A square matrix $A$ is DIAGONALISABLE if and only if
A$A$ is symmetric
B$A$ is upper triangular
C$\det A \neq 0$
D$A$ has $n$ indep eigenvectors
Answer & Solution
Correct answer: D. $A$ has $n$ indep eigenvectors
1. DIAGONALISABLE: $A$ can be written as $A = PDP^{-1}$ for some diagonal matrix $D$ and invertible $P$.
2. THEOREM: $A$ is diagonalisable iff $A$ has $n$ LINEARLY INDEPENDENT eigenvectors (a basis of eigenvectors).
3. Sufficient condition: $A$ has $n$ DISTINCT eigenvalues (each gives an eigenvector, and distinct-eigenvalue eigenvectors are independent).
4. Symmetric matrices (option A) are diagonalisable, but the condition is not necessary in general (complex matrices). Determinant (option C) is unrelated. Triangular matrices may not be diagonalisable if they have repeated eigenvalues with deficient eigenspaces.
5. Practical method: find char polynomial roots → eigenspaces → check if dim sum equals $n$.
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.7 (Diagonalisation)._
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