What *isn’t* the square root of -1?

I stumbled across an article trying to explain what the square root of -1 (j) is. It’s pretty cool that the author’s trying to provide motivation for a not-immediately-intuitive mathematical construct. Unfortunately, the author made a goof that makes his example fall apart.

The place where he goes off the rails is:

In other words, the eigen values, are all values such that:

Ax = ex

Where A is the matrix whose eigenvalues we are trying to find, and e is the scalar eigen-value we are trying to solve for. x is any non-null (non-zero) vector.

What he should have said is that x is some non-zero vector. If x can be any vector, then to find values of e such that Ax = ex, we would need to solve (AeI)x = 0 ⇒ (AeI) = 0 ⇒ A = eI for e. No such e exists.

The author appears to be missing the fact that every eigenvalue of a linear transformation has an associated set of eigenvectors. In this case, the set of eigenvectors associated with the eigenvalue j is all multiples of [j, 1]. For these, and only these vectors does multiplying by j give the same result as rotating 90°. It doesn’t work for, for example, [1,1].

There is a sense in which mutliplying by j is a rotation, but it doesn’t do much to motivate a need for j in the first place. If you represent a complex number as a 2-vector, multiplying by j will rotate the number 90° in the complex plane. I.e, if z = x + yj = rejθ, then jz = (by Euler’s identity) ejπ/2 z = ejπ/2 rejθ = rej(θ + π/2).

If I were trying to motivate the creation of j as the square root of -1, the first example that comes to mind is AC circuit analysis. Euler’s identity makes it easy to view complex numbers as a point in a wave with a given phase and amplitude. You can generalize the notion of resistance into impedance, with capacitors and inductors having imaginary values, and solve for the steady-state solution much the same as you would with just resistors in a DC system. Sure, you can do it without using complex numbers, but it gets clumsy fast.

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