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Suppose, an upper-triangular matrix A is invertible, has all equal entries on the main diagonal, and is not symmetric. Show that A cannot have n linearly independent eigenvectors.

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  • $\begingroup$ I guess the Identity matrix is an easy counterexample... $\endgroup$ – Vintarel May 15 '14 at 20:56
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    $\begingroup$ Identity is symmetric $\endgroup$ – ajotatxe May 15 '14 at 20:56
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Assume $A$ has $n$ linearly independent eigen vector. Then $A$ can be diagonalized, i.e., there exists $P$ in which its coulmns are all eigen of $A$ and diagonal matrix $D$ consists of all eigen values of $A$ such that $A=PDP^{-1}$. On the other hands, since $A$ is a upper-triangular matrix having equal entries on the main diagonal, namely $a$, then $D=\operatorname{diag}(a,\dots,a)=aI$. It follows that $$ A=PDP^{-1}=aI $$ that contradicts with the simmetricity of $A$.

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If $A$ diagonal and isn't symmetric then there exists $a_{ij}$, $a_{ij} \neq 0$. Also since all diagonal elemets of $A$ equal (let's say $a = a_{11} = \dots = a_{nn}$) then eigen value is $a$ and has multiplicity $n$. Let $x$ be an eigen vector, then $a x_i + \sum_{m = i+1}^n a_{i m} x_m = a x_i$, and therefore $x_j = \frac{1}{a_{i j}} \sum_{m = i+1, m \neq j}^n a_{i m} x_m$. A component of eigen vector is represented via it's other components, hence there can't be $n$ linearly independent eigen vectors.

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