Let $T\in\mathbb{C}^{n\times n}$ an upper triangular matrix with distinct eigenvalues which are precisely its diagonal entries $\{t_{11},\dots,t_{nn}\}$ (by known result). Represent $T$ as \begin{equation} T=\left( \begin{array}{ccc} t_{11} & \dots & t_{1n}\\ & \ddots & \vdots \\ \mbox{0} & & t_{nn} \\ \end{array}\right) \end{equation} I am trying to show the following propositions:

  • There is a left eigenvector of $T$ corresponding to the $i$-th eigenvalue $t_{ii}$ whose first $i-1$ components are all zero.
  • There is a right eigenvector of $T$ corresponding to the $i$-th eigenvalue $t_{ii}$ whose last $n-i$ components are all zero.

The result of the first proposition is trivial if we choose $i=n$ so that the vector $\mathbf{e}_n^T$ is a left eigenvector associated with $t_{nn}$. But I cannot come up with a general proof-example for none of the propositions.


1 Answer 1


Consider the matrix

$$ T = \begin{pmatrix}T_{11} & T_{12} & \cdots\\ 0 & T_{22} &\cdots \\ 0&0 & \ddots\end{pmatrix}$$

Obviously $v_1 = e_1$ is an eigenvector with $Te_1 = T_{11}e_1$. However $e_2$ is, by itself, not an eigenvector: $Te_2 = T_{12}e_1 + T_{22}e_2$. But we can get find a new eigenvector by adding a $e_1$ component:

\begin{align} T(e_2 + \mu e_1) &\overset{!}{=} T_{22}(e_2 + \mu e_1) \\\iff T_{22}e_2 + T_{12}e_1 + \mu T_{11}e_1 &= T_{22}e_2 + \mu T_{22}e_1 \\\iff T_{12} &= \mu (T_{22} -T_{11}) \end{align}

And since the eigenvalues were assumed to be distinct, $T_{22}-T_{11}\neq 0$ and so $$ v_2 = e_2 + \frac{T_{12}}{T_{22}-T_{11}}e_1$$ Is an eigenvector to the eigenvalue $T_{22}$ that is zero in all components but the first $2$.

Now you can continue inductively to find the $k$-th eigenvector of the form

$$v_k = e_k + \alpha_1 v_1 + \ldots + \alpha_{k-1} v_{k-1}$$


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