Construct real rotation planes for a skew-adjoint map

Try

A real matrix can preserve lengths during continuous motion without having a real eigenbasis. Start with the plane matrix R having rows (0,−3),(3,0). It sends the first unit vector e to 3f and the second unit vector f to −3e. The arrows turn perpendicular to the current direction; they do not stretch along a real eigenline.

Compute R²=−9I. This identity predicts a sine/cosine motion: a point initially at e moves as cos(3t)e+sin(3t)f. Its squared length is 1 for every real time t. The instantaneous velocity is R times its current position, which you can check by differentiating the two scalar functions. At t=π/6 it reaches f, a quarter-turn from its start.

We will prove that every real skew-adjoint map consists of such mutually perpendicular rotation planes, possibly together with directions where the map is zero. The plane construction, including the zero directions, is the main theorem; it is stronger than noticing a pattern in one 2×2 matrix.

Predict

In an orthonormal real coordinate system, skew-adjoint means Sᵀ=−S. Equivalently, ⟨Sx,y⟩=−⟨x,Sy⟩ for all vectors, where ⟨x,y⟩ is their inner product and ||x||=√⟨x,x⟩ is their norm. Substituting y=x shows ⟨Sx,x⟩=0: the velocity is perpendicular to the current position. This uses real scalars, since a real number equal to its negative is zero.

Predict whether every odd-dimensional skew-adjoint matrix must have a zero direction. If all nonzero behavior occupies two-dimensional planes, one direction must remain. But that argument depends on the theorem still to be proved. It is not legitimate to assume those planes already exist just because we can draw one.

Also predict whether a matrix satisfying J²=−I must be skew-adjoint for the ordinary dot product. Consider J with rows (0,−2),(1/2,0). Its square is −I, yet Jᵀ≠−J. A chosen inner product matters; algebraic squaring alone does not establish Euclidean rotation.

Experiment

The explorer deliberately includes that distinction. Its family is Jₐ=(0,−a;1/a,0), with a=1,2,3. Starting at (1,0), its motion is x(θ)=(cosθ,sinθ/a). Before changing a or the angle, predict its ordinary squared length cos²θ+sin²θ/a². For a≠1 this varies, so the trace is not a Euclidean circle. The weighted squared length xᵀGx with G=diag(1,a²) is nevertheless 1. The companion lesson will construct and justify compatible inner products. Do not apply the Euclidean skew-adjoint theorem to Jₐ unless a=1.

A quarter turn depends on how lengths are measured

Start at e₁ = (1, 0) and apply exp(θJ), where J has rows (0, −a), (1/a, 0). Since J² = −I, the result is x = (cos θ, sin θ/a). Predict whether x and Jx are perpendicular in the ordinary dot product before changing a.

Coordinate stretch a123 Angle θ in degrees

45°

Reset metric experiment

J = (0, −2; 1/2, 0). The metric matrix G = diag(1, 4) defines g(u,v) = u₁v₁ + 4u₂v₂. Angles in the formula use radians: θ = 45π/180.

Original coordinates

Coordinates that display the metric

Ordinary squared length xᵀx

0.625

Metric squared length xᵀGx

1

Ordinary dot product xᵀJx

-0.375

Metric product xᵀGJx

0

J²x = (-0.707107, -0.353553) = −x up to rounding. The off-diagonal entries of Jᵀ + J are -1.5: this J is not skew-adjoint for the ordinary dot product, even if a particular pair of vectors happens to have zero dot product. In contrast, JᵀG + GJ = 0 for every allowed a.

The general metric repair g₀(u,v) + g₀(Ju,Jv), starting from the ordinary dot product g₀, gives diag(1.25, 5). Here it is a positive scalar multiple of G, so it gives the same orthogonality.

Try a = 2 and θ = 90°. Ordinary squared length falls to 1/4 while metric squared length remains 1. Return to 45° to see nonzero ordinary dot product. Explain why J² = −I alone cannot guarantee Euclidean length preservation. Values use floating-point arithmetic and are rounded to six decimal places; the lesson proves the general identities. Controls are temporary and do not record completion.

For a separate exact worked example, take S with rows (0,−2,0,1), (2,0,−1,0), (0,1,0,−2), (−1,0,2,0). Verify Sᵀ=−S. Introduce the four unit vectors

e=(1,0,1,0)/√2, f=(0,1,0,1)/√2, g=(1,0,−1,0)/√2, h=(0,1,0,−1)/√2.

Their pairwise dot products are zero. Calculate all four images before reading the table. This identifies both the planes and their frequencies, not just a determinant or eigenvalue list.

Plane	First image	Second image	Positive frequency	−S² on plane
span(e,f)	Se=f	Sf=−e	1	I
span(g,h)	Sg=3h	Sh=−3g	3	9I

With Q having columns e,f,g,h, the matrix QᵀSQ is block diagonal with blocks (0,−1;1,0) and (0,−3;3,0). Orthogonality QᵀQ=I is essential: an arbitrary coordinate change would not automatically preserve the Euclidean metric in its displayed coordinates.

Understand

Let S act on a finite-dimensional real inner-product space. In formulas with an abstract space, S∗ means its adjoint; in an orthonormal real basis it is Sᵀ. Assume S∗=−S. We use the already established real self-adjoint spectral theorem: a self-adjoint map has an orthonormal eigenbasis with real eigenvalues. We will apply it to a different map H, not assume the desired conclusion for S.

Construct one plane from the positive square

Set H=−S²=S∗S. This is self-adjoint, and ⟨Hx,x⟩=||Sx||²≥0. Its eigenvalues are nonnegative: testing a unit eigenvector gives its eigenvalue as that nonnegative quadratic value. Also ker H=ker S, because Hx=0 forces ||Sx||²=0, and Sx=0 plainly implies Hx=0.

If S is not zero, H has a positive eigenvalue. Otherwise its orthonormal eigenbasis and all-zero eigenvalues would imply H=0 and then S=0. Write a positive eigenvalue as λ² with λ>0, and choose a unit eigenvector e. Define f=Se/λ. We have ||Se||²=⟨He,e⟩=λ², so ||f||=1. Also ⟨Se,e⟩=0, so e and f are perpendicular. Finally

Se=λf and Sf=S²e/λ=−λe.

Thus the span P of e and f is invariant. In the ordered basis (e,f), matrix columns record the two images, so the block is Rλ=(0,−λ;λ,0), not its negative.

The orthogonal complement P⊥ is invariant too. If z is perpendicular to e and f, then ⟨Sz,e⟩=−⟨z,Se⟩=0 and ⟨Sz,f⟩=−⟨z,Sf⟩=0. A subspace whose complement is also invariant is called reducing. This complement argument is the step that permits induction without losing the orthogonal decomposition.

Finish the real block theorem

Restrict S to P⊥. It remains skew-adjoint with the inherited inner product. Its dimension is two smaller. Repeat the construction whenever the restriction is nonzero; when it is zero, choose any orthonormal basis of that remaining space. Finite-dimensional induction proves an orthonormal decomposition into positive-frequency rotation planes and a zero block. Dimension zero has an empty basis; the zero map in positive dimension has only a zero block. Neither case needs division by a frequency.

Each rotation block has rank 2 and trivial kernel. Therefore the zero block is exactly ker S, rank S is even, and an odd-dimensional space has positive nullity. Repeated frequencies present no difficulty: the construction removes one plane at a time from a reducing complement. It does not assume every positive eigenvalue of H has multiplicity one; in fact the completed proof shows each positive eigenvalue λ² has even multiplicity.

The frequency multiset and nullity are intrinsic. The eigenvalues of H are λ² repeated twice for each frequency-λ plane, and zero once for each kernel direction. Orthogonal similarity preserves those multiplicities. Conversely, two such maps with the same positive-frequency multiset and nullity have matching ordered block forms, so composing their orthogonal basis changes gives an orthogonal similarity. Frequencies do not record a preferred plane orientation: reflecting one coordinate changes Rλ to −Rλ, which is orthogonally similar. Knowing H as a matrix need not recover S itself; R₃ and −R₃ both give H=9I but are unequal.

Rotations and complex eigenvectors

Define exp(tS)=Σ from k=0 to infinity of tᵏSᵏ/k!. This finite-matrix power series converges absolutely: in a submultiplicative matrix norm its terms are bounded by |t|ᵏ||S||ᵏ/k!, whose scalar sum is finite. The same estimates on a bounded t interval justify termwise differentiation. These elementary power-series facts, rather than a new spectral theorem, underlie the following formula.

On a plane, Rλ²=−λ²I. Separate the even and odd powers to obtain

exp(tRλ)=cos(λt)I+[sin(λt)/λ]Rλ.

Its rows are (cosλt,−sinλt), (sinλt,cosλt). This is an orthogonal rotation matrix by cos²+sin²=1. On the zero block the exponential is I, so kernel directions stay fixed. The complete block formula proves exp(tS) orthogonal for every t, preserving all inner products and lengths. On a plane starting at e, the motion is cos(λt)e+sin(λt)f and satisfies x′=Sx.

There is a useful converse. If exp(tA) is orthogonal for every t near zero, differentiate exp(tA)ᵀexp(tA)=I at zero. Since its derivative at zero is A, the product rule gives Aᵀ+A=0. A finite list of times or a visually circular sampled trajectory is insufficient for this derivative conclusion.

Over C, the same real map extends by complex linearity. On a rotation plane, e−if is an eigenvector with eigenvalue iλ, and e+if has eigenvalue −iλ. These two vectors are independent and span the complexified plane. Along with the zero-block basis they give a full complex eigenbasis. Thus a real skew-adjoint map is diagonalizable over C, but a nonzero one is not diagonalizable over R. Indeed, a real eigenvector v with real eigenvalue μ satisfies μ||v||²=⟨Sv,v⟩=0, so μ=0. A real eigenbasis would then force S=0. Real block diagonalization and real diagonalization are distinct conclusions.

Apply

Change the exact four-dimensional example to rows (0,−3,0,1), (3,0,−1,0), (0,1,0,−3), (−1,0,3,0). Keep the same e,f,g,h. Determine the two frequencies directly, then predict the motion from the standard first coordinate vector at t=π/4.

Review the changed planes and trajectory

Now Se=2f, Sf=−2e, Sg=4h and Sh=−4g. The initial standard vector is (e+g)/√2. Its trajectory is [cos(2t)e+sin(2t)f+cos(4t)g+sin(4t)h]/√2. At π/4 this is (f−g)/√2=(−1/2,1/2,1/2,1/2). Its squared norm is 1. The entire trajectory has norm 1 because the two plane components are orthogonal and each retains squared norm 1/2.

As a separate proof audit, append one zero coordinate to this 4×4 matrix. Explain why the resulting five-dimensional map has rank 4 and a one-dimensional kernel, why the fifth coordinate stays fixed under its exponential, and why none of its nonzero rotation frequencies becomes a real eigenvalue. These conclusions require the block theorem, not merely a plot of one starting point.

Continue

The real theorem identifies what skew-adjointness guarantees in a declared inner product: perpendicular invariant rotation planes plus a stationary kernel. The explorer’s Jₐ family also shows why the inner product must be named. An ellipse in ordinary coordinates may be a circle for a compatible weighted metric, without making its matrix Euclidean skew-symmetric.

Continue to complex structures and compatible metrics to construct that metric and interpret multiplication by i on real planes. Return to orthogonal decomposition and Riesz representation for orthogonal complements, or the self-adjoint spectral theorem for the only diagonalization prerequisite used in constructing H’s eigenvectors.