Hello, I'm trying to work through this question.
I define linearly independent as:
$a_1*v_1+a_2*v_2+...+a_n*v_n = 0$ iff every $a_i=0$.
I also know that an eigenvector is a vector $v$ such that:
$T(v)=\lambda*v$
where $\lambda$ is the eigenvalue.
Thanks for your help.
Hints:
Remember that eigenvectors of linear mappings are by definition non-zero.
Take one eigenvector $\;v_i\;$ from each different eigenvalue $\;\lambda_i\;$ and prove this $\;n\;$ eigenvectors are linearly independent and thus they are a basis for $\;V\;$.
Calculate the matrix representation of $\;T\;$ wrt the basis $\;\{v_1,...,v_n\}\;$ . Further hint:
$$Tv_i=\lambda_iv_i=0\cdot v_1+\ldots+0\cdot v_{i-1}+\lambda_iv_i+0\cdot v_{i+1}+\ldots+0\cdot v_n$$
Calculate now the determinant of $\;V\;$ wrt the above representation, remembering that the determinant is invariant wrt choice of basis.
Since $v_1,\ldots,v_n$ are $n$ linearly independant in a linear space $V$ with dimension $n$ the the family $\mathcal B=(v_1,\ldots,v_n)$ is a basis of $V$ but since these vectors are eigenvectors of $T$ then the matrix of $T$ relative to the basis $\mathcal B$ is diagonal: $$[T]_{\mathcal B}=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$$ hence $$\det (T)=\det([T]_{\mathcal B})=\prod_{k=1}^n\lambda_k$$
Look at diagonalizing the matrix representation of $T$. So $T = QDQ^{-1}$, where $Q$ is your matrix from the eigenvectors and $D$ is the diagonal matrix of the eigenvalues. We note $det(Q) = \frac{1}{det(Q^{-1}}$. It should be fairly straight-forward to get $det(T) = \prod_{i=1}^{n} \lambda_{i}$.
