Jordan Normal Form 2 | An Example [dark version]

TL;DR

Factor det(A−λI) to get eigenvalues and their algebraic multiplicities; the exponent of each factor equals the algebraic multiplicity.

Briefing Cornell Notes

Briefing

A 4×4 matrix is worked through to find its Jordan normal form, with the key takeaway that eigenvalues and their multiplicities narrow the structure dramatically—often enough to avoid the heavier “generalized eigenspace” calculations. The process starts by forming the characteristic polynomial det(A−λI). Using the matrix’s block structure, the determinant factors neatly into (2−λ)·(4−λ)^3, revealing two eigenvalues: λ=2 with algebraic multiplicity 1, and λ=4 with algebraic multiplicity 3.

Algebraic multiplicity then immediately fixes the total Jordan block sizes: a 1×1 Jordan block for λ=2 and a 3×3 Jordan block for λ=4. What remains is how many Jordan blocks sit inside the 3×3 block for λ=4, which is governed by the geometric multiplicity (the number of Jordan blocks equals the geometric multiplicity). Three configurations are possible for a 3×3 algebraic multiplicity: geometric multiplicity 3 (three 1×1 blocks, making the matrix diagonalizable), geometric multiplicity 2 (a 2×2 block plus a 1×1 block), or geometric multiplicity 1 (a single 3×3 block).

To determine which case applies, the calculation focuses on the eigenspace for λ=4: ker(A−4I). The transcript performs Gaussian elimination on A−4I to compute the kernel dimension. After subtracting the first row from the second and then combining the third row with half of the second, the matrix is reduced to row echelon form with two pivot columns. That leaves two free variables, so the kernel has dimension 2. Since geometric multiplicity equals dim ker(A−λI), the geometric multiplicity for λ=4 is γ=2.

With γ=2, the 3×3 Jordan block must split into exactly two Jordan blocks: one 2×2 block and one 1×1 block. The λ=2 part contributes a separate 1×1 block. The only remaining freedom is the ordering of these Jordan blocks (and the placement of the boxes within the 3×3 structure), not their sizes. The example is presented as a “nice” case because the eigenvalue multiplicities and a single eigenspace computation suffice—no generalized eigenspace dimensions are needed. The result sets up the next lesson: more complicated matrices can force additional generalized eigenspace calculations when geometric multiplicity alone doesn’t fully determine the Jordan structure.

Cornell Notes

The Jordan normal form for a 4×4 matrix is determined by combining eigenvalue multiplicities with the geometric multiplicity of each eigenvalue. The characteristic polynomial factors as (2−λ)·(4−λ)^3, giving eigenvalues λ=2 (algebraic multiplicity 1) and λ=4 (algebraic multiplicity 3). Algebraic multiplicity fixes block sizes: a 1×1 block for λ=2 and a total 3×3 block for λ=4. The geometric multiplicity for λ=4 comes from dim ker(A−4I), computed via Gaussian elimination, which yields two free variables and thus γ=2. Therefore the 3×3 portion splits into a 2×2 Jordan block plus a 1×1 Jordan block; only the order of blocks remains flexible.

How does the characteristic polynomial determine the possible Jordan block sizes?

Compute det(A−λI) and factor it. Each eigenvalue’s exponent in the characteristic polynomial equals its algebraic multiplicity. Algebraic multiplicity fixes the total size of the Jordan blocks for that eigenvalue: λ=2 has exponent 1, so it contributes a 1×1 Jordan block. λ=4 has exponent 3, so it contributes a combined 3×3 Jordan structure (the sizes of individual blocks must add up to 3).

Why does geometric multiplicity control how the 3×3 Jordan structure splits?

For an eigenvalue with algebraic multiplicity 3, the Jordan blocks inside the 3×3 structure can be arranged in three ways: three 1×1 blocks (geometric multiplicity 3), one 2×2 plus one 1×1 (geometric multiplicity 2), or one 3×3 block (geometric multiplicity 1). Geometric multiplicity equals dim ker(A−λI), which equals the number of Jordan blocks for that eigenvalue.

How is geometric multiplicity for λ=4 computed in this example?

Form A−4I and row-reduce it using Gaussian elimination to row echelon form. The transcript ends with two pivot columns, meaning there are two free variables. The kernel dimension is therefore 2, so dim ker(A−4I)=2. That gives geometric multiplicity γ=2 for λ=4.

What Jordan block configuration follows from γ=2 when the algebraic multiplicity is 3?

If λ=4 has algebraic multiplicity 3 and geometric multiplicity 2, the 3×3 Jordan structure must split into two blocks whose sizes sum to 3. The only option is a 2×2 block plus a 1×1 block. The λ=2 eigenvalue independently contributes its 1×1 block.

What freedom remains after eigenvalues, algebraic multiplicities, and geometric multiplicities are known?

Once block sizes are fixed (1×1 for λ=2; 2×2 and 1×1 for λ=4), the remaining ambiguity is the ordering of the Jordan blocks and the placement of the boxes within the overall Jordan arrangement. The transcript emphasizes that only the order changes, not the block dimensions.

Review Questions

Given a characteristic polynomial factor (λ−a)^m(λ−b)^n, what do m and n tell you about Jordan block sizes for eigenvalues a and b?
If dim ker(A−λI)=k for an eigenvalue with algebraic multiplicity 3, which Jordan block size patterns are possible and how do you match them to k?
Why does computing ker(A−4I) avoid generalized eigenspace calculations in this particular example?

Key Points

1
Factor det(A−λI) to get eigenvalues and their algebraic multiplicities; the exponent of each factor equals the algebraic multiplicity.
2
Algebraic multiplicity fixes the total Jordan block size for each eigenvalue (e.g., exponent 3 implies a combined 3×3 Jordan structure).
3
Geometric multiplicity equals dim ker(A−λI) and determines how many Jordan blocks appear inside that combined structure.
4
For algebraic multiplicity 3, geometric multiplicity 3 means three 1×1 blocks, geometric multiplicity 2 means a 2×2 plus a 1×1 block, and geometric multiplicity 1 means a single 3×3 block.
5
Gaussian elimination on A−λI (row echelon form) reveals the number of free variables, which directly gives dim ker(A−λI).
6
In this worked example, ker(A−4I) has dimension 2, forcing the λ=4 part to be a 2×2 Jordan block plus a 1×1 block.
7
After block sizes are fixed, only the ordering of Jordan blocks (not their sizes) remains flexible.

Highlights

The characteristic polynomial factors as (2−λ)·(4−λ)^3, immediately locking in algebraic multiplicities 1 for λ=2 and 3 for λ=4.

Algebraic multiplicity determines block sizes, but the split inside the 3×3 structure depends on geometric multiplicity.

Row-reducing A−4I leaves two free variables, so dim ker(A−4I)=2 and γ=2 for λ=4.

With algebraic multiplicity 3 and geometric multiplicity 2, the λ=4 Jordan structure must be exactly one 2×2 block and one 1×1 block.

The final Jordan form is determined up to the ordering of blocks and the arrangement of boxes within the 3×3 portion.