Move an operator across an inner product

Try

For A=[[1,2],[0,3]] in standard real coordinates, calculate ⟨Ax,y⟩ and collect the coefficients of x. They are exactly the coordinates of Aᵀy, so ⟨Ax,y⟩=⟨x,Aᵀy⟩. This transfer rule is the purpose of the adjoint. It expresses how an output measurement pulls back to an input measurement, while respecting the inner products on the two spaces.

Predict

Predict the complex version by testing the one-dimensional map x↦ix. We have ⟨ix,y⟩=ix ȳ. To write this as ⟨x,cy⟩, the second argument must use c=−i, because it will be conjugated. Thus the adjoint of multiplication by i is multiplication by −i. Ordinary transpose would miss this change. The standard matrix adjoint is conjugate transpose, denoted A*, provided the coordinate bases are orthonormal.

Experiment

For A=[[1,i],[0,2]], use A*=[[1,0],[−i,2]]. Take x=(1,1) and y=(i,1). Compute Ax=(1+i,2), then ⟨Ax,y⟩=(1+i)(−i)+2=3−i. Also A*y=(i,3), and ⟨x,A*y⟩=−i+3. Repeat with x=(i,0). Write both routes side by side; this catches both a missed conjugation and an accidental reversal of argument convention.

Understand

For T:V→W between finite-dimensional inner-product spaces, define T*:W→V by ⟨Tx,y⟩_W=⟨x,T*y⟩_V. For each fixed y, the left side is a linear functional of x, so Riesz representation gives a unique T*y. Conjugate-linearity in the second argument shows T* is itself linear. In orthonormal bases, evaluating on basis vectors gives the conjugate-transpose formula. The defining identity also proves (ST)*=T*S*, (T*)*=T, and (aT)*=conj(a)T*. For a nonorthonormal basis a Gram matrix intervenes; a coordinate transpose alone need not represent the adjoint.

Apply

An output vector y lies in ker T* exactly when ⟨Tx,y⟩=0 for every x. Therefore ker T*=(im T)⊥. Taking orthogonal complements gives im T=(ker T*)⊥, and exchanging T with T* gives im T*=(ker T)⊥. These identities explain the geometry of the four fundamental subspaces. They also prove that T*T has the same kernel as T: ⟨T*Tx,x⟩=⟨Tx,Tx⟩=‖Tx‖². Thus a zero value forces Tx=0, while the converse is immediate. This positive operator will be central to singular values.

Check the weighted-coordinate warning directly over R. For G=diag(1,9), define ⟨x,y⟩=xᵀGy. If A=[[0,1],[0,0]], its adjoint is G⁻¹AᵀG=[[0,0],[1/9,0]]. Indeed AᵀG=G A†, which is exactly the equality obtained by expanding ⟨Ax,y⟩=⟨x,A†y⟩ for arbitrary x and y. Ordinary transpose gives the wrong factor. This also explains why “symmetric matrix” and “self-adjoint operator” agree only when the coordinate geometry is accounted for. In an orthonormal basis G=I and the familiar transpose formula returns.

Two coordinate systems require two Gram matrices

Keep the physical map fixed while changing the input and output bases independently. Let A be its m-by-n coordinate matrix. Write the input inner product as ⟨x,z⟩=z*Gx and output inner product as ⟨u,y⟩=y*Hu, with Hermitian positive definite Gram matrices G and H. Here * on a coordinate matrix means conjugate transpose; Gᵢⱼ is the inner product of input basis vector j with basis vector i under our linear-first convention. Positive definiteness makes both Gram matrices invertible.

If B represents the adjoint, its defining identity for all coordinate x,y is y*HAx=(By)*Gx=y*B*Gx. Testing all coordinate basis pairs gives HA=B*G, equivalently A*H=GB, hence B=G⁻¹A*H. Conversely this formula satisfies the defining identity for every pair, proving it is the correct adjoint. B is n-by-m; a rectangular A has no meaningful same-space normality test AA*=A*A unless dimensions and metrics are addressed. The return adjoint, using exchanged metrics, is H⁻¹B*G=A.

For one orthogonal but unnormalized basis fᵢ, put wᵢ=⟨fᵢ,fᵢ⟩>0. Expansion gives v=Σ⟨v,fᵢ⟩fᵢ/wᵢ, so Aᵢⱼ=⟨Tfⱼ,fᵢ⟩/wᵢ. The matrix of the adjoint has Bᵢⱼ=conj(Aⱼᵢ)wⱼ/wᵢ. This recovers conjugate transpose only when all relevant weights agree; orthogonality alone does not imply normalization. With different orthogonal input/output bases, the factor is the output weight for column j divided by the input weight for row i.

Change coordinates without changing the physical map

The fixed map T has rows (1,2), (0,1), (1,0). Its physical witness vectors are x=(1,1), y=(1,−1,2). New input basis columns form S with rows (1,s), (0,k); output basis columns form R with rows (1,r,0), (0,1,0), (0,0,l). Predict which quantities should stay unchanged.

Input shear s -101 Input scale k 12 Output shear r -101 Output scale l 12

Defining identity: left 4.0000, correct right 4.0000, naive-transpose right 16.0000. The naive physical pullback is (7.0000, 9.0000).

Matrix	Rows
A=R⁻¹TS	(1.0000, 3.0000); (0.0000, 2.0000); (0.5000, 0.5000)
G=SᵀS	(1.0000, 1.0000); (1.0000, 5.0000)
H=RᵀR	(1.0000, 1.0000, 0.0000); (1.0000, 2.0000, 0.0000); (0.0000, 0.0000, 4.0000)
B=G⁻¹AᵀH	(0.0000, -0.5000, 2.0000); (1.0000, 1.5000, 0.0000)
Naive Aᵀ	(1.0000, 0.0000, 0.5000); (3.0000, 2.0000, 0.5000)

Maximum entry discrepancy in GB=AᵀH: 0. Independent physical-coordinate reconstruction discrepancy: 0. Difference between the naive and correct adjoint matrices: 2. One matching witness alone would not prove equality of operators.

Reset

This real-valued experiment has 36 bounded basis choices and fixed two-/three-dimensional matrices. The complex proof requires conjugate transpose. State is temporary and awards no mastery.

Prove that the formula survives another basis change

Suppose old input coordinates equal K times new input coordinates, and old output coordinates equal L times new output coordinates. Then A′=L⁻¹AK, G′=K*GK and H′=L*HL. Substitute these expressions into (G′)⁻¹(A′)*H′. The inverse and conjugate-transpose factors cancel in order, leaving K⁻¹G⁻¹A*HL=K⁻¹BL. That is exactly the coordinate change rule for the same return map. The matrices change; the physical operator and its inner-product identity do not.

In the experiment the physical map T has rows (1,2), (0,1), (1,0). For physical x=(1,1), y=(1,−1,2), both sides of the adjoint identity equal4. With default bases, A has rows (1,3), (0,2), (1/2,1/2), and B has rows (0,−1/2,2), (1,3/2,0). Using Aᵀ instead produces a physical pullback (7,9), whose product with physical x is16 instead of4. Resetting all shears to zero and both scales to one restores orthonormal coordinates, where the transpose works. A coincidentally matching witness in another example would not establish equality of the matrices; the defining identity must hold for every x,y.

Keep dual pullback, adjoint and a degenerate form separate

An algebraic linear functional written cᵀx pulls back along A to the coefficient column Aᵀc, even over C. This is a statement about dual coordinates, not vector coordinates for an inner product. Identifying functionals with vectors uses the Gram matrix and complex conjugation; the adjoint formula therefore has different factors. Do not replace one construction by the other merely because both reverse an arrow.

If the form is degenerate, the Riesz uniqueness argument and Gram inverse are unavailable. For G=diag(1,0) over R and the zero map on R², the formal identity GB=0 allows every B with zero first row. There is no unique adjoint determined by this form. This failure is outside the positive-definite inner-product theorem, not a reason to use an arbitrary inverse or pseudoinverse without new conditions.

For a written repair, derive the Gram formula before inspecting the ledger. Change one basis and predict how A, B and the corresponding Gram matrix move together. Then explain why a naive transpose can fail while the physical map and all true inner products remain unchanged.

Continue

For a final domain check, write the spaces beside every arrow: T goes from V to W, while T* returns from W to V. Consequently T*T acts on V and TT* acts on W. Their matrices need not even have the same size when T is rectangular, so an equality between them is not a meaningful normality test in that setting.

The algebraic dual map acts on functionals and is represented by ordinary transpose in dual bases; an adjoint acts on vectors after inner products identify them with measurements. Over C that identification is conjugate-linear, which explains the conjugation in the formula. Next, adjoints will distinguish normal operators from arbitrary ones. All range equalities here use finite-dimensional orthogonal decomposition; infinite-dimensional ranges may fail to be closed.

Review Linear Algebra I when a prerequisite needs practice. Continue to Strengthen triangularization with orthonormal geometry.