A matrix is diagonalizable when algebraic and geometric multiplicities are equal

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A matrix is diagonalizable when algebraic and geometric multiplicities are equal.
My professor proved this in class today, but I did not fully understand his explanation and proof. Can someone please help?
 

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HallsofIvy
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I thought I had answered this before. The "algebraic multiplicity" is the number of eigenvalues, counting "multiplicity". If follows that, including complex eigenvalues, since every polynomial can be factored into linear factors over the complex numbers, that the "algebraic multiplicity" of an n by n matrix is n. The "geometric multiplicity" is the number of independent eigenvalues. Thus, if the algebraic multiplicity is the same as the geometric multiplicity, the geometric multiplicity is also n and there exist n independent eigenvectors. But the "underlying" vector space of an n by n matrix has dimension n so those eigenvectors form a basis for that vector space. Writing the linear transformation this matrix represents in that vector space gives a diagonal matrix.
 
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Jack, find the eigenvalues and eigenvectors of A by hand(!)

[tex]A = \begin{pmatrix}2 &1 &0\\0 &2 &0\\0 &0 &2\end{pmatrix}[/tex]
 

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