Q.Let $D$ be a $n × n$ real, diagonal matrix. Show that there exists a $x ∈ R ^n$ such that span of $\{D^k x : 0 ≤ k ≤ n − 1\} = R ^n$ if and only if the eigen-values of D are distinct.
My approach
First, let's assume that the eigenvalues of $D$ are distinct. In this case, we can write $D$ as follows:
$D = P Λ P^{-1}$
where $Λ$ is a diagonal matrix containing the distinct eigenvalues of $D$, and $P$ is a matrix whose columns are the eigenvectors of $D$.
Let $x$ be the vector whose entries are all $1$, i.e., $x = [1 1 ... 1]^T$. Then we can compute $D^k$ . $x$ for $k = 0, 1, ..., n-1$ as follows:
$D^k . x = P Λ^k P^{-1} . x$
= $P [λ_1^k 0 ... 0; 0 λ_2^k ... 0; ...; 0 ... λ_n^k] P^{-1} . x$
= $P [λ_1^k; λ_2^k; ...; λ_n^k] P^{-1} . x$
Since the eigenvalues of $D$ are distinct, the columns of $P$ are linearly independent. Thus, the span of ${D^k .x : 0 ≤ k ≤ n − 1}$ is the same as the span of ${[λ_1^k; λ_2^k; ...; λ_n^k] : 0 ≤ k ≤ n − 1}$. This span is all of $R^n$, since the Vandermonde matrix whose rows are ${[1; λ_1; λ_1^2; ...; λ_1^(n-1)], [1; λ_2; λ_2^2; ...; λ_2^(n-1)], ..., [1; λ_n; λ_n^2; ...; λ_n^(n-1)]}$ is invertible, and the columns of this matrix are precisely ${[λ_1^k; λ_2^k; ...; λ_n^k] : 0 ≤ k ≤ n − 1}$.
Now let's assume that there exists a vector $x$ such that span ${D^k .x : 0 ≤ k ≤ n − 1} = R^n$. Let $λ_1, λ_2, ..., λ_n$ be the eigenvalues of $D$ (not necessarily distinct), and let $P$ be the matrix whose columns are the eigenvectors of $D$. Then we can write $D$ as follows:
$D = P Λ P^{-1}$
where $Λ$ is the diagonal matrix containing the eigenvalues of $D$, including multiplicities.
Let $y$ be a vector in $R^n$ such that $Py = x$. Then for any $k$, we have
$D^k . x = P Λ^k P^{-1} . x = P Λ^k P^(-1) Py = P Λ^k y$.
Thus, span ${D^k .x : 0 ≤ k ≤ n − 1}$ = span ${P Λ^k y : 0 ≤ k ≤ n − 1}$. Since $P$ is invertible, we have span ${P Λ^k y : 0 ≤ k ≤ n − 1} = R^n$ if and only if span ${Λ^k y : 0 ≤ k ≤ n − 1} = R^n$. But $Λ$ is diagonal, so we have
span ${Λ^k y : 0 ≤ k ≤ n − 1} = span {[λ_1^k y_1; λ_2^k y_2; ...; λ_n^k y_n] : 0 ≤ k ≤ n − 1}$.
Thus, in order for span of ${D^k .x : 0 ≤k≤n-1}=ℝ ^n$
Please cheak this...if there are any mistake then give another solution or hints...