Suppose $T\in\mathcal{L} (\mathbb{R^5})$ is defined by $$T(x_1,\dots,x_5) = (x_1+\dots+x_5,\dots,x_1+\dots+x_5)$$ Proof if T is diagonalizable.
Proof:
$$T(x_1,\dots,x_5) = (x_1+\dots+x_5,\dots,x_1+\dots+x_5)$$ $$\lambda x_1 = (x_1+\dots+x_5,\dots,x_1+\dots+x_5)$$ $$\lambda x_2 = (x_1+\dots+x_5,\dots,x_1+\dots+x_5)$$ $$\lambda x_3 = (x_1+\dots+x_5,\dots,x_1+\dots+x_5)$$ $$\lambda x_4 = (x_1+\dots+x_5,\dots,x_1+\dots+x_5)$$ $$\lambda x_5 = (x_1+\dots+x_5,\dots,x_1+\dots+x_5)$$
Hence $\lambda = 0$ and $x_1 = \dots = x_n \implies \lambda = n$
Hence $n = \lambda = 5$ and $\lambda = 0$ be the two eigenvalue of T.Let $\lambda = 5$ correspond to vector $(1,1,1,1,1)$ and let $\lambda = 0$ is just the null space of T, and it is spanned by the four vectors $(1,−1,0,0,0,),(1,0,−1,0,0), (1,0,0,−1,0)$ and $(1,0,0,0,−1)$. We can show that these four vectors are linearly independent, so this eigenspace contains 4 linearly independent vectors. Hence there are a total of five linearly independent eigenvectors, and henceTis diagonalizable.
Is my proof correct?