The eigenvectors of a general upper triangular matrix

Question

I am searching for the eigenvectors of an upper $n\times n$ triangular matrix $U$ with distinct and non-zero entries. What I know is that the diagonal entries are the eigenvalues of $U$ and to determine $v_{\lambda_{i}}$ I took the simplest case that is for $n=2$, we can see that an upper triangular matrix has the first column of the canonical basis of $\mathbb{R}^{n}$ as an eigenvector to $\lambda_{1}$ of $U$. Moreover, I believe $v_{\lambda_{2}}$ can be founded as a linear combination of $e_{2}$ and $e_{1}$. Using the basic form of the eigenvalue problem, I am able to derive the second eigenvector of $U$ as follow :
$$
U(e_{2}+\alpha e_{1})=u_{2,2}(e_{2}+\alpha e_{1})
$$
$$
u_{2,2}+u_{1,2}e_{1}+\gamma u_{1,1}e_{1}=u_{2,2}e_{2}+\gamma u_{2,2}e_{1}
$$
$$
u_{1,2}=\gamma(u_{2,2}-u_{1,1})
$$
assuming distinct eigenvalues $(u_{1,1}\neq u_{2,2})$ we get :
$$
v_{\lambda_{2}}=e_{2}+\frac{u_{1,2}}{u_{2,2}-u_{1,1}}e_{1}
$$
For the $j-1$ eigenvector corresponding to $\lambda_{j-1}$, what I can see is that it must have its first $j-1$ entries as non-zero so I expect $v_{\lambda_{j-1}}$ to be formed as a linear combination of the past $j-2$ eigenvectors such that :
$$
v_{j-1}=e_{j-1}+\alpha_{1} v_{1}+\ldots+\alpha_{j-2}v_{j-2}
$$
I am not able to see what pattern the coeffecients $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{j-1}\}$ have except for the simple case of $n=2$ therefore I hope someone can assist me in generalizing $U$ for its full size to determine the formula for the $j^{th}$ eigenvector

Answer 1

You know that the eigenvalues are the diagonal elements, Let's assume moreover that they are distinct. Reduce now to finding the kernel of an upper triangular $n\times n$ matrix $A$ with exactly one $0$ on the diagonal. Assume that the $k$-th diagonal element is $0$. Let $v$ a vector such that $A v = 0$. Then the components with index $k+1$ up to $n$ of $v$ are all $0$. Only the first $k$ components of $v$ may be non-zero. To determine then, consider the leading $k\times k$ block of $A$. The truncated (first $k$ components) $\bar v$ is orthogonal to the first $k-1$ rows of this block. Now, there is a formula to find a vector ortogonal to $k-1$ of the rows of a $k\times k$ matrix, it involves using algebraic complements ( similar to the vector product that is orthogonal to two vectors).