# Finding eigenvalues of a linear transformation

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.

• Try to clarify the definition of $T$. What are $x_1+$ and $+x_n$? – user66407 Oct 17 '15 at 17:13
• I believe that there are a bit too many commas there – AdLibitum Oct 17 '15 at 17:33
• Yes, there were two commas that had no business being there. Thanks. – SharkFin Oct 17 '15 at 17:34
• I don't think I see how this is linear when you have that $c$ there. Doesn't this have $T(0,0,\dots,0)=(c,0,\dots,0) \neq (0,0,\dots,0)$? – Ian Oct 17 '15 at 18:17
• @Ian This is a typo introduced by the latest edit (not by OP). I removed it. OP please check that it is correct now. – Winther Oct 17 '15 at 18:21

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 eigenvalues of $A$ are $0$ with multiplicity of $n$ and $n-1$ with multiplicity of $1$.
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.
• Not till you convince yourself that the characteristic polynomial of that matrix is $(1-\lambda)^n$. – AdLibitum Oct 17 '15 at 17:43
• HINT#2: Ask yourself what the rank of that matrix is, and when you have answered that, ask yourself what the dimension of $\ker(T)$ is. – AdLibitum Oct 17 '15 at 17:46