Abstract Linear Algebra 46 | Example of Schur Decomposition

TL;DR

Schur decomposition produces an upper triangular matrix R via a unitary similarity transformation R = U*AU.

Briefing Cornell Notes

Briefing

Schur decomposition turns any complex square matrix into an upper triangular “Schur normal form” using only unitary similarity transformations. In this worked example, a specific 3×3 real matrix is reduced step-by-step until the final triangular form is obtained, illustrating how eigenvalues appear on the diagonal and how unitary matrices make the process stable and structured.

The matrix chosen is A = [ -3 -4 0; 4 5 0; 3 5 1 ]. Because Schur decomposition applies over the complex numbers, the task is to find a unitary matrix U such that U* A U becomes upper triangular. The key payoff is that similar matrices share eigenvalues, so the diagonal of the resulting triangular matrix must match the eigenvalues of A. The example anticipates a “best-case” outcome where the diagonal entries are all the eigenvalue 1 (with multiplicity three), while nonzero entries may remain above the diagonal.

The algorithm begins by selecting an eigenpair. Applying A to the vector (0,0,1)ᵀ returns (0,0,1)ᵀ, so (0,0,1) is an eigenvector for eigenvalue λ₁ = 1. After normalizing (it already has length one), the method extends this eigenvector to an orthonormal basis of ℂ³. The simplest choice is to take y₂ = (1,0,0)ᵀ and y₃ = (0,1,0)ᵀ, producing an orthonormal set {x₁, y₂, y₃}. These basis vectors become the columns of the first unitary matrix U₁. Since U₁ is built from orthonormal vectors, U₁* is just its conjugate transpose (here, only transposition matters).

Next, the algorithm applies the unitary similarity transformation in stages: first it computes U₁* A U₁, which rearranges rows/columns and yields a block structure where the top-left entry is λ₁ = 1 and the bottom-right 2×2 block becomes a new matrix A₂. The process then repeats on A₂. For the 2×2 case, an eigenvector is found using the vector (1,−1)ᵀ, again giving eigenvalue λ₂ = 1. After normalization, x₂ = (1/√2)(1,−1)ᵀ, the remaining orthonormal direction is chosen as y₃ = (1/√2)(1,1)ᵀ (a 90° rotation in the plane). These vectors define the second unitary matrix U₂.

Combining the transformations produces the final upper triangular Schur form R = U* A U. The diagonal entries are 1,1,1, confirming that 1 is the only eigenvalue (consistent with the characteristic polynomial check mentioned). The computed R has the expected triangular structure, with a concrete set of off-diagonal entries involving √2, including a notable −8 in the lower part of the first row. The example closes by emphasizing why this matters: Schur decomposition achieves triangularization using unitary matrices, a cornerstone for the spectral theorem for normal matrices.

In short, the calculation demonstrates the mechanics of Schur decomposition: pick an eigenvector, extend to an orthonormal basis to build a unitary, apply unitary similarity to isolate a smaller block, and iterate until the matrix becomes upper triangular with eigenvalues on the diagonal.

Cornell Notes

Schur decomposition finds a unitary matrix U so that U*AU becomes an upper triangular matrix R (Schur normal form). In the example, the 3×3 matrix A has (0,0,1)ᵀ as an eigenvector with eigenvalue 1, so the first diagonal entry of R is 1. After extending that eigenvector to an orthonormal basis, a unitary U₁ is built and used to reduce A to a 2×2 block A₂. Repeating the same idea on A₂ again yields eigenvalue 1, and the final unitary similarity produces R with diagonal (1,1,1). This matters because unitary triangularization underpins the spectral theorem for normal matrices.

Why does the diagonal of the Schur form R match the eigenvalues of A?

Schur form is produced by unitary similarity: R = U*AU. Similar matrices share eigenvalues, and for an upper triangular matrix, the eigenvalues are exactly the diagonal entries. So once U*AU is triangular, the diagonal must list the eigenvalues of A.

How is the first eigenpair chosen in the example, and what does it imply for the algorithm?

The vector (0,0,1)ᵀ is tested: multiplying A by it returns the same vector, so it is an eigenvector with eigenvalue λ₁ = 1. The algorithm then normalizes this eigenvector (already unit length) and uses it to build the first unitary matrix U₁, ensuring the first diagonal entry of the reduced form is 1.

What role does extending an eigenvector to an orthonormal basis play?

Schur decomposition requires a unitary transformation. After selecting an eigenvector x₁, the method chooses additional vectors y₂, y₃ that are orthogonal to each other and to x₁, and normalized to length 1. Putting {x₁, y₂, y₃} as columns creates a unitary matrix U₁ whose conjugate transpose is easy to compute and whose similarity transformation isolates a smaller block.

How does the algorithm reduce the 3×3 problem to a 2×2 problem?

After computing U₁*AU₁, the result has a block structure: the top-left entry is λ₁ = 1, and the bottom-right 2×2 block becomes a new matrix A₂. This happens because the chosen unitary basis aligns one basis vector with the eigenvector, forcing zeros below the diagonal in the transformed matrix.

Why does the second step again yield eigenvalue 1?

For the 2×2 block A₂, the vector (1,−1)ᵀ is used and is shown to reproduce itself under multiplication by A₂ up to the same scalar factor, giving eigenvalue λ₂ = 1. The corresponding normalized eigenvector is x₂ = (1/√2)(1,−1)ᵀ, and an orthonormal complement is chosen as y₃ = (1/√2)(1,1)ᵀ to form U₂.

What is the final output and how is it verified?

The final Schur normal form is R = U*AU, where U is assembled from U₁ and U₂ (with the appropriate block/embedding). Verification comes from the triangular structure and the diagonal: the diagonal entries are 1,1,1, matching the claim that 1 is the only eigenvalue (also cross-checked via the characteristic polynomial).

Review Questions

In an upper triangular matrix, how do the eigenvalues relate to the diagonal entries?
What conditions must the vectors y₂ and y₃ satisfy when extending an eigenvector to build a unitary matrix?
Why does applying U*AU with a unitary U preserve eigenvalues?

Key Points

1
Schur decomposition produces an upper triangular matrix R via a unitary similarity transformation R = U*AU.
2
Eigenvalues of A appear on the diagonal of R because similarity preserves eigenvalues and triangular matrices have eigenvalues on their diagonal.
3
The algorithm starts by finding an eigenvector x₁ and normalizing it to length one.
4
An eigenvector is extended to an orthonormal basis; the basis vectors become columns of a unitary matrix U₁.
5
Applying U₁*AU₁ creates a block form that reduces the problem size from 3×3 to 2×2.
6
The same procedure applied to the remaining block yields the final unitary transformation and completes the triangularization.
7
For normal matrices, Schur decomposition is a pathway to the spectral theorem, since it uses unitary transformations to reach structured forms.

Highlights

The example uses (0,0,1)ᵀ as an eigenvector, immediately fixing the first Schur diagonal entry as λ₁ = 1.

After the first unitary similarity, the matrix becomes block-reduced: a 1×1 eigenvalue block plus a 2×2 remaining block.

The 2×2 block again has eigenvalue 1, so the final Schur form has diagonal entries (1,1,1).

Triangularization is achieved using only unitary matrices, keeping the transformation well-behaved and setting up results like the spectral theorem for normal matrices.

Topics

Schur Decomposition
Unitary Similarity
Eigenvalues
Upper Triangular Form
Spectral Theorem