Recipe

Find the minimal polynomial of a matrix

Use when: you have a square matrix $A$ and need the unique monic polynomial of smallest degree with $p(A)=0$

Goal: the minimal polynomial $m_A$ , confirmed to annihilate $A$ with no smaller degree working

Taught in The minimal polynomial

Quick recipe

List candidate linear factors from $A$ 's diagonal entries
Test the smallest-degree candidate first
Compute $p(A)$ directly and check for the zero matrix
Confirm no smaller monic polynomial could also have worked
If asked to confirm uniqueness, test a specific alternative directly

1
For each distinct number on $A$ 's main diagonal, write down the linear factor $(x-a)$ ; these are your starting candidates.

Example

For $A=\begin{pmatrix}2&0\\0&2\end{pmatrix}$ , the diagonal entries are both $2$ , giving one candidate factor, $(x-2)$ . For the Jordan block $A=\begin{pmatrix}2&1\\0&2\end{pmatrix}$ , the diagonal entries are also both $2$ , the same single candidate.
- $A$ is diagonal or upper/lower triangular — the diagonal entries are exactly its eigenvalues, so they're exactly the right candidates.
- $A$ isn't triangular — the eigenvalues aren't visible on the diagonal, so solve $\det(A-xI)=0$ for the characteristic polynomial and use its roots as your candidates instead.
Why: if $\lambda$ is an eigenvalue with eigenvector $\mathbf v$ , then $p(A)\mathbf v=p(\lambda)\mathbf v$ for any polynomial $p$ , so $p(A)=0$ forces $p(\lambda)=0$ , meaning every eigenvalue must be a root of whatever you're testing.
2
Build the monic polynomial from your candidate factors at the lowest total degree, and substitute $A$ into it.

Example

For $A=2I$ : test $p(x)=x-2$ , giving $p(A)=A-2I$ . For the Jordan block: the same first candidate, $p(x)=x-2$ , giving $p(A)=A-2I$ again.

Why: the minimal polynomial is defined as the smallest-degree monic annihilator, so testing the cheapest candidate first finds it as soon as possible.
3
Multiply out (or just substitute, at degree $1$ ) and see whether every entry comes out to zero.

Example

$A-2I=\begin{pmatrix}2&0\\0&2\end{pmatrix}-\begin{pmatrix}2&0\\0&2\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$ for the diagonal case, a success; but $A-2I=\begin{pmatrix}0&1\\0&0\end{pmatrix}\ne0$ for the Jordan block, a failure.
- the result is the zero matrix — this candidate is $m_A$ ; stop here.
- the result is nonzero — raise the exponent on the failing factor by one, and repeat this step on the new candidate.
Why: a monic polynomial is the minimal polynomial exactly when it's the first, lowest-degree candidate that actually annihilates $A$ .
4
Check that every strictly smaller monic degree, down to a nonzero constant, was already ruled out by a test in Step 3.

Example

For the Jordan block, a nonzero constant never annihilates a nonzero matrix, and degree $1$ already failed above, so degree $2$ is genuinely minimal.

Why: the definition of minimal polynomial is about degree, not just about working, so a working candidate only counts once every smaller degree is eliminated.
5
Substitute the alternative candidate polynomial into $A$ and show it does NOT vanish, ruling it out at that same degree.

Example

For $A=2I$ with $m_A(x)=x-2$ , test $q(x)=x-3$ : $q(A)=A-3I=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\ne0$ , so $q$ fails.

Why: uniqueness of the minimal polynomial means no other monic polynomial at that same minimal degree can also annihilate $A$ , and one failing example already disproves it.

Common mistakes

Assuming a non-triangular matrix's minimal-polynomial candidates are visible on the diagonal, when they aren't.
Trusting a candidate polynomial by inspection instead of actually substituting $A$ into it in Step 3.
Testing a higher exponent on a factor before ruling out the smaller exponent for that same factor.
Declaring a working candidate minimal without checking that every smaller degree genuinely failed first.