Finding eigenvalues of a linear transformation

Question

Find all eigenvalues associated to the linear transformation $T \in L(\mathbb{K}^n)$ defined by $$T(x_1,\ldots,x_n)=(x_1 + \cdots + x_n, \ldots, x_1 + \cdots + x_n)$$ So, to find all eigenvalues one would normally only find the matrix associated to the linear transformation, and easily find the eigenvalues from there, right? Well I have no idea how to "put" this linear transformation into matrix form.

Answer 1

The transformation matrix of $T$ is $$ A=\pmatrix{1\cdots1\\\hspace{-9 mm}\vdots&\hspace{-11 mm}\vdots\\1\cdots1}=\pmatrix{1\\\vdots\\1}\pmatrix{1\cdots1}=vv^T $$ Then the characteristic polynomial of $A$ is $$ p(\lambda)=|\lambda I-A|=|\lambda I-vv^T|=\lambda^n-Tr(vv^T)\lambda^{n-1}=\lambda^n-n\lambda^{n-1}=\lambda^{n-1}(\lambda-n) $$ for $vv^T$ is a Rank-$1$ matrix and all principal minors above $2$ are $0$.

Thus of $A$ has an eigenvalue of $0$ with its multiplicity of $n-1$ and $n$ with the multiplicity of $1$.

Answer 2

Denote $e_i$ the $i$-th vector of the standard basis $\cal E$. Then $$ T(e_i)=(1,...,1)=e_1+\cdots +e_n $$ for all $i=1,...,n$. Thus the matrix associated to $T$ with respect to the basis $\cal E$ is the matrix which has $1$ in every place.