Higher dimensional real matrix representation of imaginary unit?

Question

The imaginary unit can be represented as real $2 \times 2$ matrix:

$$ i = \begin{pmatrix} 0 & -1 \\ 1 & 0\\ \end{pmatrix}, \qquad i^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1\\ \end{pmatrix} $$

From my shallow knowledge of group representation theory, I believe that it should also be possible to find higher dimensional real representations of $i$. I tried to find a $3 \times 3$ representation, which is basically asking if there is matrix square root like

$$ \sqrt{\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0\\0 & 0 & -1 \end{pmatrix}} \sqrt{\begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0\\0 & 0 & -1 \end{pmatrix}} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0\\0 & 0 & -1 \end{pmatrix} $$

but couldn't find any (by try and error). Nevertheless I was able to find a $4 \times 4$ matrix which can be used as representation of $i$:

$$ \begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\0& 0 & 0 & -1\\ 0 & 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0\\0& 0 & 0 & -1\\ 0 & 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0\\0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix} $$

So can it be that $i$ can only be represented as real matrix of even dimensions $N=2n$, ($n>0$)? Can this be proven and maybe a construction be given on how to construct higher dimensional real matrix representations of the imaginary unit? And can it be shown, that the $2 \times 2$ representation is really the lowest dimensional representation of $i$ as a real matrix?

Answer 1

Suppose $T$ is an $n\times n$ matrix such that $T^2=-I$. Note then then $\mathbb{R}^n$ becomes a vector space over $\mathbb{C}$ (extending the $\mathbb{R}$-vector space structure), by letting $a+ib\in\mathbb{C}$ act by $aI+bT$. But then by picking a basis for this vector space over $\mathbb{C}$, we have $\mathbb{R}^n\cong \mathbb{C}^m$ for some $m$. As a real vector space, $\mathbb{C}^m\cong \mathbb{R}^{2m}$, and so it follows that $n=2m$ is even.

Conversely, for any $m$, $\mathbb{C}^m$ is a $2m$-dimensional real vector space, and multiplication by $i$ is an $\mathbb{R}$-linear map whose square is $-1$. So such a matrix does exist whenever $n$ is even. Explicitly, this is just the obvious generalization of your $4\times 4$ matrix, a block diagonal matrix with the $2\times 2$ diagonal matrix for $i$ on the diagonal blocks.

Answer 2

I hope Catalin Zara's comment has convinced you that you can only find a $n\times n$ matrix $M$ with $M^2=-1$, if $n=2k$. Here I will construct such $M$ in full generality.

Let us make a few definitions. $$ \Sigma_{(p,q)} = \begin{pmatrix} 1_{p\times p} &0\\ 0 & -1_{q\times q} \end{pmatrix}, \qquad Y=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} $$ where $\Sigma_{(p,q)}$ is $(p+q)\times (p+q)$ , while $Y$ is $2\times 2$. Now, if you use Jordan normal form for real matrices, then with a little bit of work, you find that if $M$ satisfies $M^2=-1$, then its real Jordan normal form is necessarily of the form $J = \Sigma_{(p,q)}\otimes Y$ where $p+q=k$ and $\otimes$ is the Kronecker product. As a result, $M^2=-1$ if and only if $$ M = P^{-1}(\Sigma_{(p,k-p)}\otimes Y) P $$ where $P$ is an invertible real matrix and $0\leq p\leq k$. You can think of $p$ as "signature" of $M$, which classifies matrices with $M^2=-1$, up to a similarity relation.

Answer 3

Forward:

I'm not specifically trying to get into the semantics of being overly pedantic on formal proofs. I am providing this not as a direct answer or solution to the problem. This is only to demonstrate the relationships and properties of what we do know, what we can observe from the numbers based on their properties and their relationships based on rudimentary or fundamental transformations derived from basic Identities.

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Recognizing Patterns within Relationships:

When it comes to values, vectors, matrices, polynomials, functions, etc. We can use many of the various Identities within mathematics to simplify our assumptions. I will give a list of a few well-known algebraic - trigonometric identities and their basic properties.

• Algebraic:

  • Additive Identities:

    • $A + 0 = A$
    • $0 + A = A$

  • Additive Inverse Identities:

    • $A - 0 = A$
    • $0 - A = -A$

From the algebraic Identities above, we can see that Addition is Commutative where Subtraction or its Additive Inverse is not. This is fairly straight forward and elementary.

Within Addition a linear transformation, translation to be precise. The order of operations does not matter. $A + 0 == 0 + A \implies A$

Within Subtraction a linear transformation order does matter since subtraction is not Commutative. $A - 0 \neq 0 - A :: A - 0 = A, 0 - A = -A$

What we need to consider from the basic properties of addition above is that for these two identities: $A + 0 \quad \&\&\quad A - 0$

Is that when the initial value of $A \neq 0$ being the $LHS$ operand, then the applied operators $\boldsymbol{\oplus}, \boldsymbol{\ominus}$ of the $RHS$ operand onto $LSH$ does not translate. Here, $0$ doesn't transform $A$.

These operations are Not Generative.

Within the cases of the other two Identities: $0 + A \quad \&\&\quad 0 - A$

Is that when the $LHS$ is $0$, then applying the value of the $RHS$ operand of $A$ onto $LHS$ provided $A \neq 0$, then a transformation does occur. This is Generative.

We know that subtraction can be defined in terms of addition through the multiplicative inverse identity:

• Multiplicative Identities:
  • $A * 0 = 0$
  • $A * 1 = A$
  • $A * -1 = -A$
  • $1 * 1 = 1$
  • $1 * 0 = 0$
  • $1 * -1 = -1$
  • $-1 * -1 = 1$

From these we can also establish other well-known Identities, Here' are a few:

• Common Identities:
  • $\sin^2 x + \cos^2 x = 1$
  • $e^{i \pi} + 1 = 0$
    • $e^{i \pi} = -1$
    • $(e^{i \pi})^2 = 1$
  • $\frac{sin(\pi/4)}{\cos(\pi/4)} = \tan(\pi/4) = 1$

Here we can see a direct relationship between linear and trigonometric properties we can see direct patterns within the trig functions with respect to the linear expressions where $\tan(\theta)$ is primarily the slope m within the slope-intercept form of the line: $y = mx+b$.

To simplify things to a base case, we can set b to 0. Then the linear equation above reduces to $y = mx$.

We can establish that $m = \tan(\theta)$ from the slope formula $\frac{rise}{run}$ from any two coordinate points $P_1 = \left<x_1,y_1 \right>, P_2 = \left<x_2, y_2\right>$ on that line defined as:

$$ m = \frac{P_1}{P_2} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{dy}{dx} = \frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta)$$

From the above, we can establish that for some specific angles of $\theta$: We have specific patterns that show up in the slope m or within the $\tan$ function.

When we apply an angle of $0^\circ$ we have the following form: $$ \tan(0^\circ) = \frac{\sin(0^\circ)}{\cos(0^\circ)} = \frac{0}{1}$$

When we apply an angle of $90^\circ$ we have the following form: $$ \tan(90^\circ) = \frac{\sin(90^\circ)}{\cos(90^\circ)} = \frac{1}{0}$$

Now, before others start to make the comment or complain that we have division by 0 and that it's undefined. Do not think of this in terms of direct division. It's not undefined as one would think, but that's not the argument here.

The direct applications of applying transformations, and the emerging results: Identities are being shown. Here we have 2 identities in ratio proportional form:

• $\tan(0^\circ) = \frac{0}{1}$
• $\tan(90^\circ) = \frac{1}{0}$

This is what I like to call the Orthogonality between $\sin\&\&\cos \ \| \ 0\&\&1$ respectively.

When we take the unit vector $\overrightarrow {\left<1,0\right>}$ and rotate it by $0^\circ$ There is no translation.

This is congruent to, $1 + 0$, $1 - 0$, $1 * 1$ which are identities. The result of this rotation with respect to the slope or tangent, resolves to $0$, it converges (decreasing aspect of the even function of cosine). We end up with the same vector in the same orientation.

When we take the unit vector $\overrightarrow {\left<1,0\right>}$ and rotate it by $90^\circ$ There is a translation.

This is congruent to, $0 + 1$, $0 - 1$, $1 * A: A > 1$ which are also identities. The result of this rotation with respect to the slope or tangent resolves to $\pm\infty$, it diverges (increasing aspect of the odd function of cosine). We translated the unit vector above we end up with two distinct unit vectors respectively.

• $ \overrightarrow {\left<0,1\right>} $
• $ \overrightarrow {\left<0,i\right>} $

Now, considering the fact that there is a direct relationship between the Cosine function and the Dot Product and knowing that we can find the Dot product between any two vectors in any Dimensional Space, 2D or greater and knowing that the Cosine is Even, Decreasing and Converges.

I think this makes perfect sense of why we see this only occurring in Even Dimensions. This corresponds perfectly with the symmetry of the polynomials with even integer exponential orders.

In some ways we can think of this as the following:

• Cosine governs: $x^2, x^4, x^6, ...$

Then when it comes to the Cross Product, we only see this within the 3rd dimension, and in some special cases within the Hamiltonians (Octonions) within the 7th dimension. This is indicative of the partial symmetry from the oddness of the Sine function with respect to being the vertical displacement of the tangent. Similarly, above as with the cosine and the Even Polynomials we can see the pattern that:

• Sine governs: $x^1, x^3, x^5, ... $

And this even and oddness of the cosine and sine functions, and their properties of decreasing and increasing as well as convergence and divergence is quite telling especially when their ratio is both the Tangent and the slope or gradient of Linearity.

The reason we don't see $i$ within the context of the Real numbers (Horizontal Numbers) is because $dx$ is symbolic of $\cos(\theta)$ in pure horizontal translation when $dy$ or $\sin(\theta)$ resolves to $0$.

When we reverse their roles by rotating by $90^\circ$, Then $dx$ goes to $0$ and $dy$ becomes constant and since $dy$ is related to $\sin(\theta)$ and being that it is an ODD function. This is why we end up with two-unit vectors of $\left<0,1\right>$ and $\left<0,i\right>$ respectively. One within the real plane and the other within the complex.

The OP had asked this:

 "So can it be that i can only be represented as real matrix of even dimensions N=2n(n>0)?"

I think this table of the orders if $i$ might help to see the patterns:

i^N	Deg	Res
$i^0$	$0^\circ$	$1$
$i^1$	$90^\circ$	$i$
$i^2$	$180^\circ$	$-1$
$i^3$	$270^\circ$	$-i$
$i^4$	$360^\circ$	$1$

And through modulus arithmetic these values will repeat for all other positive integer exponents of $i$

We can see from the Even Exponents of $i$ the following list of results: $i^0$, $i^2$, $i^4$ $\implies$ $1, -1, 1$. Real Values.

We can see from the Odd Exponents of $i$ the following list of results: $i^1$, $i^3$ $\implies$ $i,-i$. Complex Values.

And these patterns repeat.

I believe or think it has to do with the direct relationship of the evenness of Cosine, the fact that it is decreasing from an initial value of $\left<0,1\right>$ or $\left<0,i\right>$ and converges, and how it relates to and defines the Dot Product between two vectors, the $dx$ component of the $\tan$. I also think this is indicative of why we are able to differentiate and integrate in the first place.

For any and all arbitrary values that are not $0$ are Orthogonal, Perpendicular, Normal, Tangent to $0$.

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Observational Takeaway:

Everything within mathematics regardless of the field or branch we are working with or in, regardless of if we are looking at it discretely or even abstractly the bottom line is that it all can be derived from (not differentiation) $0 + 1 = 1$. The ability to add one to zero, to apply a transformation to zero translating it that causes a state change. This translation, state change is what defines the relational properties of numbers and values in the first place. Numbers themselves are rotational. Once we have a value or magnitude of one even agnostic of its relational direction, we automatically have the construct of the unit circle.

Proof of the unit circle being an inherit property or construct that is embedded within the value of one.

$$\sin^2(\theta) + \cos^2(\theta) = 1$$

Based on the definition of the Circle:

$$(X-h)^2 + (Y-k)^2 = r^2$$

Where $\left<X,Y\right>$ is a point that lies on the circumference, $\left<h,k\right>$ is the center of the circle and $r$ is its radius.

From this general form or definition of the circle, we can apply the center of the circle $\left<h,k\right>$ to be at the origin, and we can set its radius to be $1$.

$$(X-0)^2 + (Y-0)^2 = 1$$ $$X^2 + Y^2 = 1$$

The above definition of the unit circle located at the origin is also a specific case of the well-known Pythagorean Theorem:

$$A^2 + B^2 = C^2$$

Where within the context of the unit circle, $C^2 = 1$

And this is where we get the Trigonometric Pythagorean Identity from.

$$\cos^2(\theta) + \sin^2(\theta) = 1$$

Maybe these relationships and properties that are being derived from basic identities might give some clarity towards what you are asking, towards your assumptions from what you are observing. Perhaps it could lead you into even answering, solving, proving or disproving your own questions, concerns. Think critically and challenge everything, put it to the test.