Diagonalisability criteria

In one sentence

Diagonalisable means enough eigenvectors, which is the same as the minimal polynomial having no repeated factor, which is the same as geometric equalling algebraic multiplicity everywhere — and distinct eigenvalues are sufficient but never necessary.

Why the exam asks it

Because there are four equivalent criteria and each one makes a different question easy, so a candidate who knows only the eigenvector count will do three times the work. And because diagonalisability depends on the field, which options exploit relentlessly.

The idea

For $A$ an $n \times n$ matrix over a field $F$, these are equivalent:

1. $A$ is similar to a diagonal matrix over $F$.
2. $F^n$ has a basis of eigenvectors of $A$.
3. The minimal polynomial $m_A$ splits over $F$ into distinct linear factors.
4. For every eigenvalue, geometric multiplicity equals algebraic multiplicity, and $\chi_A$ splits over $F$.

Criterion 3 is usually the fastest to apply, and criterion 4 is usually the fastest to disprove with — computing one rank of $A - \lambda I$ settles it.

Sufficient but not necessary

$n$ distinct eigenvalues $\Rightarrow$ diagonalisable, because eigenvectors for distinct eigenvalues are independent and there are $n$ of them.

The converse is false in the most obvious way possible: the identity matrix is diagonal and has one eigenvalue. Any option saying "diagonalisable, so the eigenvalues are distinct" is wrong, and it is set often because the sufficient condition is the one people memorise.

The field matters

$\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has $\chi_A = x^2 + 1$. Over $\mathbb{R}$ there are no eigenvalues at all, so it is not diagonalisable. Over $\mathbb{C}$ its eigenvalues are $\pm i$, distinct, so it is diagonalisable. Same matrix, different answers, and the stem tells you which field.

The general statement: $A$ is diagonalisable over $F$ only if $\chi_A$ splits over $F$. Over $\mathbb{C}$ that is automatic, so over $\mathbb{C}$ the only obstruction is a repeated factor in $m_A$.

Families that are always diagonalisable

• Real symmetric matrices, and more generally normal matrices ($AA^* = A^*A$) over $\mathbb{C}$ — orthogonally or unitarily, by the spectral theorem.
• Idempotents ($A^2 = A$): $m_A$ divides $x^2 - x = x(x-1)$, distinct linear factors.
• Involutions ($A^2 = I$) in characteristic $\ne 2$: $m_A \mid (x-1)(x+1)$.
• Matrices of finite order over $\mathbb{C}$, since $m_A \mid x^k - 1$, which has distinct roots.

Each of these is a one-line argument through criterion 3, and recognising the pattern "$A$ satisfies a polynomial with distinct roots" is worth more than the list itself.

The trap in "symmetric"

Real symmetric is diagonalisable. Complex symmetric is not — symmetry without conjugation buys nothing. The witness:

$$A = \begin{pmatrix}1 & i\\ i & -1\end{pmatrix}, \qquad A^{T} = A, \qquad A^2 = 0 .$$

It is symmetric, nonzero, and nilpotent, so it is as far from diagonalisable as a matrix can be. The condition that generalises real symmetry to $\mathbb{C}$ is Hermitian ($A^* = A$), not symmetric.

Simultaneous diagonalisation

Two diagonalisable matrices are simultaneously diagonalisable $\iff$ they commute. Commuting alone is not enough — both must be diagonalisable to begin with — and diagonalisable alone is certainly not enough.

Where intuition breaks

"Diagonalisable, so $n$ distinct eigenvalues." The identity. The implication runs the other way only.

"Diagonalisable, so invertible." The zero matrix is diagonal. Diagonalisability says nothing about $0$ being an eigenvalue.

"Invertible, so diagonalisable." $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has determinant $1$ and is not diagonalisable. The two properties are independent, in all four combinations.

"$A^2$ is diagonalisable, so $A$ is." $A = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ has $A^2 = 0$, which is diagonal. Tempting because most operations that preserve structure seem reversible; taking square roots of matrices is not one of them.

"$A$ and $B$ diagonalisable, so $A + B$ is." $\begin{pmatrix}0&1\\0&1\end{pmatrix}$ and $\begin{pmatrix}0&0\\0&-1\end{pmatrix}$ are each diagonalisable — each has eigenvalues $0$ and $\pm 1$, distinct — and they sum to $\begin{pmatrix}0&1\\0&0\end{pmatrix}$, which is not. The same holds for products.

"Symmetric, so diagonalisable." Over $\mathbb{R}$ yes, over $\mathbb{C}$ no — $\begin{pmatrix}1&i\\i&-1\end{pmatrix}$. The option will say "symmetric" and rely on you supplying "real" for free.

"All eigenvalues equal, so $A$ is a scalar matrix." Only if it is also diagonalisable. Otherwise, any Jordan block.

"Not diagonalisable over $\mathbb{R}$, so not over $\mathbb{C}$." The rotation matrix. Diagonalisability can only improve when the field grows, since more polynomials split.

The exam's angle

1. Read the field. Over $\mathbb{C}$, only repeated factors of $m_A$ obstruct; over $\mathbb{R}$, non-real eigenvalues do as well.
2. Look for a polynomial the matrix satisfies. $A^2 = A$, $A^2 = I$, $A^k = I$, $A^2 = 0$ — each settles the question instantly through distinct-or-repeated roots.
3. To disprove, compute one rank. If $\operatorname{rank}(A - \lambda I) > n - (\text{algebraic multiplicity})$, there are too few eigenvectors and you are done.
4. Keep the four independence examples ready: diagonalisable and not invertible (zero), invertible and not diagonalisable ($\begin{pmatrix}1&1\\0&1\end{pmatrix}$), both, neither.
5. For sums and products, assume nothing. Only commuting diagonalisable matrices behave, and the option will rarely say "commuting".

The night before

• Equivalent: similar to diagonal; a basis of eigenvectors; $m_A$ has distinct linear factors; geometric $=$ algebraic for every eigenvalue with $\chi_A$ split.
• $n$ distinct eigenvalues $\Rightarrow$ diagonalisable. Converse false — the identity.
• Diagonalisability depends on the field. Rotation: not over $\mathbb{R}$, yes over $\mathbb{C}$.
• Real symmetric, Hermitian, normal, idempotent, involution, finite order: all diagonalisable.
• Complex symmetric is not: $\begin{pmatrix}1&i\\i&-1\end{pmatrix}$ is symmetric and nilpotent.
• Diagonalisable and invertible are independent in all four combinations.
• $A^2$ diagonalisable does not give $A$ diagonalisable; sums and products of diagonalisable matrices need not be.
• Simultaneously diagonalisable $\iff$ commuting, for diagonalisable matrices.