I have to prove this result:

If $A \in M_n (F)$ has $n$ distinct eigenvalues then $A$ is diagonalizable.

My attempt at proof :

Let $\lambda_1,\lambda_2,\ldots,\lambda_n$ be the distinct eigenvalues of $A$ and hence of $L_A$.

Corresponding to each $i$, (from $1$ to $n$) let $v_i$ be an eigenvector of $A$.

Then for each $i$, $v_i$ will be an eigenvector of $L_A$.

$L_A$:$F^n \to F^n$, is a linear operator,

Now $\{v_1,v_2,\ldots,v_n\}$ will form a linearly independent subset of $F^n$ and the $\dim(F^n)$ =$n$, hence $\{v_1,v_2,\ldots,v_n\}$ will form a basis for $F^n$.

So, $\{v_1,v_2,\ldots,v_n\}$ is a basis for $F^n$ consisting of the eigenvectors of $L_A$,

so $L_A$ is diagonalizable and hence $A$ is diagonalizable.

Is this proof correct ?

  • $\begingroup$ Absolutely correct. Check however if you know how to show the intermediary results (such as : if vectors correspond to different eigenvalues, they are linearly independent). $\endgroup$ – Ewan Delanoy Oct 27 '13 at 14:58
  • $\begingroup$ @EwanDelanoy Thank you ! $\endgroup$ – johny Oct 27 '13 at 15:19

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