Why the Parameter Values Matter
A matrix family can have a simple Jordan form for most parameter values and a different one at a small number of exceptional values. An eigenvalue may split into distinct eigenvalues as the parameter changes, or several eigenvalues may meet and form a larger Jordan block. Even when the eigenvalue stays fixed, a parameter can break or restore the links in a Jordan chain. The task is to find the values where this happens and verify the Jordan structure on each resulting case.
Work over \(\mathbb{C}\), unless a family is explicitly stated to be real and we are discussing its real eigenvalues. For a matrix \(\mathbf{A}\), an eigenvalue \(\lambda\) is a scalar for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some nonzero vector \(\mathbf{v}\). The eigenspace is \(E_\lambda=\ker(\mathbf{A}-\lambda\mathbf{I})\). The Jordan block \(J_s(\lambda)\) has \(\lambda\) on its diagonal, ones immediately above the diagonal, and zeros elsewhere. A Jordan form is a block diagonal matrix of such blocks similar to \(\mathbf{A}\).
The characteristic polynomial identifies the eigenvalues and their algebraic multiplicities. It does not by itself determine the Jordan block sizes: for example, an eigenvalue of algebraic multiplicity three might belong to one block of size three, a block of size two together with a block of size one, or three blocks of size one. The successive kernel dimensions distinguish these possibilities.
Kernel Dimensions Reveal the Blocks
Proof. In a Jordan form, the part belonging to \(\lambda\) is a direct sum of blocks \(J_s(\lambda)\). On each such block, \((J_s(\lambda)-\lambda\mathbf{I})^k\) has a kernel of dimension \(\min(k,s)\): its nilpotent part shifts coordinates along the block, leaving exactly the first \(\min(k,s)\) coordinates in the kernel. Blocks for other eigenvalues contribute no kernel to this power, because their difference from \(\lambda\) is invertible. Since the dimension of the kernel of a block diagonal matrix is the sum of the kernel dimensions of its blocks, \(d_k\) is the sum of \(\min(k,s)\) over the blocks belonging to \(\lambda\). Increasing \(k-1\) to \(k\) increases a block’s contribution by one exactly when \(s\ge k\). Thus \(d_k-d_{k-1}\) counts blocks of size at least \(k\). Subtracting the number of blocks of size at least \(k+1\) from the number of blocks of size at least \(k\) gives the count of blocks of size exactly \(k\), as stated. \(\square\)
This result makes the kernel calculation a precise test, not just a pattern-recognition trick. Each new difference \(d_k-d_{k-1}\) counts how many chains still extend to length \(k\). When the differences reach zero, no block is longer.
Compute the characteristic polynomial as a function of the parameter. Note where eigenvalues coincide or where their multiplicities change.
For each case, identify the eigenvalues, then compute the dimensions of the kernels of successive powers of \(\mathbf{A}(t)-\lambda\mathbf{I}\).
Use the differences of kernel dimensions to count block sizes. If useful, exhibit eigenvectors or a Jordan chain and verify the defining equations.
A Link in a Jordan Chain Can Vanish
Worked Example: A Three-Dimensional Nilpotent Family
Consider the real matrix family, viewed over \(\mathbb{C}\), \(\mathbf{N}(t)=\begin{bmatrix}0&1&0\\0&0&t\\0&0&0\end{bmatrix}\). Determine its Jordan form for \(t=0\) and \(t\ne0\).
The matrix is strictly upper triangular, so its only eigenvalue is \(0\). Its action on the standard basis is \(\mathbf{N}(t)\mathbf{e}_1=\mathbf{0}\), \(\mathbf{N}(t)\mathbf{e}_2=\mathbf{e}_1\), and \(\mathbf{N}(t)\mathbf{e}_3=t\mathbf{e}_2\). For \(t\ne0\), the vectors \(t\mathbf{e}_1,t\mathbf{e}_2,\mathbf{e}_3\) form a chain: applying \(\mathbf{N}(t)\) to each vector gives, in order, \(\mathbf{0},t\mathbf{e}_1,t\mathbf{e}_2\). They are independent because they are nonzero multiples of three different standard basis vectors. In this ordered basis, the matrix is \(J_3(0)\).
For the kernel dimensions, direct multiplication gives \(\mathbf{N}(t)^2=\begin{bmatrix}0&0&t\\0&0&0\\0&0&0\end{bmatrix}\) and \(\mathbf{N}(t)^3=\mathbf{O}\). If \(t\ne0\), solving \(\mathbf{N}(t)\mathbf{x}=\mathbf{0}\) gives \(x_2=0\) and \(x_3=0\), so \(d_1=1\). Solving \(\mathbf{N}(t)^2\mathbf{x}=\mathbf{0}\) gives \(x_3=0\), so \(d_2=2\), and \(d_3=3\). The differences are \(1,1,1\), which give one block of size three.
When \(t=0\), \(\mathbf{N}(0)\mathbf{x}=\mathbf{0}\) requires only \(x_2=0\), so \(d_1=2\). Its square is the zero matrix, so \(d_2=d_3=3\). The differences are \(2,1,0\): there are two blocks of size at least one, but only one of size at least two. Thus the blocks have sizes two and one, giving \(J_2(0)\oplus J_1(0)\).
Check: At \(t=0\), the matrix sends \(\mathbf{e}_2\) to \(\mathbf{e}_1\) and sends both \(\mathbf{e}_1\) and \(\mathbf{e}_3\) to zero. This is exactly one chain of length two and one independent eigenvector, as the kernel calculation predicts.
Distinct Eigenvalues and Their Meeting Point
A useful way to check a case with distinct eigenvalues is to exhibit one eigenvector for each eigenvalue. The following elementary fact explains why this suffices when there are \(n\) distinct eigenvalues in an \(n\)-dimensional space.
Proof. Let \(\mathbf{v}_1,\ldots,\mathbf{v}_q\) be nonzero eigenvectors with pairwise distinct eigenvalues \(\lambda_1,\ldots,\lambda_q\). Suppose \(\sum_{i=1}^q c_i\mathbf{v}_i=\mathbf{0}\). Fix an index \(i\) and apply the matrix \(\prod_{j\ne i}(\mathbf{A}-\lambda_j\mathbf{I})\) to this equation. Every term except the \(i\)-th vanishes, because its eigenvector is annihilated by its corresponding factor. The remaining term is \(c_i\prod_{j\ne i}(\lambda_i-\lambda_j)\mathbf{v}_i=\mathbf{0}\). Each scalar factor \(\lambda_i-\lambda_j\) is nonzero, and \(\mathbf{v}_i\ne\mathbf{0}\), so \(c_i=0\). This holds for every \(i\), proving independence. If \(q=n\), these \(n\) independent vectors form a basis of \(\mathbb{C}^n\), so the matrix is diagonal in that eigenvector basis. \(\square\)
Worked Example: Three Eigenvalues Merge at Zero
For \(\mathbf{A}(t)=\begin{bmatrix}0&1&0\\0&t&1\\0&0&2t\end{bmatrix}\), find the Jordan form for \(t\ne0\) and \(t=0\).
Because the matrix is upper triangular, its characteristic polynomial is \((-\lambda)(t-\lambda)(2t-\lambda)\), so its eigenvalues are \(0,t,2t\). For \(t\ne0\), these are distinct. Corresponding eigenvectors are \(\mathbf{v}_0=(1,0,0)^\top\), \(\mathbf{v}_t=(1,t,0)^\top\), and \(\mathbf{v}_{2t}=(1,2t,2t^2)^\top\). Direct multiplication verifies them: \(\mathbf{A}(t)\mathbf{v}_0=\mathbf{0}\), \(\mathbf{A}(t)\mathbf{v}_t=(t,t^2,0)^\top=t\mathbf{v}_t\), and \(\mathbf{A}(t)\mathbf{v}_{2t}=(2t,4t^2,4t^3)^\top=2t\mathbf{v}_{2t}\). The proposition shows that these vectors form a basis, so the Jordan form is diagonal with entries \(0,t,2t\).
At \(t=0\), the matrix becomes \(\mathbf{A}(0)=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}=J_3(0)\). Its characteristic polynomial is \((-\lambda)^3\). Its kernel consists of vectors with \(x_2=x_3=0\), so \(d_1=1\). Its square is \(\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}\), giving \(d_2=2\); its cube is zero, giving \(d_3=3\). The successive differences are \(1,1,1\), confirming one block of size three.
Check: At \(t=0\), \(\mathbf{A}(0)\mathbf{e}_3=\mathbf{e}_2\), \(\mathbf{A}(0)\mathbf{e}_2=\mathbf{e}_1\), and \(\mathbf{A}(0)\mathbf{e}_1=\mathbf{0}\). This is a Jordan chain of length three. For \(t\ne0\), the three displayed eigenvectors are independent by the proposition, in agreement with the diagonal form.
Jordan Structure Can Change Without Eigenvalues Moving
Worked Example: A Two-Dimensional Family
Let \(\mathbf{B}(t)=\begin{bmatrix}0&1\\t&0\end{bmatrix}\). Find its Jordan form over \(\mathbb{C}\), and describe what happens over \(\mathbb{R}\).
The characteristic polynomial is \(\det(\mathbf{B}(t)-\lambda\mathbf{I})=\lambda^2-t\). If \(t\ne0\), the two roots over \(\mathbb{C}\) are distinct, so the matrix is diagonalizable. For either root \(\lambda\), the vector \((1,\lambda)^\top\) is an eigenvector because \(\mathbf{B}(t)(1,\lambda)^\top=(\lambda,t)^\top=\lambda(1,\lambda)^\top\), using \(\lambda^2=t\).
At \(t=0\), the matrix is \(J_2(0)\). Its kernel is the set of vectors \((x_1,0)^\top\), so \(d_1=1\), while \(\mathbf{B}(0)^2=\mathbf{O}\) gives \(d_2=2\). The differences \(1,1\) give one block of size two. Thus the repeated eigenvalue at the exceptional value does not produce two size-one blocks.
For real \(t>0\), the eigenvalues are the distinct real numbers \(\sqrt{t}\) and \(-\sqrt{t}\). For real \(t<0\), they are distinct complex conjugates, so diagonalization holds over \(\mathbb{C}\), not over \(\mathbb{R}\). In both cases the Jordan blocks over \(\mathbb{C}\) have size one.
Check: For \(t\ne0\), each displayed vector satisfies the eigenvalue equation because its second component produces \(t=\lambda^2\). At \(t=0\), \(\mathbf{B}(0)(1,0)^\top=\mathbf{0}\) and \(\mathbf{B}(0)(0,1)^\top=(1,0)^\top\), verifying the length-two chain.
Common Pitfall: Treating a Zero Link as Nonzero
In the three-dimensional nilpotent family, a tempting but wrong calculation is to claim that \(\mathbf{N}(0)\) still has a chain of length three because the first superdiagonal contains a \(1\). That overlooks the second link: at \(t=0\), \(\mathbf{N}(0)\mathbf{e}_3=\mathbf{0}\), not \(\mathbf{e}_2\). In fact, \(\mathbf{N}(0)^2=\mathbf{O}\), so no vector can generate a chain of length three. The correction is to substitute the exceptional parameter into every entry before computing kernels or chains: the resulting kernel dimensions \(2,3,3\) give block sizes two and one.
A reliable analysis keeps eigenvalues and block sizes distinct. First find which eigenvalues occur at each parameter value. Then compute the kernels of the shifted powers for each eigenvalue. A repeated root of the characteristic polynomial signals a case to examine, but does not alone determine whether the repeated eigenvalue belongs to one large block or several smaller blocks.
Check Your Understanding
For each question, justify the claimed Jordan structure with an eigenvalue calculation and kernel dimensions or an explicit eigenvector check.
- For \(\begin{bmatrix}0&1\\0&t\end{bmatrix}\), find the eigenvalues and determine whether \(t=0\) changes the Jordan block sizes.
- For \(\begin{bmatrix}0&1&0\\0&0&t\\0&0&0\end{bmatrix}\), calculate \(d_1,d_2,d_3\) at \(t=0\) and \(t\ne0\).
- Explain why three distinct eigenvalues of a \(3\times3\) complex matrix imply three Jordan blocks of size one.
- For \(\begin{bmatrix}0&1\\t&0\end{bmatrix}\), state the exceptional parameter value for its Jordan block sizes over \(\mathbb{C}\), and distinguish it from the change in whether its eigenvalues are real.
- If \(d_1=2,d_2=3,d_3=4\) for one eigenvalue and \(d_4=4\), determine the Jordan block sizes for that eigenvalue.
