Rotation in 2D by an angle $t$ can be performed using $$R=\begin{pmatrix}\cos(t) &-\sin(t) \\ \sin(t) & \cos(t)\end{pmatrix}$$ matrix. But, if I want to rotate a point or vector in 4D, is there any rotation matrix in explicit form?
I have read rotation about planes in 4D (Rotating two planes while the other two planes remains constant), but I am interested in rotation around an axis. More specifically the Quaternion 4D rotation matrix.
Kindly help me out.
Best Answer
The reason you haven't read about "rotation about an axis" in 4D is that there is no such thing. Perhaps I should say "the reason is that any rotation that leaves fixed a single axis also leaves fixed another axis orthogonal to it, and hence leaves fixed an entire plane." You were probably hoping to find something with the property that $$ Rv = v $$ for any $v$ in the "axis" $L$, and the property that $$ Rw \cdot w = \cos \theta $$ for any vector $w$ orthogonal to $L$, i.e., all vectors $w$ in the plane orthogonal to $L$ are rotated by an angle of $\theta$. But, alas, there is no such thing.
The trick is to understand that our standard way of describing rotations in 3-space is slightly flawed: instead of saying that "we rotated 20 degrees about the $z$ axis", for instance, we could say "we rotated in the $xy$-plane by $20$ degrees" (i.e., naming the plane of rotation rather than the single axis that happens to be orthogonal to the plane). If we did the latter, then generalizing would work fine: we could talk (in 4-space) about rotating in the $xy$-plane, which leaves two perpendicular dimensions ($z$ and $w$) fixed. Such rotations are relatively easy to write down, but they do not describe all possible rotations in 4-space: there are also ones that rotate by some amount in a plane spanned by vectors $v_1, v_2$, and by some (possibly different) amount in a plane spanned by vectors $u_1, u_2$, with the $u$s orthogonal to the $v$s.
It's not too tough to write down proofs of these claims, at least if you know about eigenvalues and eigenvectors; if you want, I can expand my answer to do so. But I think the main contribution that I can make is to help you understand that you're looking for something that (provably) does not exist, and perhaps you should attempt to gain a different understanding.
Post-comment additions: @Widawensen asked that I expand a bit on my remarks, so I'll do so, giving proofs for a few things I've said.
First: Throughout, $R$ will be a rotation in $4$-space, or the $4 \times 4$ matrix of this rotation with respect to the standard basis.
Claim 1: If, for some nonzero vector $v$, we have $Rv = v$, then there's another vector $w$ orthogonal to $v$ with $Rw = w$. In other words: if a rotation fixes all points on a single axis (the span of the vector $v$), then it actually fixes all points on an entire plane.
Proof: A rotation matrix preserves lengths, so its eigenvalues must all be complex (or real numbers) of absolute-value 1. We know, since $Rv = v$ and $v$ is nonzero, that one of the eigenvalues is 1. So the eigenvalues are $1, a, b, c$. Either all of $a,b, c$ are real (in which case they're all $\pm 1$), or some of them are complex. In the real case, since the determinant of a rotation is $+1$, we must have $abc = +1$, hence either 0 or two are $-1$.
Case 1: all four eigenvalues are $+1$: then $R$ is the identity, and any vector orthogonal to $v$ can be chosen as $w$ and we're done.
Case 2: The evalues are $+1, +1, -1, -1$. Since $+1$ is an eigenvalue twice, the eigenspace $E$ corresponding to $+1$ is two-dimensional$^{**}$, and it evidently contains $v$. Extending $v$ to an orthogonal basis $ \{v, w\}$ of $E$ via Gram-Schmidt, we get our vector $w$ and we're done.
If there are complex eigenvalues, the situation is similar to the second case: Because the characteristic polynomial is a real polynomial (no imaginary numbers!) the complex eigenvalues occur in complex conjugate pairs, so we have
Claim 2: Rotations that fix all points of some plane in 4-space do not actually exhaust all rotations of 4-space: there are rotations of 4-space that move the points of every 2D plane.
Proof: Look at the matrix \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & c & -s \\ 0 & 0 & s & c \end{bmatrix} where $c = s = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$.
The eigenvalues of this matrix are $-1, -1, c+is, c-is$. In particular, the number $1$ is not an eigenvalue, so there is no vector $v$ with $Rv = v$.
Claim 3: Every rotation $R$ in 4-space is a product $R = ST$, where $S$ is a rotation that fixes all points of a plane $A$, and $T$ is a rotation that fixes all points of a plane $B$, and the planes $A$ and $B$ are orthogonal, in the sense that for any vectors $a \in A$ and $b \in B$, we have $A \cdot b = 0$.
Proof: Again this is by cases, based on the eigenvalue structure. The five possible eigenvalue sets are $(1,1,1,1)$, $(1,1,-1, -1)$, $(-1, -1, -1, -1)$, $(1, 1, a+bi, a-bi)$ (where $b \ne 0$ and $a^2 + b^2 = 1)$, $(a+bi, a-bi, c+di, c-di)$, where $b, d \ne 0$ and $a^2 + b^2 = c^2 + d^2 = 1$. We'll examine these one at a time:
Case 1: $(+1, +1, +1, +1)$. $R$ is the identity, and can be factored as $I$ \cdot $I$, where the first $I$ fixes all points of the $xy$-plane, and the second fixes all points of the $zw$-plane, so that these two planes serve the roles of $A$ and $B$.
Case 2: $(+1, +1, -1, -1)$. The eigenspace for $+1$ is 2-dimensional, so pick an orthonormal basis $v_1, v_2$ for it. Extend this, by Gram Schmidt, to an orthonormal basis for 4-space, $v_1, v_2, v_3, v_4$. Because the eigenspaces for $+1$ and $-1$ must be orthogonal, the span of $v_3, v_4$ must be the eigenspace for $-1$. Let $M$ be the matrix whose columns are $v_1, v_2, v_3, v_4$, and $D$ be the diagonal matrix with diagonal entries $-1, -1, +1, +1$. Then $S = M D M^t$ negates $v_1$ and $v_2$ and fixes $v_3$ and $v_4$. Picking $T = I$, we have $R = S T$, and we're done.
Case 3: $(+1, +1, a+bi, a-bi)$. The eigenspace for $+1$ is 2-dimensional, so pick an orthonormal basis $v_1, v_2$ for it. Extend this, by Gram Schmidt, to an orthonormal basis for 4-space, $v_1, v_2, v_3, v_4$. Let $M$ be the matrix whose columns are $v_1, v_2, v_3, v_4$. Look at $$ Q = M R M^t $$ This matrix fixes $e_1$ and $e_2$ and has the same eigenvalues as $R$ (because $M$ is a rotation). Since it's a rotation, its columns must be orthogonal. That means that $Q$ must have the form $$ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & * & * \\ 0 & 0 & * & * \end{bmatrix} $$ where the $2 \times 2$ block at the lower right is actually an 2D rotation. That means it has the form \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} for some $\theta$ and has eigenvalues $\cos \theta \pm i ~ \sin \theta$. But we know that its eigenvalues must be $a \pm bi$. So the actual form of $Q$ must be $$ Q = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & a & -b \\ 0 & 0 & b & a \end{bmatrix} $$ (or its transpose).
This matrix factors into $I \cdot Q$, where $I$ fixes the $zw$-plane, and where $Q$ fixes all points of the $xy$ plane. So we have
$$ M R M^t = IQ $$ Conjugating by $M$, we get \begin{align} M^t (M R M^t) M &= M^t (IQ) M \\ (M^t M) R (M^t M) &= M^t (IMM^tQ) M \\ R &= (M^t IM) (M^tQ M) \end{align} which shows that $R$ is a product of something that fixes the $v_3, v_4$ plane (namely $M^t I M = I$) and something that fixes the $v_1, v_2$ plane (namely $M^t Q M = R$). The key point is that $M^t M = M M^t = I$, because its columns forma an orthonormal basis for 4-space.
Case 3: $(-1, -1, -1, -1)$. In this case, $R$ is just the negative identity, and factors nicely as $$ R = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. $$
Case 4: Like case 2. $R$ is (up to conjugation by a matrix $M$ whose columns are its eigenvectors) just the matrix of "Claim 2" above.
Case 5: Similar, except that there's no single plane that's either pointwise fixed or pointwise negated: instead, there are two perpendicular planes that are each rotated by some non-multiple-of-$\pi$ amount.
Footnote: $^{**}$ FShrike points out that it's not obvious that just having $+1$ as an eigenvalue twice means that there are two eigenvectors with eigenvalue $1$. After all, consider the matrix $\pmatrix{1 & 1 \\ 0 & 1}$, right? In general, all we can say is that the generalized eigenspace for $\lambda = +1$ is two-dimensional. That gives us an eigenvector $v$ (which we might as well adjust to have length 1) for the eigenvalue $+1$, and a second independent vector $u$ for which (1) $Au \in span \{v, u\}$, (2) $(A-I)u$ may be nonzero, but (3) $(A-I)^2 u$ is zero. By applying Gram-Schmidt to the pair $\{v, u\}$, we get a pair $\{v, w\}$, where the three properties above also hold for $w$.
By item 1, we know that $Aw = cv + dw$ for some $c, d$, so $(A-I)w = cv + (d-1)w$. Applying this a second time gives us \begin{align} (A-I)^2 w &= c (A-I)v + (d-a)(A-I)w \\ &= 0 + (d-1)[cv + (d-1)w]\\ &= c(d-1)v + (d-1)^2w \end{align} which must be $0$, and because $v$ and $w$ are independent, we see that $(d-1)^2 = 0$, i.e. $d = 1$. Now look at $Aw = cv + w$. Because $A$ is orthogonal (which means length-preserving), the left-hand side has length $\|Aw\| = \|w\| = 1$. The right-hand side (remember that $v$ and $w$ are orthogonal and unit length, because of Gram Schmidt) has length $\sqrt{c^2 + 1}$. Hence $c = 0$, and we've shown that the second generalized eigenvector, $w$, is actually an eigenvector, so we're done.
[As @FShrike notes, if you know the spectral theorem, this is all a good deal easier.]