# Upper Triangular Matrices of Linear Operators

Recall from The Matrix of a Linear Map page that if $V$ and $W$ are finite-dimensional vector spaces over the field $\mathbb{F}$ such that $\mathrm{dim} (V) = n$ and $\mathrm{dim} (W) = m$ and if $B_V = \{ v_1, v_2, …, v_n \}$ and $B_w = \{ w_1, w_2, …, w_m \}$ are bases of $V$ and $W$ respectively, then if $T \in \mathcal L (V, W)$ is such that $T(v_k) = a_{1,k}w_1 + a_{2,k}w_2 + … + a_{m,k} w_m$ then the matrix of the linear map $T$ with respect to the bases $B_V$ and $B_w$ is:

(1)

We will now look more specifically at the matrices of a linear operator $T : V \to V$ where $V$ is finite-dimensional such that $\mathrm{dim} (V) = n$. Let $B_V = \{ v_1, v_2, …, v_n \}$ be a basis of $V$. Then since $B_V$ is a basis of $V$ we have that for some scalars $a_{1, k}, a_{2,k}, …, a_{n,k} \in \mathbb{F}$, the image of each basis vector $v_k$ for $k = 1, 2, …, n$ under the linear operator $T$ is:

(2)

The corresponding matrix for this linear operator with respect to this basis $B_V$ is going to be a square $n \times n$ matrix:

(3)

Of course, the matrix $\mathcal M (T)$ above dependents on our choice of basis $B_V$. Now suppose that we want to find a basis $B_v$ for which the matrix $\mathcal M (T)$ is relatively simple. There are two types of “simple” matrices. The first type are diagonal matrices and the second type are triangular matrices and at the moment, we’ll want upper triangular matrices. Let’s look at a few propositions.

Proposition 1: If $V$ is a finite-dimensional complex vector space, then there exists a basis $B_V$ of $V$ such that $\mathcal M (T, B_V) = \begin{bmatrix} \lambda & * & \cdots & *\\ 0& * & \cdots & *\\ \vdots & \vdots & \ddots & \vdots\\ 0 & * & \cdots & *\\ \end{bmatrix}$. |

**Proof:**Let $V$ be a finite-dimensional vector space over the complex numbers with $\mathrm{dim} (V) = n$. From The Existence of an Eigenvalue on Finite-Dimensional Complex Vector Spaces page, we know that every linear operator $T \in \mathcal L (V)$ has an eigenvalue, call it $\lambda$. Let $v \in V$ be a corresponding nonzero eigenvector to this eigenvalue $\lambda$, that is $T(v) = \lambda v$. Since $V$ is finite-dimensional, we can extend the set $\{ v \}$ to a basis of $V$ of $n$ vectors $\{ v, v_1, v_2, …, v_{n-1} \}$. Then we note that $T(v) = \lambda v + 0 v_1 + … + 0 v_n$ since $\{ v, v_1, …, v_{n-1} \}$ is a linearly independent set as a basis, and thus such a matrix $\mathcal M (T, B_V)$ in the form above exists. $\blacksquare$

Proposition 2: If $V$ is a finite-dimensional nonzero vector space, $T \in \mathcal L(V)$, and $B_v = \{ v_1, v_2, …, v_n \}$ is a basis of $V$ then the following statements are equivalent:a) The matrix $\mathcal M (T, B_V)$ is upper triangular.b) $T(v_k) \in \mathrm{span} (v_1, v_2, …, v_k)$ for $k = 1, 2, …, n$.c) The set of vectors $\mathrm{span} (v_1, v_2, …, v_k)$ is invariant under $T$ for $k = 1, 2, …, n$. |

**Proof of $a) \implies b)$:**Suppose that $\mathcal M (T, B_V)$ is upper triangular. Then $\mathcal M (T, B_V) = \begin{bmatrix} a_{1,1} & a_{1, 2} & \cdots & a_{1,n}\\ 0 & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & a_{n,n} \end{bmatrix}$. From how the matrix of $T$ with respect to the basis $B_V$ is constructed, we see that $T(v_k) = a_{1,k}v_1 + a_{2,k}v_2 + … + a_{k,k}v_k$ for $k = 1, 2, …, n$, and so $T(v_k) \in \mathrm{span} (v_1, v_2, …, v_k)$ for $k = 1, 2, …, n$.

**Proof of $b) \implies c)$:**Suppose that $T(v_k) \in \mathrm{span} (v_1, v_2, …, v_k)$ for $k = 1, 2, …, n$. Then we have that $T(v_1) \in \mathrm{span} (v_1) \subset \mathrm{span} (v_1, v_2, …, v_k)$, $T(v_2) \in \mathrm{span} (v_1, v_2,) \subset \mathrm{span} (v_1, v_2, …, v_k)$, …, and $T(v_k) \in \mathrm{span} (v_1, v_2, …, v_k)$.

- Let $v \in \mathrm{span} (v_1, v_2, …, v_k)$ for $k = 1, 2, …, n$. Then $T(v) \in \mathrm{span} (v_1, v_2, …, v_k)$ and so $\mathrm{span} (v_1, v_2, …, v_k)$ is invariant under $T$.

**Proof of $c) \implies b)$:**Suppose that $\mathrm{span} (v_1, v_2, …, v_k)$ is invariant under $T$ for $k = 1, 2, …, n$. Then for $v_k \in \mathrm{span} (v_1, v_2, …, v_k)$ we have that $T(v_k) \in \mathrm{span} (v_1, v_2, …, v_k)$ for $k = 1, 2, …, n$. $\blacksquare$

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