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Let $E$ be a vector space and $u\in{\cal L}(E)$.

Given $\beta$ a basis of $E$, we denote by $M_\beta(u)$ the matrix of $u$ relative to $\beta$.

If we suppose that, for every basis $\beta$, the matrix $M_\beta(u)$ is upper triangular, then it is easy to prove that there exists some scalar $\lambda$ such that $u=\lambda\,id_E$.

Indeed, if $x$ is any nonzero vector, we can consider suitable vectors $e_2,\cdots,e_n$ such that $\beta=(x,e_2,\cdots,e_n)$ is a basis of $E$. Since $M_\beta(u)$ is upper triangular, we see that $x$ is an eigenvector.

At this point, a classical lemma shows that the conclusion holds.

My question :

How could it be proved that the same results holds when replacing "upper triangular" with "triangular" ?

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    Does triangular just mean upper triangular or lower triangular? – 2017-01-17
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    Of course, but depending of the basis, it can be sometimes upper and sometimes lower – 2017-01-17
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    OH! I see what you were getting at now. – 2017-01-17

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For any given basis $(e_1,\ldots,e_n)$, we can show that $e_1$ or $e_n$ is an eigenvector. So we obtain one eigenvector $f_1$.

Now, choose a basis $(g_1,\ldots,g_n)$ such that $f_1\notin \text{span}\{g_1,g_n\}$. Thus, we obtain a second eigenvector $f_2$ (which is $g_1$ or $g_n$) linear independent with $f_1$. We can repeat the procedure in order to obtain $f_1,f_2,\ldots,f_{n-1}$ linear independent eigenvectors.

Now, let $f\notin\text{span}\{f_1,f_2,\ldots,f_{n-1}\}$ and consider the basis $(f+f_1,\ldots,f+f_{n-1},f)$. Again, $f+f_1$ or $f$ is an eigenvector. Notice that neither $f$ or $f+f_1$ belongs to $\text{span}\{f_1,f_2,\ldots,f_{n-1}\}$.

Thus, we obtained a basis formed by eigenvectors. Now, I leave to you the proof that the eigenvalues associated to these eigenvectors must be equal in order to satisfy the hypothesis.