$T^n=1$ then the vector space has $T-$eigenvectors.

I have been working on this problem:-

Let $T:V\rightarrow V$ be a linear transformation on an $n$-dimensional vector space $V$ over $\mathbb{C}$ such that $T^{n}=1$. Prove that $V$ has a basis of $T-$eigenvectors. I feel I need to use idea of diagonalizable of linear transformation but not sure how wisely to used. any idea will appreciated.

• Can you say anything about the zeros of $x^n-1$? – copper.hat Oct 23 '15 at 14:33
• You mean root of unity. yeah in this case we have $n$ distinct root ( complex). not sure is this will be the answer for your question. – henry Oct 23 '15 at 14:35
• If the multiplicities of the zeros the minimal polynomial are all one, then the matrix is diagonalisable. – copper.hat Oct 23 '15 at 14:37
• Good point.. .. in case the matrix is diagonalizable it mean they have $T−eigenvectors$? thanks – henry Oct 23 '15 at 14:39
• Yes. If a matrix is diagonal, then the vectors $e_n$ are eigenvectors, so you can find eigenvectors of the original matrix using the similarity transform that diagonalises the matrix. So, there is a basis of eigenvectors. – copper.hat Oct 23 '15 at 14:43

The minimal polynomial of $T$ divides $x^n - 1$. Over $\mathbb{C}$, the polynomial $x^n - 1$ splits into $n$ distinct linear factors and thus the minimal polynomial of $T$ also splits into distinct linear factors implying that $T$ is diagonalizable.
• I like to see the way you connected the minimal polynomial of $T$ and $T−eigenvectors as a basis? thanks – henry Oct 23 '15 at 14:40 • You can read about it in most linear algebra texts and on the web - for example, on math.uconn.edu/~kconrad/blurbs/linmultialg/minpolyandappns.pdf. – levap Oct 23 '15 at 14:44 Whenever you have a polynomial$p(\lambda)=\prod_{j=1}^{k}(\lambda-\lambda_j)$with no repeated roots for which$p(T)=0$, then$T$is diagonalizable, meaning that it has a basis of eigenvectors. The key observation is that $$(A-\lambda_l)\prod_{j\ne l}(A-\lambda_j)=0.$$ Therefore, if you define$p_{l}(\lambda)=\prod_{j\ne l}(\lambda-\lambda_j)$, you see that the range of$p_{l}(A)$is in the null space of$A-\lambda_l I$; in other words, everything in the range of$p_{l}(A)$is either$0$or is an eigenvector of$A$with eigenvalue$\lambda_l$. To show that you have a basis of such eigenvectors uses the fact that $$1 = \frac{p_{1}(\lambda)}{p_{1}(\lambda_l)}+\frac{p_2(\lambda)}{p_2(\lambda_2)}+\cdots+\frac{p_k(\lambda)}{p_k(\lambda_k)}.$$ Therefore, $$I = \frac{1}{p_1(\lambda_1)}p_1(A)+\frac{1}{p_2(\lambda_2)}p_2(A)+\cdots+\frac{1}{p_k(\lambda_k)}p_k(A),$$ which means that every vector$x$can be written terms of eigenvectors of$A\$: $$x = \frac{1}{p_1(\lambda_1)}p_1(A)x+\frac{1}{p_2(\lambda_2)}p_2(A)x+\cdots+\frac{1}{p_k(\lambda_k)}p_k(A)x.$$