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I went through a linear algebra course and I'm a bit confused..

I think I understand the geometric interpretation of imaginary numbers - multiplying by $i$ results in rotation by $90$ degrees in so that $1$ becomes $i$ and so forth. And this is where that $i^2 = -1$ comes from.

And then there's the matrix representation of $i$, which I understand emerged from a later generalization of complex numbers. I interpret the matrix representation as transform function which basically projects the imaginary axis to the real axis. I've thought of it as something very similar to vectors, with the difference that with vectors I write:

$P = x\mathbf{\hat{i}} + y\mathbf{\hat{j}}$ where $\mathbf{\hat{i}} = (1, 0)$ and $\mathbf{\hat{j}} = (0, 1)$

..and with complex numbers I can write:

$C = a + bi$ where $i$ = $2\times 2$ matrix, which represents the same $90$ degree transform logic by transformation.

Correct? Or at least close?

Anyways, as I understand, both of these interpretations of $i$ are actually later than $i=\sqrt{-1}$ itself. Is there an earlier interpretation? How did those who invented imaginary number prove that $i = \sqrt{-1}$ in the first place?

Thanks!

Thomas Russell
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user315648
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    No one proved $i = \sqrt{-1}$, in fact it's better to say that $i$ was defined so that $i^2 = -1$. But yes, complex number algebra can be recast as a product of 2x2 matrix multiplication and addition via the following construction. Let $z = x+iy$, then we could think of $z$ as the matrix $\left(\begin{array}{cc} x & -y \ y & x\end{array}\right)$. This is really a rotation matrix in disguise because $x = r\cos(\theta)$ and $y = r\sin(\theta)$. So it's clear to see that multiplication of complex numbers is exactly like a rotation (with a scaling) in this way. – Cameron Williams Apr 25 '13 at 18:24
  • For more than a century after Bombelli, there was basically only formal manipulation. – André Nicolas Apr 25 '13 at 18:32
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    To get into why $i^2$ is defined to be $-1$ harkens back to algebra. Mathematicians were exploring solutions to polynomial equations and continued to come up against issues with certain kinds of polynomials. The classical polynomial being $x^2 + 1 = 0$. It's easy to see that no real number solves this because if you square any real number, you must get a nonnegative number. So mathematicians, wanting to develop formal solutions to such an equation defined the unit $i$ as being the solution to $i^2 + 1 = 0$. In fact, such a simple definition allows us to solve for the zeros of any polynomial. – Cameron Williams Apr 25 '13 at 18:34
  • I think the term 'imaginary number' is the root of much pedagogical misunderstanding. Time for a 'rebranding'... – copper.hat Apr 25 '13 at 18:48

3 Answers3

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As Cameron Williams' helpful comment clarifies: It wasn't a matter of "proving" that $i =\sqrt{-1}$, but more a matter of defining $i$ to represent the solution to $i^2 = -1$, as a means, for example, for solving polynomials like $x^2 + 1 = 0$.

You might find the following post helpful for help understanding $i$ in different contexts:

Community
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amWhy
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As already commented, $i^2=-1$ was never proven, but lies at the origin of the creation of the imaginary numbers. The polynomial equation $x^2+1=0$ has no real root, therefore $\mathbb{R}$ is not algebraically closed. In order to create an algebraically closed field extension of $\mathbb{R}$, you have to add a solution of this equation. If we call the solution $i$, it must of course have the property of $i^2=-1$. In order to see that adding $i$ is indeed sufficient to get to an algebraic closure, you might want to read up on the Fundamental theorem of Algebra (http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra).

A use of the matrix representation of complex numbers is to see how multiplication with a complex numbers acts on vectors in $\mathbb{R}^2 \simeq \mathbb{C}$.

Pedro
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Leon Hendrian
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$i$ wasn't proved, as much as it was conceived - invented, if you will, to solve a set of problems that previously did not have solutions. The most famous such problem (albeit not the first that was treated with this approach) is a second degree polynomial equation with no real roots

$ax^2+bx+c=0$ such that $\left(\frac{b}{a}\right)^2 - \frac{c}{a} < 0$

The equation $x^2 + 1=0$ is just one example of one such equation but once one defines the quantity $i$ to solve it, one can also specify solutions to any second-degree equation, and not just those with real roots. And not only that - you can also find all three solutions to cubic equations were only one root is real. And much more, of course...

When mathematicians started doing this it was not a matter of proving that $i$ solves $x^2+1=0$, so much as defining it to do so.

Tomas Aschan
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    This is not historically accurate. $i$ was conceived to solve cubic equations, not quadratic equations. In a quadratic equation, one can simply take the appearance of $i$ in the solution to mean that there is no solution, and this is how it was historically understood. It wasn't until the Cardano formula for the solution of the cubic gave real solutions in terms of complex arithmetic that $i$ was understood as a legitimate number that could be operated upon consistently. – MJD Apr 25 '13 at 18:41
  • @MJD: Note that I never stated that the second-degree equations were the first problems to be solved by means of imaginary numbers - but they're certainly the most famous ones. However, I see why make the interpretation - I'll edit to clarify. – Tomas Aschan Apr 25 '13 at 18:53
  • It wasn't even as tidy as that. Imaginary numbers were referred to as "sophistical numbers" (numbers for the sake of argument) which Cardano employed but was not terribly comfortable with. Bombelli demonstrated that one could construct a consistent arithmetic with them, but it wasn't until Euler's time that imaginary numbers became an object of serious pure mathematical study (and well into the Nineteenth Century before people in what we'd call mathematical physics found a use for them). – colormegone Apr 25 '13 at 18:54