Preserve geometry and diagonalize compatible operators

Try

A rotation preserves lengths while a stretch changes them. The matrix U=[[0,−1],[1,0]] satisfies UᵀU=I, and the complex diagonal matrix diag(i,−1) satisfies U*U=I. Both preserve the standard inner product. Calculate the images of two perpendicular unit vectors in each case. Their coordinates change, but their lengths and mutual inner product do not. In a complex space such maps are called unitary; over R they are called orthogonal.

Predict

Predict whether commuting diagonalizable operators always admit a common orthonormal eigenbasis. Ordinary diagonalizability alone is insufficient: even one nonnormal operator may have no orthonormal eigenbasis. The useful theorem requires normality as well as commutation over C. Conversely, matrices diagonal in the same orthonormal basis certainly commute, because scalar diagonal entries multiply independently. Identify exactly where each hypothesis enters the forward proof.

Experiment

Use A=diag(2,2,5) and B=[[1,1,0],[1,1,0],[0,0,7]]. They commute because B only mixes coordinates inside the repeated 2-eigenspace of A. The vectors (1,1,0)/√2, (1,−1,0)/√2 and (0,0,1) are common orthonormal eigenvectors. List their eigenvalue pairs as (2,2),(2,0),(5,7). Now insert a nonzero entry connecting the first and third coordinates of B and compute AB−BA to see why mixing different A-eigenspaces breaks commutation.

Understand

U is unitary when U*U=UU*=I. In equal finite dimensions, U*U=I already implies injectivity and hence invertibility, so the second identity follows. Such a map preserves inner products and lengths, and every eigenvalue has magnitude one by applying norm preservation to an eigenvector. For commuting normal operators A and B, each eigenspace of A is B-invariant. Because these eigenspaces form an orthogonal direct sum and all are B-invariant, B is block diagonal in that decomposition; so is B*. Each restricted block is normal. Apply the spectral theorem inside each block to obtain a common orthonormal eigenbasis. The argument extends to a finite commuting family by repeating the refinement.

Apply

Every square matrix A also has a polar decomposition A=UP with P=(A*A)¹ᐟ² positive semidefinite and U unitary. Define a map from im P to im A by Px↦Ax. It is well-defined because ker P=ker A, and it preserves inner products since ⟨Ax,Ay⟩=⟨Px,Py⟩. Extend orthonormal bases of these equal-dimensional images to full bases; map the added vectors correspondingly to extend the isometry to a unitary U. Then UP=A. If A is invertible, P is invertible and U=AP⁻¹ is unique. If A is singular, the extension on ker P is not determined, though the isometry on im P is fixed. For rectangular maps one generally uses a partial isometry instead of a square unitary factor.

For A=diag(−2,5), polar decomposition gives P=diag(2,5) and U=diag(−1,1). The positive factor stretches, and the orthogonal factor reflects one direction. Compute their product and note that a negative eigenvalue of A becomes a positive stretch plus a sign change, not a negative singular value. This is the simplest exact example connecting polar decomposition to SVD.

Reconstruct a nonnormal map from its Gram matrix

Start with perpendicular unit vectors v₁=(3,4)/5 and v₂=(−4,3)/5. Stretch them by 3 and 1, then rotate by a quarter-turn. In standard coordinates P has rows (43,24)/25 and (24,57)/25, and A has rows (−24,−57)/25 and (43,24)/25. Direct multiplication gives AᵀA=P². The longest output of a unit input has length 3, attained at ±v₁. A itself need not be symmetric or normal: stretch axes are eigenvectors of AᵀA, not generally eigenvectors of A.

Separate stretching, orientation and lost information

Predict the longest output before changing t. Input axes are v₁=(3,4)/5 and v₂=(−4,3)/5. The matrix applies stretches 3 and t along these axes, followed by a quarter-turn. The positive factor below is recovered from AᵀA, not copied from its diagonal description.

Second stretch t 01234 Unit input v₁v₂(v₁+v₂)/√2 Kernel extension sign (available only at t=0) +1−1

Rank 2; operator norm 3; selected output length 3.000000. Reconstruction discrepancy 2.22e-16 (floating-point arithmetic).

Matrix	First row	Second row
A	-0.960000, -2.280000	1.720000, 0.960000
AᵀA	3.880000, 3.840000	3.840000, 6.120000
Positive P	1.720000, 0.960000	0.960000, 2.280000
Canonical W	0.000000, -1.000000	1.000000, 0.000000
Orthogonal extension U	0.000000, -1.000000	1.000000, 0.000000

The support is the whole input space; W=U is unique. At t=3 every input direction maximizes length; otherwise choose the larger stretch axis.

Reset

This bounded two-dimensional experiment uses temporary state. Numerical agreement supports these examples; the general uniqueness proof is separate.

Recover the positive square root without guessing its axes

For a nonzero positive semidefinite real symmetric two-by-two Gram matrix G, let δ=√det(G) and s=√(tr(G)+2δ). Then P=(G+δI)/s. Cayley–Hamilton gives G²−tr(G)G+det(G)I=0. Expanding (G+δI)² therefore yields (tr(G)+2δ)G, so P²=G. Every eigenvalue of G+δI is nonnegative and s is positive, so P is positive semidefinite. The positive square-root uniqueness theorem identifies it as the required factor. If G=0, use P=0 separately: the displayed division is undefined. For invertible A recover U=AP⁻¹; for singular A never invert P.

The canonical factor is a partial isometry

Let A map a finite-dimensional real or complex inner-product space V into W, and P=(A* A)¹ᐟ². Orthogonal decomposition gives V=im(P) ⊕ ker(P). On im(P), define J(Px)=Ax, and on ker(P) define J=0. The existing well-definedness and inner-product calculation show that J is an isometry on im(P). Consequently A=JP, J*J is the orthogonal projection onto im(P), and JJ* is the projection onto im(A). To justify the last statement, J maps its support isometrically onto im(A); its adjoint is the inverse there and kills the orthogonal complement of that image.

This J is unique among factors satisfying A=JP and ker(J)=ker(A): the equation fixes J on im(P), while the kernel condition fixes zero on its complement. This is the canonical partial isometry. A square unitary extension instead sends kernel vectors to the orthogonal complement of im(A), and need not be unique. At t=0 in the experiment, changing the extension sign changes U but not UP; J remains fixed. For a rectangular map, a norm-preserving extension V→W exists exactly when dim(V)≤dim(W): necessity follows from injectivity, and sufficiency follows by extending the two support bases with enough orthonormal output vectors. Such an extension satisfies U*U=I, but need not satisfy UU*=I. Do not call a rectangular isometric embedding unitary.

Norm, maximizing directions and evidence

For every x, ‖Ax‖²=⟨P²x,x⟩=‖Px‖². Expanding x in an orthonormal eigenbasis of P shows that the norm of A is the largest eigenvalue of P. Equality for a unit vector holds exactly when it lies in the corresponding eigenspace. In the experiment, t below 3 selects v₁, t above 3 selects v₂, and t=3 makes every unit vector a maximizer. At t=0, the second axis is lost; an arbitrary unitary extension does not restore information in A.

Write the positive-factor reconstruction, its zero/rank hypotheses and the maximizing direction before reading the ledger. Then set t=0 and compare both extensions on the lost input v₂. Explain why UP stays fixed even though Uv₂ changes, and why the canonical Jv₂ is zero.

Continue

Polar decomposition separates a change of geometry from a geometry-preserving motion. It does not generally use the same axes on input and output. The next lesson makes those two sets of axes explicit through singular-value decomposition. Real orthogonal maps may require rotation blocks rather than real eigenvectors, while their complexifications have eigenvalues on the unit circle. Keep this field distinction when interpreting simultaneous diagonalization.

Review Linear Algebra I when a prerequisite needs practice. Continue to Find the input and output axes of any linear map.