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I must prove:

If $V$ is a finite dimensional vector space with $\dim(V)=n$ and $T$ is a linear transformation over $V$ with $n$ distinct eigenvalues, then $T$ is diagonalizable.

I know in order to show that $T$ is diagonalizable we must show that there is an ordered basis for $V$ such that $[T]_\beta$ is a diagonal matrix. I'm slightly confused how to start this problem in general. I initially stated that the set of vectors ${u_1, u_2, ..., u_n} \in V$ are the eigenvectors associated with the eigenvalues. Since these are vectors in $V$ they can be written as linear combinations of some ordered basis of $V$ but I'm not sure if that is what I need to focus on?

Any help would be very much so appreciated! Thanks!

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    $\begingroup$ In fact, the set of eigenvectors will be a basis for $V$. Hint on showing they're linearly independent: if $c_1 u_1 + \cdots + c_n u_n = 0$, what happens if you apply $(T - \lambda_2 I) (T - \lambda_3 I) \cdots (T - \lambda_n I)$ to both sides? $\endgroup$ – Daniel Schepler May 4 '17 at 21:30
  • $\begingroup$ If you apply $(T-\lambda_2I)(T-\lambda_3I)...(T-\lambda_nI)$ to both sides of the equation $c_1u_1 + ... + c_nu_n=0$? $\endgroup$ – Boosh May 4 '17 at 21:54
  • $\begingroup$ Yes, that's what I meant. $\endgroup$ – Daniel Schepler May 4 '17 at 22:02
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Suppose that $T$ has $n$ distinct eigenvalues say $c_1, c_2,\ldots ,c_n$. Let $v_1, v_2,\ldots ,v_n $ be the corresponding eigenvectors. First thing you can show that eigenvectors $v_1, v_2,\ldots ,v_n $ corresponding to distinct eigenvalues of $T$ are linearly independent. Then since $\dim(V) = n$, $S$ = $\{v_1, v_2,\ldots ,v_n \}$ is an ordered basis of $V$ which consists of eigenvectors of $T$. Hence $T$ is diagonalizable.

Hint for proving result: Eigenvectors corresponding to distinct eigenvalues of $T$ are linearly independent.

Suppose that $T$ has $n$ distinct eigenvalues say $c_1, c_2,\ldots ,c_n$. Let $v_1, v_2,\ldots ,v_n $ be the corresponding eigenvectors.

Then $T(v_i) =c_i v_i$ for $i = 1, 2, \ldots ,n$. We shall prove that $S$ = $\{v_1, v_2,\ldots ,v_n \}$ is linearly independent. We proceed by induction on $n$. If $n = 1$, then $S = \{v_1\}$ is linearly independent as $v_1\neq 0$. Suppose that the set $\{v_1, v_2,\ldots ,v_k \}$ is linearly independent, where $k<m$. We shall prove that $\{v_1, v_2,\ldots ,v_{k+1} \}$ is linearly independent.

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  • $\begingroup$ Then using the ordered basis, would it be warranted to state "$T(v_j) = c_jv_j = D_{jj}v_j = $ sum from $i=1$ to $n$ of $D_{ij}v_i$ where $D=[T]_\beta$? And would this be enough to prove that T is diagonalizable? $\endgroup$ – Boosh May 4 '17 at 22:34
  • $\begingroup$ @Boosh Yes, If S = $\{v_1, v_2,\ldots v_n \}$ be a basis of $V$ consisting of eigenvectors of $T$. There must exist scalars $c_i \in F$ such that $T(v_i) = c_i v_i$, for $i = 1, 2, \ldots n$ . And matrix of $T$ with respect to this basis would be diagonal matrix with diagonal elements as $c_i$. That is what we want to claim . $\endgroup$ – srijan May 4 '17 at 22:48

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