What the Minimal Polynomial Records
Jordan form separates a matrix into blocks, each associated with an eigenvalue. The minimal polynomial condenses this block structure into one polynomial: for each eigenvalue, its exponent records the size of that eigenvalue’s largest Jordan block. The word “largest” matters. Several blocks for one eigenvalue contribute only the size of the largest, not the sum of their sizes.
We work over \(\mathbb{C}\), so every eigenvalue and every Jordan block under discussion lies in the scalar field. Let \(\mathbf{A}\in\mathbb{C}^{n\times n}\) be a finite-dimensional square matrix. For a polynomial \(p(x)=c_0+c_1x+\cdots+c_dx^d\), its value at \(\mathbf{A}\) is \(p(\mathbf{A})=c_0\mathbf{I}+c_1\mathbf{A}+\cdots+c_d\mathbf{A}^d\). A polynomial annihilates \(\mathbf{A}\) if \(p(\mathbf{A})=\mathbf{O}\).
The final statement in the definition can be checked directly. If \(\mathbf{A}=\mathbf{P}\mathbf{J}\mathbf{P}^{-1}\), then for every polynomial \(p\), \(p(\mathbf{A})=\mathbf{P}p(\mathbf{J})\mathbf{P}^{-1}\), by applying the similarity relation to each power of the matrix. Since \(\mathbf{P}\) is invertible, \(p(\mathbf{A})=\mathbf{O}\) exactly when \(p(\mathbf{J})=\mathbf{O}\). Thus it is enough to find the minimal polynomial from a Jordan form.
When Does a Polynomial Annihilate One Block?
Write \(J_s(\lambda)=\lambda\mathbf{I}_s+\mathbf{N}_s\), where \(\mathbf{N}_s\) has ones immediately above the diagonal and zeros elsewhere. This matrix is nilpotent: \(\mathbf{N}_s^s=\mathbf{O}\), whereas \(\mathbf{N}_s^k\ne\mathbf{O}\) for \(0\le k<s\). For \(k=0\), the power is \(\mathbf{I}_s\); for \(1\le k<s\), the entry in row 1, column \(k+1\) of \(\mathbf{N}_s^k\) is \(1\).
Proof. By the definition of multiplicity, write \(p(x)=(x-\lambda)^kq(x)\), where \(q(\lambda)\ne0\). Substitution gives \(p(J_s(\lambda))=\mathbf{N}_s^k q(\lambda\mathbf{I}_s+\mathbf{N}_s)\). To analyze the second factor, expand \(q\) in powers of \(x-\lambda\). This gives \(q(\lambda\mathbf{I}_s+\mathbf{N}_s)=q(\lambda)\mathbf{I}_s+\mathbf{N}_s\mathbf{S}\) for some polynomial matrix \(\mathbf{S}\) in \(\mathbf{N}_s\). The matrix \(\mathbf{N}_s\mathbf{S}\) is nilpotent: it is a polynomial in \(\mathbf{N}_s\) with zero constant term, so its \(s\)th power is zero. Since \(q(\lambda)\ne0\), the matrix \(q(\lambda)\mathbf{I}_s+\mathbf{N}_s\mathbf{S}\) is invertible. Indeed, after factoring out \(q(\lambda)\), its inverse is given by the finite geometric sum for \(\mathbf{I}_s+q(\lambda)^{-1}\mathbf{N}_s\mathbf{S}\). Therefore \(p(J_s(\lambda))=\mathbf{O}\) exactly when \(\mathbf{N}_s^k=\mathbf{O}\). The stated powers of \(\mathbf{N}_s\) show this holds exactly when \(k\ge s\). \(\square\)
The zero polynomial also annihilates every block, but it has no finite factor multiplicity and is not the minimal polynomial. The criterion is stated for nonzero polynomials for this reason. It says more than that \((J_s(\lambda)-\lambda\mathbf{I}_s)^s=\mathbf{O}\): it rules out every polynomial whose \(x-\lambda\) factor occurs fewer than \(s\) times.
From the Blocks to the Minimal Polynomial
Proof. Similarity invariance reduces the question to \(\mathbf{J}\). A polynomial \(p\) annihilates a block diagonal matrix exactly when it annihilates each diagonal block, because evaluating \(p\) on that matrix gives the block diagonal matrix whose blocks are the corresponding evaluations of \(p\). Fix an eigenvalue \(\lambda\). The Jordan-block criterion says that \(p\) must contain the factor \((x-\lambda)^s\) for every block \(J_s(\lambda)\). This is equivalent to requiring multiplicity at least \(s_{\lambda}\), the largest such block size. Doing this for every eigenvalue shows that \(p(\mathbf{J})=\mathbf{O}\) exactly when \(p\) is divisible by every \((x-\lambda)^{s_{\lambda}}\). These powers for distinct eigenvalues are relatively prime. Their product therefore divides \(p\): for two relatively prime factors \(f,g\), if both divide \(p=fq\), the relative primality of \(f\) and \(g\) implies \(g\mid q\), so \(fg\mid p\); applying this repeatedly gives the claim for all the factors. The displayed product itself annihilates every block, since its factor for that block’s eigenvalue has exponent at least the block size. It is monic, and every nonzero annihilating polynomial is divisible by it. It is therefore the monic annihilating polynomial of least degree, as required. \(\square\)
Worked Example: Repeated Blocks for One Eigenvalue
Determine the minimal polynomial of the following Jordan matrix. Its blocks have eigenvalue \(2\), with sizes \(3\) and \(1\).
Multiplying \(\mathbf{N}\) by itself gives \(\mathbf{N}^2=\begin{bmatrix}0&0&1&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}\), and then \(\mathbf{N}^3=\mathbf{O}\). Thus \((\mathbf{J}-2\mathbf{I}_4)^3=\mathbf{O}\), while \((\mathbf{J}-2\mathbf{I}_4)^2\ne\mathbf{O}\). The largest block size is \(3\), so the theorem gives \(m_{\mathbf{J}}(x)=(x-2)^3\). The block of size \(1\) does not increase the exponent.
Check: the computed powers show that \((x-2)^3\) annihilates \(\mathbf{J}\), and that \((x-2)^2\) does not. The block criterion also rules out any annihilating polynomial with a smaller multiplicity of \(x-2\).
Worked Example: Two Eigenvalues and Two Largest Blocks
Find the minimal polynomial for a matrix with Jordan form \(\mathbf{J}=\operatorname{diag}(J_2(1),J_2(-2))\). In full,
For the first block, \(J_2(1)-\mathbf{I}_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), whose square is zero but which is not itself zero. For the second, \(J_2(-2)+2\mathbf{I}_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), and its square is also zero while the matrix itself is nonzero. Thus the largest block size is \(2\) for each eigenvalue, giving \(m_{\mathbf{J}}(x)=(x-1)^2(x+2)^2\).
Check: the first factor squared evaluates to the zero matrix on the \(1\)-block, and the second factor squared evaluates to the zero matrix on the \(-2\)-block. Since each relevant first power is nonzero, neither exponent can be reduced. The product therefore annihilates both diagonal blocks and is minimal.
Worked Example: Checking an Explicit Candidate
Let \(\mathbf{J}=\operatorname{diag}(J_2(3),[-1])= \begin{bmatrix}3&1&0\\0&3&0\\0&0&-1\end{bmatrix}\). Find its minimal polynomial and check it by evaluating the polynomial on each block.
The largest block for \(3\) has size \(2\), and the block for \(-1\) has size \(1\). The formula gives \(m_{\mathbf{J}}(x)=(x-3)^2(x+1)\). Expanding the scalar polynomial, \((x^2-6x+9)(x+1)=x^3-5x^2+3x+9\).
On the \(3\)-block, put \(\mathbf{N}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\), so \(J_2(3)=3\mathbf{I}_2+\mathbf{N}\) and \(\mathbf{N}^2=\mathbf{O}\). Therefore \((J_2(3)-3\mathbf{I}_2)^2(J_2(3)+\mathbf{I}_2)=\mathbf{N}^2(4\mathbf{I}_2+\mathbf{N})=\mathbf{O}\). On the \(-1\) block, the factor \((-1+1)\) is zero, so the whole polynomial evaluates to zero. Thus \(m_{\mathbf{J}}(\mathbf{J})=\mathbf{O}\).
To check that the exponent \(2\) at \(3\) is necessary, evaluate the candidate with that exponent lowered by one. On the \(3\)-block this gives \((J_2(3)-3\mathbf{I}_2)(J_2(3)+\mathbf{I}_2) =\mathbf{N}(4\mathbf{I}_2+\mathbf{N})=4\mathbf{N} =\begin{bmatrix}0&4\\0&0\end{bmatrix}\ne\mathbf{O}\). So the lower exponent does not annihilate the matrix, as required.
Why the Largest Block, Not the Total Size?
For a fixed eigenvalue, every block imposes a lower bound on the multiplicity of its factor: a block of size \(s\) requires at least \(s\) copies of \(x-\lambda\). All these requirements are met by taking their maximum. They are not added, because a single factor \((x-\lambda)^{s_{\lambda}}\) already annihilates every block for that eigenvalue. Factors for distinct eigenvalues must both be present, which is why their required powers are multiplied in the minimal polynomial.
This gives a useful contrast with the characteristic polynomial. Its multiplicity for an eigenvalue counts the sizes of all blocks for that eigenvalue, whereas the exponent in the minimal polynomial is the size of only the largest block. The two polynomials answer different questions: the minimal polynomial records the highest power needed to annihilate the matrix, not the total number of coordinates assigned to an eigenvalue.
Common Pitfall
Adding block sizes gives the wrong exponent. For the matrix in the first example, one might incorrectly add the block sizes \(3+1\) and claim that the minimal polynomial is \((x-2)^4\). But direct multiplication gives \((\mathbf{J}-2\mathbf{I}_4)^3=\mathbf{O}\), so \((x-2)^3\) already annihilates the matrix. The block criterion confirms it is minimal because the size-\(3\) block is not annihilated by an exponent smaller than \(3\). The size-\(1\) block adds no further requirement.
A second error is to use only the eigenvalues and omit the block sizes. Knowing that the eigenvalue is \(2\) does not decide whether the factor is \(x-2\), \((x-2)^2\), or a higher power. The exact exponent follows from the largest Jordan block, and the block criterion explains why that exponent is necessary as well as sufficient.
Check Your Understanding
Use the Jordan-block criterion and show how the block sizes determine each exponent.
- A Jordan form has blocks \(J_3(4)\) and \(J_1(4)\). What is the minimal polynomial?
- A Jordan form has blocks \(J_2(0)\), \(J_4(0)\), and \(J_1(5)\). Determine its minimal polynomial.
- For \(J_3(1)\), why does \((x-1)^2\) fail to annihilate the block?
- A matrix has two blocks for eigenvalue \(-2\), of sizes \(2\) and \(5\). What exponent of \(x+2\) is required, and why is their sum not required?
- Explain why the minimal polynomial is unchanged when a matrix is replaced by a similar matrix.
