Upper Triangular Matrices for Operators on Complex Vector Spaces

We recently saw that if $V$ is a finite-dimensional nonzero vector space over the complex numbers $\mathbb{C}$ then every operator $T \in \mathcal L (V)$ has at least one eigenvalue. The following theorem will tell us that for every operator $T$ of $V$ there exists a basis $B_V$ of $V$ such that $T$ has an upper triangular matrix.

Theorem 1: If $V$ is a finite-dimensional vector space over the complex numbers and if $T \in \mathcal L(V)$ then there exists a basis $B_V$ such that $\mathcal M (T, B_V)$ is upper triangular matrix.

• Proof: To prove Theorem 1, we will use mathematical induction. Suppose that $V$ is a finite-dimensional vector space over $\mathbb{C}$.

• If $\mathrm{dim} (V) = 1$, then the matrix $\mathcal M (T, B_v)$ is a $1 \times 1$ matrix and is trivially upper triangular.

• Suppose that Theorem 1 holds for all complex vector spaces with dimension less than $\mathrm{dim} (V)$, and let $\lambda$ be an eigenvalue of $T$. Let $U = \mathrm{range} (T - \lambda I)$. Since $\lambda$ is an eigenvalue of $T$ we know that $T - \lambda I$ is not injective and so $T - \lambda I$ is not surjective, so $\mathrm{dim} (U) < \mathrm{dim} (V)$. We also note that $U$ is invariant under $T$ since if $u \in U$ then:

(1)

\begin{align} \quad T(u) = T(u) - \lambda u + \lambda u \\ \quad T(u) = (T - \lambda I)(u) + \lambda u \end{align}

• Since $(T - \lambda I)(u) \in U$ and $\lambda u \in U$ we have that $T(u) \in U$ so $U$ is invariant under $T$.

• Therefore $T \mid_U : U \to U$ is an operator. Since $\mathrm{dim} (U) < \mathrm{dim} (V)$ as we saw earlier, we have that by our induction hypothesis, the operator $T \mid_U : U \to U$ has an upper triangular matrix. Since $T \mid_U$ has an upper triangular matrix with respect to some basis $\{ u_1, u_2, ..., u_j \}$ of $U$, and we have by one of the theorems on the Upper Triangular Matrices of Linear Operators page that for each $j = 1, 2, ..., m$ that:

(2)

\begin{align} \quad T(u_j) = (T \mid_U) (u_j) \in \mathrm{span} (u_1, u_2, ..., u_j) \end{align}

• We then take the basis $\{ u_1, u_2, ..., u_m \}$ of $U$ and extend it to a basis $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_n \}$ of $V$, and so for each $k$ we have that:

(3)

\begin{align} \quad T(v_k) = T(v_k) - \lambda v_k + \lambda v_k \\ \quad T(v_k) = (T - \lambda I)(v_k) + \lambda v_k \end{align}

• Note that since $U = \mathrm{range} (T - \lambda I)$ we have that $(T - \lambda I)(v_k) \in U = \mathrm{span} (u_1, u_2, ..., u_m)$ and so $T(v_k) \in \mathrm{span} (u_1, u_2, ..., u_m, v_1, v_2, ..., v_k)$ for each $k$.

• Therefore $T(u_j) \in \mathrm{span} (u_1, u_2, ..., u_j)$ for each $j$ and $T(v_k) \in \mathrm{span} (u_1, u_2, ..., u_m, v_1, v_2, ..., v_k)$ for each $k$. From the theorem mentioned from the Upper Triangular Matrices of Linear Operators page linked above, we conclude that $T$ must have an upper triangular matrix with respect to the basis $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_n \}$. $\blacksquare$