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Let $T : V\to V$ be a linear transformation such that dimension of $\operatorname{Range}(T)= k \leq n$, where the dimension of $V$ is $n$. Show that $T$ can have at most $(k+1)$ distinct eigenvalues.

I can realise that the rank will correspond to the number of non-zero eigenvalues (counted up to multiplicity) and the nullity will correspond to the 0 eigenvalue (counted up to multiplicity) but I cannot design an analytical proof of this.

Thanks for any help .

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If this is homework, please add the [homework] tag; people will still help, so don't worry. –  Zev Chonoles May 19 '12 at 4:29
No , this is not homework . –  Ester May 19 '12 at 4:32
Great! I'd just like to add that I really appreciate your including your thoughts about the problem, and asking politely for help - these things are unfortunately rare. –  Zev Chonoles May 19 '12 at 4:34
I think this is probably a duplicate. –  Arturo Magidin May 19 '12 at 4:36
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2 Answers 2

up vote 9 down vote accepted

Since the nullity of $T$ is $n-k$, that means that the geometric multiplicity of $\lambda=0$ as an eigenvalue of $T$ is $n-k$; hence, the algebraic multiplicity must be at least $n-k$, which means that the characteristic polynomial of $T$ is of the form $x^{N}g(x)$, where $N$ is the algebraic multiplicity of $0$, hence $N\geq n-k$ (so $n-N\leq k$), and $\deg(g) =n-N$. Thus, $g$ has at most $n-N$ distinct roots, none of which are equal to $0$, and that means that the characteristic polynomial of $T$ has exactly: $$1 + \text{# distinct roots of }g \leq 1 + n-N \leq 1 + k$$ distinct eigenvalues.

Note that in fact we can say a bit better that $T$ has at most $\min\{k+1,n\}$ distinct eigenvalues (when the rank is $n$).

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Thanks for giving a complete proof . I really appreciate it . What did you mean by duplicate in your last comment ? –  Ester May 19 '12 at 4:58
@Sopu: I meant that I think that this question (or something essentially equivalent to it) has been posted before on this site. But I was not able to find it yet. –  Arturo Magidin May 19 '12 at 5:00
@ArturoMagidin Yours answers are self-contained. Thank you very much sir. –  srijan May 19 '12 at 5:50
@ArturoMagidin I have edited. Thanks again. –  srijan May 19 '12 at 5:54
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Here is an outline of one way to solve the problem.

  • Eigenvectors for distinct eigenvalues are linearly independent.
  • Eigenvectors for nonzero eigenvalues are in the range of $T$.
  • The range of $T$ cannot contain a linearly independent set with more than $k$ vectors.
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I have got it .Thank you very much for your help . –  Ester May 19 '12 at 4:56
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