alleryStart learning →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)._
Keep practising Linear Algebra
Browse all Linear Algebra questions for GATE MA, or explore more Mathematics topics.
Related questions
Let $A$ be a $3 \times 3$ matrix with eigenvalues 1, 2, 3. Then $\det(A)$ equalsIf $A$ is invertible, the eigenvalues of $A^{-1}$ areThe TRACE of $A = \begin{pmatrix} 2 & 5 \\ 7 & 3 \end{pmatrix}$ isGRAM-SCHMIDT process converts a basis intoTwo vectors $u, v$ are ORTHOGONAL iffAn ORTHOGONAL matrix $Q$ satisfiesFor a SYMMETRIC real matrix $A$, all eigenvalues areCAYLEY-HAMILTON THEOREM states that for any $n \times n$ matrix $A$ with characteristic po