What is the proof that if $A$ ($m\times m$ Matrix) is normal i.e $(AA^{\ast} = A^{\ast}A)$ then $A$ is non defective i.e (for each eigenvalue of $A$, its algebraic multiplicity is equal to the geometric multiplicity).
2026-04-01 20:23:54.1775075034
Proof if $A$ is normal then it is nondefective
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It follows from:
A normal matrix is diagonalizable.
For a diagonalizable matrix algebraic and geometric multiplicity coincide.
Both are quite standard results.