One of the most interesting functions in mathematical analysis and calculus is the exponential function, .

For one, this is the only non-trivial function that is it’s own derivative (the trivial function would be ). That is, it is the solution to this very simple differential equation:

, with initial conditions

(Reminder: You solve it by separating the variables and then integrating. See here for a demonstration).

This “definition” of the exponential has interesting implications when you turn away from a simple equation and look toward systems of equations.

Consider this matrix equation:

… where is in the place of the vector , is in the place of the vector , goes from 1 to n, and is just any matrix.

The *form* of this matrix equation is identical to the differential equation written above. Indeed, you might want to just say that it’s a generalization to arbitrary dimensions. It might also invite you to suppose, as was done previously, a solution involving exponentials. Except this time you must account for the vectorized nature of the equation. So try this:

, where is unknown at this point.

Differentiating with respect to t, we get:

.

Plugging into our matrix equation:

.

Since the exponentials can’t attain zero, we can divide them out, getting the following:

This particular equation is the so-called eigenvalue problem. Waxing ecstatic about the eigenvalue problem is all too easy, because it’s quite a profound little tool, both in mathematics and in science. It gives rise to the ideas of operator calculus, wherein one thinks of a complicated system of equations as an operator acting on a simpler system of equations. This allows you to reduce the system so that it’s (hopefully) solveable.

In the case of the eigenvalue problem, we’ve transformed a system of differential equations into a matrix operation that simply scales any vector it’s applied to by . Solving this problem involves application of what universities sometimes call ‘linear algebra’. Solving it involves knowledge which I’ll slightly gloss over, but which you can look up easily in a linear algebra text or online.

Subtracting everything to one side, we get:

Using the fact that , for any vector , where is the identity matrix, we can write this equation as:

Removing the common factor (which requires the previous step; can’t subtract a vector from a matrix!), the equation resolves to:

Now, we wish this equation to have a non-trivial solution–that is, one where .

There’s a particular theorem in linear algebra (sometimes called the Invertible Matrix Theorem) which will guide us here. First, a quick definition.

A matrix is said to be **invertible** if there exists a matrix such that . The matrix is called the inverse of and is written .

The Invertible Matrix Theorem states that a matrix is invertible if and only if the matrix equation has only the trivial solution . Which means if we want a non-trivial solution out of our equation, we need to ensure that , which we’ll just call , is not invertible. To ensure this, let’s take a look at a hypothetical inverse for .

Assume, for a moment, that is the following:

You may recall that if has an inverse, it must be:

This rule generalizes to arbitrary finite dimensions. Which means if , then the matrix is not invertible, because you’re dividing by zero. Horrah! (Here, is the determinant, in case you have not seen it before).

So we get this equation:

The left side of this equation turns out to be a polynomial in *a*. Solving for its zeroes gives you your different scale factors, called **eigenvalues** (these are your *a*‘s). What do you do with them? Plug ’em back into the equation and solve for your ‘s!

Since you already know an expression for in terms of , you can now solve for various values of .

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