Let us begin with a brief recap of Lecture 23, which introduced a number of new concepts which may seem complex at first, but are in fact rather simple. Let be an -dimensional vector space; the dimension could be arbitrarily large, so visualizing geometry in can be arbitrarily hard. However, no matter what is, we have a naturally associated -dimensional vector space We can visualize any -dimensional vector space easily, by picturing it as a line. Algebraically, we associate to a given set of vectors of vectors in a point on the line namely the point The set is linearly independent if and only if the point is not the zero point on the line This is pure algebra. If is a Euclidean space, meaning that we can measure lengths and angles in using a scalar product, then the line is a Euclidean line, i.e. a line on which we have a notion of distance defined. The volume of the parallelepiped is then the distance from the point to zero on the line There are only two points at distance one from the origin on the line which we call and These are analogous to the numbers and on the number line Choosing one of these, say is what it means to choose an orientation on The oriented volume of the parallelepiped is then the unique number such that For example, if choosing means the number line is oriented left to right, and choosing means the number line is oriented right to left. In the former case, the oriented length of the line segment is while in the latter case it is That’s it, more or less.

In this lecture, we use the above machinery to define a function

which tells us when a given linear operators is invertible. This function is called the **determinant**. All the complexity (i.e. the hard stuff) is concentrated in the definitions from the past two lectures, and if you understand these, then the definition of the determinant is exceedingly simple.

Let be a given linear operator. We associate to the linear operator defined by

**Definition 1:** The determinant of is the unique eigenvalue of

Let us unpack this definition. The key point is that the vector space is -dimensional. This means that every operator in is simply multiplication by a number (see Problem 1 on Assignment 6). We are defining to be the number by which the operator scales its argument. That is, is the number such that

for all

**Theorem 1: **An operator is invertible if and only if

*Proof: *Suppose first that is invertible. Let be a linearly independent set in the -dimensional vector space Since is invertible, is also a linearly independent set in , and the linear independence of this set is equivalent to the statement that

is not the zero tensor. Since the above nonzero tensor is equal to

we have

Conversely, suppose that Let be a linearly independent set in The statement that is invertible is equivalent to the statement that is linearly independent, which is in turn equivalent to the statement that

is not the zero tensor. But

is the nonzero tensor scaled by the nonzero number , hence it is nonzero.

— Q.E.D.

If you already know something about determinants of matrices, you are probably wondering how Definition 1 relates to this prior knowledge; let us explain this now. Let be an orthonormal basis in and let

be the corresponding unit volume tensor in We then have that

this follows from the fact that is by definition of the eigenvalue of the operator together with the fact that is an orthonormal basis of Now, we can explicitly evaluate the above inner product in terms of the matrix elements of relative to the basis of Here is the computation:

We recall that is the set of permutations of the numbers , i.e. bijections For example, in the case this computation reads

This is the formula for the determinant of the matrix

which you likely learned in high school algebra (if you didn’t learn it then, you just learned it now).

To reiterate, for any operator the determinant is the constant such that

for all Taking the scalar product with a unit volume tensor on either side, this becomes

The scalar product on the LHS is the oriented volume of the parallelepiped while the scalar product on the right is the oriented volume of the parallelepiped Assuming that the latter volume is nonzero — i.e. that is a linearly independent set in — we thus have

which says that is the scalar factor by which the oriented volume of transforms when each of the vectors transforms by As an application of this geometric interpretation of the determinant, let us calculate the area of an ellipse whose semimajor and semiminor axes have lengths and , respectively. The basic observation is that such an ellipse is the image of the unit disc in the Euclidean plane under the linear transformation defined by

where and is the standard basis. The determinant of the operator is

Let be an extremely small positive number, and tile the unit disc with translated copies of the tiny square so that the tiling approximates the area of the disc up to an exceedingly small error. The oriented area of this tiny square is

The image of the tiny square under the transformation is the tiny parallelogram and the oriented area of this tiny parallelogram is

We conclude that the area of the parallelogram tiling which approximates the ellipse is the area of the square tiling which approximates the circle scaled by and hence that the area of the ellipse equals the area of the disc scaled by Since the area of the unit disc is the area of the ellipse in question is

I believe there is a typo in the expression above “This is the formula for the determinant of the 2 \times 2 matrix”.

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Corrected, thanks.

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In the second-to-last sentence of the first paragraph, I believe “choosing -omega = -1” should be changed to “choosing omega = -1.”

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Corrected, thanks.

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