Can we prove $T : R^n \rightarrow R^n$ $T(x_1,x_2,...x_n) = ((x_1+x_2+...x_n),...,(x_1+x_2+...x_n))$ ,is diagonalizable by induction (by finding e.v)?

Question

Let $T : {\Bbb R}^n \to {\Bbb R}^n$ be defined by

$$ T(x_1, x_2, \dots, x_n) = ((x_1 + x_2 + \cdots + x_n), \dots,(x_1 + x_2 + \cdots + x_n))$$

It is possible to prove that $T$ is diagonalizable by induction (to find eigenvalues and eigenvectors) for every $n \geq 2$?

It is obvious, we need to find $n$ eigenvectors to prove that T is diagonalizable for every $n ≥ 2$. If we look at the equation $T(v) = λv$ we can see that $n$ is a sum of $x_1,...x_n$ so it is an eigenvalue to the eigenvector $(1,....1)$.If we will take the standard base for $R^n$ we will get that representation matrix is $[T]_ {\displaystyle {\mathcal {E}}}$ is $\begin{pmatrix} 1& ....& 1 \\ .........\\ 1 & ......& 1 \end{pmatrix} $ so the eigenvalues are $n$ and $0$ but how calculate it ? We can't use the charectsitcstic polynomial because calculating $det$ of that matrix takes too long. We can use induction to prove it on a smaller matrices (also on calculating the eigenvectors)?

Answer 1

Let $$ e_i=(0,0,\ldots,0,1,0,\ldots,0) $$ be the vector in $\mathbb R^n$ with all entries equal to 0, except the $i-$entry which is equal to 1.

Then $$ T(e_i-e_{i+1})=0, \quad i=1,\ldots,n-1, $$ and $T(e_1+\cdots+e_n)=n(e_1+\cdots+e_n).$

So, the $e_i-e_{i+1}$ and $e_1+\cdots+e_n$ are all eigenvectors of $T$, and they are linearly independent. Hence, $T$ is diagonalizable.

Another faster way to show that $T$ is diagonalizable is to observe that $T^2-nT=0$.