# Linear independent eigenvectors and eigenvalues

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!

• 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. – copper.hat Nov 20 '13 at 3:19
• @Akaichan Do you still need help with this or is Copper.Hat's comment enough. – Git Gud Nov 25 '13 at 21:41

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: $$\begin{bmatrix} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n\end{bmatrix}.$$