Strengthen triangularization with orthonormal geometry

Try

Compare A=[[1,5],[0,2]] with D=diag(1,2). Both are upper triangular and have the same eigenvalues. Yet A has coupling between its coordinate directions, while D scales two perpendicular directions independently. Compute A*A and AA*: they differ, whereas D*D=DD*. The equality T*T=TT* defines normality. It is a geometric condition involving the inner product, not simply a statement that a matrix has enough eigenvectors.

Predict

Predict whether every diagonalizable matrix has an orthonormal eigenbasis. The eigenvectors of A above can be chosen as (1,0) and (5,1), which are not perpendicular, and their eigenspaces are one-dimensional. No rescaling makes them perpendicular. Thus ordinary diagonalization is weaker than unitary diagonalization. Over complex finite-dimensional spaces, normality is exactly the condition that supplies an orthonormal eigenbasis.

Experiment

First compute A*A=[[1,5],[5,29]] and AA*=[[26,10],[10,4]]. Compare their first diagonal entries: the extra squared row contribution 25 has no matching first-column contribution. Next consider an arbitrary upper-triangular R. Make a table comparing the first row squared norm |r₁₁|²+Σⱼ>₁|r₁ⱼ|² with the first column squared norm |r₁₁|². If R is normal, equality forces every term in the sum to vanish. This elementary comparison is the engine of the spectral proof.

Understand

Complex Schur triangularization says every operator has an upper-triangular matrix in some orthonormal basis. Choose a unit eigenvector, which exists because the characteristic polynomial splits over C, and extend it orthonormally. The first column is (λ,0,…,0), leaving a trailing block B. Inductively triangularize B by a unitary change and extend that change by fixing the first basis vector. This proves Schur without assuming normality. If T is normal, its Schur matrix R is also normal because unitary similarity preserves adjoints and products. The row/column calculation forces the first row tail to zero; the trailing block remains normal. Induction makes R diagonal.

Apply

Conversely a diagonal matrix commutes with its conjugate transpose, and unitary similarity preserves that equality. Therefore a complex operator is normal if and only if it is unitarily diagonalizable. This proof supplies orthogonality of distinct eigenspaces and the norm formula ‖T‖₂=max|λ|. To verify the norm formula, express a unit vector in the orthonormal eigenbasis; the squared output norm is a weighted sum Σ|λⱼ|²|cⱼ|², bounded by the largest squared magnitude, with equality at a corresponding eigenvector. A poorly conditioned nonorthonormal eigenbasis would not justify the same argument for a general diagonalizable matrix.

Normality also implies ker(T−λI)=ker(T*−conj(λ)I). For a normal S, compare ‖Sx‖² and ‖S*x‖² using S*S=SS*. Their equality makes their kernels equal. Apply this to S=T−λI, which remains normal by expanding the products. This supplies a direct route to orthogonality of distinct eigenspaces and serves as an independent consistency check on the diagonal-basis proof.

Continue

Over R, a normal matrix may require two-dimensional rotation blocks rather than a real diagonal matrix. The real rotation by 90 degrees is normal but has no real eigenvalues. Real symmetric operators do admit real orthonormal eigenbases; the next lesson proves that specialization and develops its variational interpretation. Schur form is also an important numerical target because unitary changes preserve Euclidean lengths, unlike arbitrary similarity transformations.

Review Linear Algebra I when a prerequisite needs practice. Continue to Read a self-adjoint operator through real energy values.

For the real skew-adjoint subclass, construct the rotation planes and prove the block theorem. Then build a compatible complex structure and track its metric. These constructions distinguish a change of coordinates from a change of inner product.