Show that $F:\ \mathbb{R}^{100} \rightarrow \mathbb{R}^{100}$ is diagonalizable

Question

Let $F:\ \mathbb{R}^{100} \rightarrow \mathbb{R}^{100}$ be a linear operator such that $F(e_1)=e_1$ and $F(e_k)=ke_k+e_{k-1}$ for $k\geq2$, where $\mathcal{B}=(e_1,e_2,...,e_{100})$ is the basis of $\mathbb{R}^{100}$. Show that $F$ is diagonalizable.

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I thought that this property is probably true for any $n\in \mathbb{R}$, not only $n=100$, and proving it for any $n$ would of course solve the problem.

Here of course I thought of induction proof.

For $n=1$ it is true.

Now we assume that it is true for certain $k$ and prove the the it for $k+1$.

We already know that $F:\ \mathbb{R}^{k} \rightarrow \mathbb{R}^{k}$ must has $k$ eigenvalues, we need to prove that rising the dimension adds another one.

Unfortunately it is as far as I could get on my own, so I'm asking for your help.

Answer 1

Hint : The matrix of $F$ in the given basis is triangular and all the diagonal elements are distinct.