28
$\begingroup$

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 .

$\endgroup$
29
$\begingroup$

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 $n$).

$\endgroup$
10
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.