From Kernel Dimensions to Jordan Blocks
A Jordan form is a block diagonal matrix, with each block associated with an eigenvalue. When the characteristic polynomial and the dimensions of successive kernels are known, those data determine how many blocks of each size occur. The essential information is the growth of the kernels: each increase records how many Jordan chains are long enough to continue one more step.
We work over \(\mathbb{C}\), so the characteristic polynomial of an \(n\times n\) matrix \(\mathbf{A}\) splits into linear factors. For a real matrix, the same procedure gives its Jordan form over \(\mathbb{C}\). The characteristic polynomial is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_n)\); its roots are the eigenvalues, and the exponent of \((\lambda-\lambda_0)\) at a root \(\lambda_0\) is the algebraic multiplicity \(\operatorname{am}(\lambda_0)\). The sign of the polynomial’s leading coefficient depends on the convention, but does not change its roots or their multiplicities.
Here \(\ker\mathbf{B}\) is the set of vectors \(\mathbf{x}\) satisfying \(\mathbf{B}\mathbf{x}=\mathbf{0}\), and dimension is taken over \(\mathbb{C}\). For a fixed eigenvalue, the kernel dimensions eventually stop increasing. The Jordan existence theorem, established earlier in the study of Jordan form, guarantees a decomposition into Jordan blocks; the calculations below explain how their sizes are recovered from these dimensions.
The Counting Rule
Proof. Consider a Jordan block of size \(s\) for \(\lambda\). Subtracting \(\lambda\mathbf{I}\) from the block leaves a nilpotent block with ones immediately above the diagonal. Its \(k\)th power has kernel dimension \(\min(k,s)\): in the standard coordinate basis, the block maps each basis vector to the preceding one, and its \(k\)th power kills exactly the first \(\min(k,s)\) basis vectors. If \(s\leq k\), the whole block is in the kernel; if \(s>k\), precisely the first \(k\) basis vectors are.
On a Jordan block for a different eigenvalue \(\mu\ne\lambda\), the matrix \(\mathbf{J}_s(\mu)-\lambda\mathbf{I}\) is invertible. Indeed, it is \((\mu-\lambda)\mathbf{I}+\mathbf{N}\), where \(\mathbf{N}^s=\mathbf{O}\); the finite geometric sum gives an inverse because \(\mu-\lambda\ne0\). Its powers therefore have zero-dimensional kernel. Since the Jordan form is block diagonal, the dimension \(d_k\) is the sum of \(\min(k,s)\) over the blocks for \(\lambda\). A block of size \(s\) contributes \(1\) to \(d_k-d_{k-1}\) when \(s\geq k\), and contributes \(0\) when \(s<k\). Thus \(d_k-d_{k-1}\) counts exactly the blocks of size at least \(k\). Subtracting the number of blocks of size at least \(k+1\) from the number of size at least \(k\) leaves exactly the blocks of size \(k\). Substitution gives \(c_k=(d_k-d_{k-1})-(d_{k+1}-d_k)=2d_k-d_{k-1}-d_{k+1}\). The sum of the block sizes for \(\lambda\) is its algebraic multiplicity \(m\), so no block has size larger than \(m\) and \(d_m=m\). \(\square\)
This rule gives a practical sequence of steps. First, find the roots and algebraic multiplicities of \(p_{\mathbf{A}}\). For each eigenvalue, calculate \(d_1,d_2,\ldots\) until the sequence reaches its algebraic multiplicity. Take successive differences to obtain the numbers of blocks of size at least \(k\), then take differences again to obtain the numbers of each exact size. Finally, assemble the blocks and check that their sizes sum to the matrix dimension.
For each root \(\lambda\), record its algebraic multiplicity \(m\) from the characteristic polynomial.
Find \(d_k(\lambda)=\dim\ker(\mathbf{A}-\lambda\mathbf{I})^k\), starting with \(d_0=0\), until \(d_k=m\).
Compute \(b_k=d_k-d_{k-1}\), then \(c_k=b_k-b_{k+1}\). The values \(c_k\) are the counts of blocks of exact size \(k\).
Write \(c_k\) copies of the size-\(k\) block for each eigenvalue. Check each eigenvalue’s total block size against its algebraic multiplicity, and all block sizes together against \(n\).
Worked Reconstructions
Worked Example: Recovering Four Blocks from Kernel Data
Suppose \(\mathbf{A}\in\mathbb{C}^{7\times7}\) has characteristic polynomial \(p_{\mathbf{A}}(\lambda)=-(\lambda-3)^7\), and the kernel dimensions for \(\lambda=3\) are \(d_0=0\), \(d_1=4\), \(d_2=6\), and \(d_3=7\). Find the Jordan block sizes.
The algebraic multiplicity is \(m=7\). Since the kernel has reached dimension \(7\) by \(k=3\), all later dimensions are \(7\). The successive increases are
Thus four blocks have size at least \(1\), two have size at least \(2\), and one has size at least \(3\). The exact-size counts are \(c_1=b_1-b_2=4-2=2\), \(c_2=b_2-b_3=2-1=1\), and \(c_3=b_3-b_4=1-0=1\). There are two blocks of size \(1\), one of size \(2\), and one of size \(3\). The Jordan form is therefore
Check: the block sizes sum to \(1+1+2+3=7\), as required by the algebraic multiplicity. Their predicted kernel dimensions are \(4\) at \(k=1\), \(2+2+1+1=6\) at \(k=2\), and \(1+1+2+3=7\) at \(k=3\). These agree with the given data.
Worked Example: Calculating Kernel Dimensions from a Matrix
Let \(\mathbf{A}=J_3(0)\oplus J_2(0)\), so the matrix is
Compute the kernel dimensions and recover the blocks. For \(\mathbf{x}=(x_1,x_2,x_3,x_4,x_5)^{\top}\), the equation \(\mathbf{A}\mathbf{x}=\mathbf{0}\) gives \(x_2=0\), \(x_3=0\), and \(x_5=0\); \(x_1,x_4\) are free. Hence \(d_1=2\). The two blocks have sizes \(3\) and \(2\), so their second powers contribute kernel dimensions \(2\) and \(2\), giving \(d_2=4\); their third powers contribute \(3\) and \(2\), giving \(d_3=5\). Subsequent powers still have kernel dimension \(5\). Thus the full sequence through the first stabilization is \(d_0=0,d_1=2,d_2=4,d_3=5\).
The increases are \(b_1=2-0=2\), \(b_2=4-2=2\), and \(b_3=5-4=1\), followed by \(b_4=5-5=0\). Therefore \(c_1=2-2=0\), \(c_2=2-1=1\), and \(c_3=1-0=1\): there is one size-\(2\) block and one size-\(3\) block.
Check: the sizes sum to \(2+3=5\), the matrix dimension. The reconstructed form is the original \(J_3(0)\oplus J_2(0)\), whose successive kernel dimensions are \(2,4,5\), as calculated.
Worked Example: Combining Data for Two Eigenvalues
Suppose a \(6\times6\) complex matrix has characteristic polynomial \((\lambda-1)^4(\lambda+2)^2\). For \(\lambda=1\), the kernel dimensions are \(d_0=0,d_1=2,d_2=3,d_3=4\). For \(\lambda=-2\), they are \(d_0=0,d_1=1,d_2=2\). Find the Jordan block sizes.
For \(\lambda=1\), the increases are \(2,1,1\), followed by \(0\). The exact-size counts are \(c_1=2-1=1\), \(c_2=1-1=0\), and \(c_3=1-0=1\). Thus there is one size-\(1\) block and one size-\(3\) block for eigenvalue \(1\). Their sizes sum to \(1+3=4\).
For \(\lambda=-2\), the increases are \(1,1\), followed by \(0\). Hence \(c_1=1-1=0\) and \(c_2=1-0=1\): there is one size-\(2\) block for eigenvalue \(-2\), with total size \(2\). The full Jordan form is
Check: the block sizes total \(1+3+2=6\), matching the matrix dimension. The total sizes associated with the two eigenvalues are \(4\) and \(2\), matching their algebraic multiplicities. Together these checks also reproduce the characteristic polynomial \((\lambda-1)^4(\lambda+2)^2\).
Consistency Checks and a Common Error
Kernel data must satisfy constraints before they are used to propose a Jordan form. For a fixed eigenvalue, \(d_0=0\); the dimensions cannot decrease; and the increases \(b_k=d_k-d_{k-1}\) cannot increase as \(k\) grows. Each \(b_k\) counts blocks that are still growing at power \(k\), so a block counted at level \(k+1\) must also be counted at level \(k\). The sequence eventually reaches the algebraic multiplicity \(m\), and the block sizes recovered from the data must sum to \(m\).
Common mistake: treating increasing kernel growth as valid. Suppose someone reports \(d_0=0,d_1=2,d_2=5,d_3=6\). Taking differences gives \(b_1=2\), \(b_2=3\), \(b_3=1\). The incorrect conclusion would be that these are the numbers of blocks of size at least \(1,2,3\). But three blocks cannot have size at least \(2\) when only two blocks have size at least \(1\). The data are inconsistent with any Jordan form: kernel growth cannot jump from \(2\) to \(3\). For instance, if the intended data were \(d_1=2,d_2=4,d_3=6\), then \(b_1=2,b_2=2,b_3=2\), giving two blocks of size \(3\), with total size \(6\). Checking the differences before reading off block sizes catches the error.
| Check | What must hold |
|---|---|
| Kernel dimensions | \(d_0=0\), and \(d_k\) is nondecreasing |
| Successive increases | \(b_k=d_k-d_{k-1}\) is nonnegative and nonincreasing |
| Exact-size counts | \(c_k=b_k-b_{k+1}\) is a nonnegative integer |
| One eigenvalue | \(\sum_k k c_k=\operatorname{am}(\lambda)\) |
| Entire matrix | The block sizes over all eigenvalues sum to \(n\) |
The characteristic polynomial supplies the total size of the blocks belonging to each eigenvalue; kernel growth distributes that total among block sizes. The two kinds of data answer different questions, so a complete reconstruction uses both. A proposed list of blocks should be checked against the kernel dimensions as well as against the algebraic multiplicities, rather than accepted merely because its sizes add to \(n\).
Check Your Understanding
For each problem, use successive differences and check the resulting block sizes against the stated algebraic multiplicity.
- An eigenvalue \(\lambda\) has algebraic multiplicity \(5\) and kernel dimensions \(d_0=0,d_1=3,d_2=4,d_3=5\). How many blocks of each size occur?
- If \(d_1=3\), can \(d_2=4\) and \(d_3=5\) be valid kernel dimensions? Explain using the successive increases.
- A matrix has eigenvalues \(0\) and \(2\), with algebraic multiplicities \(3\) and \(2\). If the respective block sizes are \(2,1\) and \(2\), what is the matrix dimension?
- For an eigenvalue with \(d_0=0,d_1=2,d_2=3,d_3=3\), find the exact number of blocks of sizes \(1\) and \(2\).
- Why must the total size of the blocks for an eigenvalue equal its algebraic multiplicity?
