaces since they both have dimension $\,2n\,$ over the reals. In this sense.

DonAntonio

5. Aug 9, 2012

### dEdt

To make sure I understand: if we're dealing with the vector space $\mathbb{C}^1$, then an operator with determinant 2 would map a region in the complex plane to a new region with twice the area?

And what happens if the determinant is complex?

6. Aug 9, 2012

### christoff

To clarify my misunderstanding, the op was asking specifically about complex vector spaces, which usually refers to vector spaces where the field is the complex numbers.

OK... if you're talking about the vector space $\mathbb{C}^1$ as a vector space over the real numbers, then that is exactly what would happen. However, this vector space is 2-dimensional, so your parallelepipeds are paralellograms in the complex plane.

If however you're talking about $\mathbb{C}^1$ as a vector space over itself, then this vector space is 1 dimensional, so your paralellepipeds are simply vectors in the complex plane.

If the determinant is complex (as would be the case for the latter situation, since your field is the complex numbers), then the operator would scale the length of the vector by a factor of the absolute value of the determinant. Say you have an operator L defined on this space, and $\det L=a+bi$. Then the absolute value is simply given by the formula you learned in first year linear algebra: $\sqrt{a^2+b^2}$.

7. Aug 9, 2012

### dEdt

Perfect, thank you.