Linear algebra is full of statements that are perfectly true in the setting where most people first learn them: finite-dimensional vector spaces over a field, written in coordinates, with matrices you can row-reduce on a page. The trouble is that the mind quietly promotes “true in the standard setting” into “true in general,” and then uses that false generality as a substitute for understanding.
A single well-chosen counterexample can fix this. Not because it humiliates the learner, but because it forces the learner to locate the exact hinge where a theorem turns. In linear algebra, that hinge is often the same one: the difference between what is guaranteed in finite dimension and what must be assumed or proved in infinite dimension, and the difference between “eigenvalue data” and “a complete basis of eigenvectors.”
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The counterexample in this article is a two-by-two matrix so simple that it looks harmless. Yet it breaks a common belief:
- If a matrix has an eigenvalue, then there is a basis of eigenvectors.
- If a matrix has only one eigenvalue, it still “ought \to” be diagonalizable, because there is nothing else going on.
Both beliefs are wrong. The right statement is more subtle, and once you see it, a large part of linear algebra snaps into place.
The counterexample: a matrix with one eigenvalue but no eigenbasis
Consider the matrix
| Matrix | Form |
|—|—|
| $J$ | $\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}$ |
This is the simplest nontrivial Jordan block. It has a single eigenvalue $\lambda = 1$. The characteristic polynomial is
So algebraic multiplicity of the eigenvalue 1 is 2.
If you have been trained to think “two-by-two matrices always diagonalize if they have real eigenvalues,” you might expect to find two independent eigenvectors and build a basis. Let us test that expectation.
An eigenvector $v\neq 0$ for $\lambda=1$ satisfies $(J-I)v = 0$. Compute
So $(J-I)(x,y)^T = (y,0)^T$. The equation $(J-I)v=0$ becomes $y=0$. That means every eigenvector is of the form $(x,0)^T$. The eigenspace is
which is one-dimensional. There is only one independent eigenvector direction.
This is the punchline: the matrix has an eigenvalue with algebraic multiplicity two, but its eigenspace has dimension one. There is no way to build a basis of eigenvectors, so the matrix is not diagonalizable.
Why this forces understanding: algebraic versus geometric multiplicity
The characteristic polynomial counts eigenvalues with multiplicity. It is an algebraic object: a determinant computation, a polynomial identity. The eigenspace dimension is geometric: it is the dimension of a kernel, a space of vectors fixed up to scalar.
For a matrix $A$ and eigenvalue $\lambda$, define:
- Algebraic multiplicity: the multiplicity of $\lambda$ as a root of $\chi_A(t)$.
- Geometric multiplicity: $\dim\ker(A-\lambda I)$.
The always-true inequality is
Diagonalizability for a matrix over an algebraically closed field is equivalent to the condition that for every eigenvalue, geometric multiplicity equals algebraic multiplicity, and that the sum of the eigenspace dimensions equals $n$.
The matrix $J$ violates that equality: algebraic multiplicity is 2, geometric multiplicity is 1.
This one failure clarifies what theorems are actually doing. When you hear “a symmetric matrix is diagonalizable,” the core content is precisely that symmetry forces the geometric multiplicities to fill the whole space. When you hear “a matrix with distinct eigenvalues is diagonalizable,” the core content is that distinct eigenvalues automatically give independent eigenvectors, and the counts match without needing multiplicity analysis.
What replaces the missing eigenvector: generalized eigenvectors
If $J$ has only one eigenvector direction, what is the second direction doing? It is not mysterious. It is the direction that is not fixed by $J$ but is “almost” fixed: applying $J$ shifts it into the eigenvector direction.
Compute $(J-I)$:
The vector $e_2=(0,1)^T$ is not an eigenvector, but applying $J-I$ sends it to the eigenvector $e_1=(1,0)^T$. Also notice $(J-I)e_1 = 0$. So we have a chain
This motivates the definition of a generalized eigenvector: a vector $v$ such that $(A-\lambda I)^k v = 0$ for some $k\ge 1$. For a true eigenvector, you can take $k=1$. For $e_2$ above, $(J-I)^2 e_2 = 0$ but $(J-I)e_2\neq 0$, so it is generalized of rank 2.
Generalized eigenvectors fill the missing directions. In fact, $\{e_1,e_2\}$ is a basis in which $J$ has the Jordan block form. The “extra structure” beyond the eigenvalue is the nilpotent part $N=J-I$, with $N^2=0$ but $N\neq 0$.
The minimal polynomial is the real diagnostic
The characteristic polynomial $(t-1)^2$ tells you eigenvalues and algebraic multiplicity. It does not tell you whether the matrix is diagonalizable. The missing information is encoded in the minimal polynomial.
The minimal polynomial $m_A(t)$ is the monic polynomial of smallest degree such that $m_A(A)=0$. It divides the characteristic polynomial.
For $J$,
so $(t-1)^2$ annihilates $J$. But $(J-I)\neq 0$, so $(t-1)$ does not annihilate $J$. Thus
Here is the key theorem:
- A matrix is diagonalizable over a field containing all its eigenvalues if and only if its minimal polynomial splits into distinct linear factors.
Distinct means no repeated roots. For $J$, the minimal polynomial has a repeated factor, so it is not diagonalizable.
This is the cleanest way to say what went wrong.
What this teaches about proofs: track the hidden hypotheses
Many linear algebra claims are “if and only if” statements with hypotheses that your brain may ignore. The counterexample forces you to put those hypotheses back.
Consider three standard statements:
- If $A$ has $n$ distinct eigenvalues, then $A$ is diagonalizable.
- If $A$ is diagonalizable, then $\chi_A$ splits and $m_A$ has no repeated factors.
- If $A$ is symmetric (over $\mathbb{R}$) or normal (over $\mathbb{C}$), then $A$ is diagonalizable with an orthonormal eigenbasis.
The first statement does not mention multiplicities because “distinct eigenvalues” forces multiplicities to be 1. The third statement is a deep structural fact about an inner product interacting with the operator. The second statement is the general diagnostic.
The counterexample $J$ is designed so that it satisfies a naive reading of “has eigenvalues,” but fails the diagnostic. It is the smallest place where you must learn to distinguish:
- Roots of $\chi_A$
- Dimensions of eigenspaces
- Structure of the nilpotent part
This distinction is not technical clutter. It is the conceptual core.
The geometric picture: why diagonals are too rigid
Diagonal matrices are rigid: they scale each coordinate axis independently. If a matrix is diagonalizable, it means there exists a basis in which the matrix acts by independent scaling along basis directions. Those directions are eigenvectors.
The matrix $J$ does not act by independent scaling. It scales by 1, but it also shears. On $\mathbb{R}^2$, $J$ sends $(x,y)$ \to $(x+y, y)$. The $y$-coordinate is preserved, but the $x$-coordinate absorbs $y$. That is shear.
A shear keeps areas and has a single invariant direction: the horizontal axis. That direction is exactly the eigenspace. Everything else is dragged along parallel lines. Expecting a full eigenbasis is expecting the shear to become a pure scaling after change of basis. It cannot, because shear is a genuine geometric feature. The nilpotent part measures that feature.
A short classification: all two-by-two failures look like this
Over an algebraically closed field, a two-by-two matrix with a repeated eigenvalue $\lambda$ has two possibilities:
- It is diagonalizable: it is similar \to $\begin{pmatrix}\lambda & 0\\0 & \lambda\end{pmatrix} = \lambda I$.
- It is not diagonalizable: it is similar \to $\begin{pmatrix}\lambda & 1\\0 & \lambda\end{pmatrix}$.
There is no third possibility. The reason is exactly geometric multiplicity: the eigenspace dimension is either 2 or 1. If it is 2, the matrix is already $\lambda I$ in some basis. If it is 1, you obtain a Jordan block.
So the counterexample is not a weird exception. It is the canonical alternative to diagonalizability.
What to carry forward
This one matrix trains several habits that pay for themselves everywhere else in linear algebra:
- When you see eigenvalues with multiplicity, check geometric multiplicity.
- When you want a basis adapted to an operator, ask what invariants obstruct it.
- When a claim feels obviously true, locate the hypothesis that makes it true.
- When diagonalization fails, look for Jordan structure: eigenvectors plus generalized eigenvectors, and a nilpotent part.
The best use of counterexamples is not to collect them. It is to let one of them reshape your mental map of the subject. The matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ does exactly that. It shows you that “having eigenvalues” is not the same as “being controlled by eigenvalues,” and it teaches you to read the fine print that theorems depend on.
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