# Determining Eigenvalues from Upper Triangular Matrices of Linear Operators

So far we have put emphasis on the importance of finding a basis $B_V$ of a finite-dimensional vector space $V$ for which the matrix of a linear operator $T$ with respect to $B_v$, $\mathcal M (T, B_V)$ is upper triangular (or diagonal). One of the reasons why we want to find such a basis is because the eigenvalues of $T$ can easily be obtained as noted in the following proposition.

Proposition 1: Let $V$ be a finite-dimensional vector space. If $T \in \mathcal L (V)$ and $\mathcal M (T, B_V)$ is an upper triangular matrix with respect to some basis $B_V$ of $V$, then the eigenvalues of $T$ are the entries on the main diagonal of $\mathcal M (T, B_V)$. |

**Proof:**Let $V$ be a finite-dimensional vector space and let $B_V = \{ v_1, v_2, …, v_n \}$ be a basis of $V$ such that for the linear operator $T \in \mathcal L (V)$ we have that the matrix $\mathcal M (T, B_V)$ is upper triangular. Then:

(1)

\begin{align} \quad \mathcal M (T) = \mathcal M (T, B_V) = \begin{bmatrix} \lambda_1 & * & \cdots & * \\ 0 & \lambda_2 & \cdots & * \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} \end{align}

- Let $\lambda \in \mathbb{F}$ and consider the matrix $\mathcal M (T – \lambda I) = \mathcal M (T) – \mathcal \lambda M(I)$:

(2)

\begin{align} \quad \mathcal M (T – \lambda I) = \begin{bmatrix} \lambda_1 & * & \cdots & * \\ 0 & \lambda_2 & \cdots & * \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} – \begin{bmatrix} \lambda & 0 & \cdots & 0 \\ 0 & \lambda & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda \end{bmatrix} = \begin{bmatrix} \lambda_1 – \lambda & * & \cdots & * \\ 0 & \lambda_2 -\lambda & \cdots & * \\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n – \lambda \end{bmatrix} \end{align}

- Now recall that we had a theorem that said $\lambda$ is an eigenvalue of $T$ if and only if $T – \lambda I$ is not invertible. Therefore we have that $T – \lambda I$ is not invertible if and only if one of the entries on the main diagonal is zero, that is, $\lambda_j – \lambda = 0$ for $j = 1, 2, …, n$, and so $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is equal to $\lambda_1$, $\lambda_2$, …, or $\lambda_n$. $\blacksquare$