$A = \left( \begin{array}{ccccc} 1 & 1 &1&1&1 \\ 1&2&2&2&2 \\ 1&2&3&3&3\\ 1&2&3&4&4\\ 1&2&3&4&5 \end{array} \right)$

How to find eigenvectors of $A$? Since $A$ is real symmetric, it has a basis consisting of eigenvectors and we can see vectors of all ones is not an eigenvector. Does that mean no eigenvector of $A$ is orthogonal to vector of all ones?


Assume that there is an eigenvector $v$ with eigenvalue $\lambda$ of $A$ that is orthogonal to $e=(1,\dots,1)^T$. Then we get $$ Av = \pmatrix{ 0 \\ \lambda v_2 \\ \vdots \\\lambda v_n}. $$ Hence $v_1=0$. This implies that the vector $v$ is orthogonal to $(0,1,1,1,1)^T$. It follows $v_2=0$, and, per induction, $v=0$.

Hence that there is no eigenvector of $A$ that is orthogonal to $e$.

This reasoning is only valid due to the special structure of the matrix. In general, the implication: $w\ne0$ is not an eigenvector $\Rightarrow$ no eigenvectors of the matrix are orthogonal to $w$ is not true.

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