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I have T as a linear transformation from V to V over the field F, V has dimension n. T has the maximum number n distinct eigenvalues, then show that there exists a basis of V consisting of eigenvectors.

I know that if I let $v_1,...,v_r$ be eigenvectors belonging to distinct eigenvalues, then those vectors are linearly independent. Can I make a basis from these linearly independent vector and prove that it spans V?

Also, what will the matrix of T be in respect to this basis?

Thank you for any input!

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    $\begingroup$ If there are $n$ linearly independent vectors in $n$-dimensional space, then they must form a basis. To see what $T$ looks like, consider what $T x_k$ looks like in the basis of eigenvectors. $\endgroup$ – copper.hat Nov 20 '13 at 3:19
  • $\begingroup$ @Akaichan Do you still need help with this or is Copper.Hat's comment enough. $\endgroup$ – Git Gud Nov 25 '13 at 21:41
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By definition, a basis for $V$ is a linearly-independent set of vectors in $V$ that spans the space $V,$ and the dimension of a finite-dimensional vector space is the number of elements in a basis for $V,$ so we know that any basis for $V$ must contain exactly $n$ linearly-independent vectors. We also know (see Theorem 5 on page 45 of Hoffman and Kunze's Linear Algebra) that every linearly-independent subset of $V$ is part of a basis for $V.$ You already know that the $n$ eigenvectors are linearly independent, so it follows that they form a basis for $V.$

To find the matrix of $T$ with respect your ordered basis $\mathscr B$ of eigenvectors, we use the fact that the ith column of that matrix is given by $[Tv_i]_{\mathscr B}$ where $[ \cdot ]_{\mathscr B}$ denotes the coordinate matrix with respect to $\mathscr B.$ We therefore compute $$[Tv_i]_{\mathscr B} = [\lambda_i v_i]_{\mathscr B} = \begin{bmatrix} 0\\ \vdots\\ 0\\ \lambda_i\\ 0\\ \vdots\\ 0\end{bmatrix}$$ where $\lambda_i$ is the eigenvalue associated with eigenvector $v_i.$ Thus, the matrix of $T$ is a diagonal matrix with the eigenvalues in the diagonal entries ordered by corresponding eigenvectors: $$\begin{bmatrix} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n\end{bmatrix}.$$

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