Does an upper triangular matrix always have $\langle 1,0,0\rangle$ as one of its normalized eigenvectors?

My question is exactly what the title asks.

An example matrix to mess around with is:

$$\left\{ \begin{matrix} 3 & 2 & 1 \\ 0 & 1 & 4 \\ 0 & 0 & 5 \end{matrix} \right\}$$

• Yes.${{{{{}}}}}$ Commented Mar 19, 2015 at 22:23
• @GitGud thanks. If you are willing to do a brief write up on why and submit it as an answer I can mark you as correct and this question can be closed. Commented Mar 19, 2015 at 22:31
• All you need to prove it is to try. Take an arbitrary upper triangular matrix $A$. Right $A$ explicitly and multiply by the vector $[1\,0\,\ldots \,0]^T$. Commented Mar 19, 2015 at 22:34

I have to emphasize that probably you mean the right eigen vector. And the answer of course is yes. The first row of the matrix $A$ multiplied with the vector gets scalar which is $a_{1,1}$. All other rows are orthogonal to the ejgenvector. Thus $$A e = a_{11} e$$
Yes if this matrix is relative to the standard basis and if this matrix is relative to another ordered basis $(v_1,v_2,v_3)$ then $v_1$ is an eigenvector associated to the eigenvalue $3$.