Question

Show that the following conditions are equivalent for a linear operator \(T\) on a finite dimensional space \(V\). 1\. \(M_{B}(T)\) is upper triangular for some ordered basis \(B\) of \(E\) 2\. A basis \(\left\\{\mathbf{b}_{1}, \ldots, \mathbf{b}_{n}\right\\}\) of \(V\) exists such that, for each \(i, T\left(\mathbf{b}_{i}\right)\) is a linear combination of \(\mathbf{b}_{1}, \ldots, \mathbf{b}_{i}\). 3\. There exist \(T\) -invariant subspaces $$ V_{1} \subseteq V_{2} \subseteq \cdots \subseteq V_{n}=V $$ such that \(\operatorname{dim} V_{i}=i\) for each \(i\).

Short answer

The conditions are equivalent: 1 implies 2, 2 implies 3, and 3 implies 1.

Step-by-step solution

Understand the Problem

The exercise asks us to show that three different conditions (about a linear operator \(T\) on a finite dimensional space \(V\)) are actually equivalent to each other. We'll need to verify that if one of these conditions is true, the others must also be true, and vice versa.

Show 1 ⇒ 2

Assume that the matrix \(M_B(T)\) is upper triangular with respect to some basis \(B = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}\). An upper triangular matrix has zeros below the diagonal, so for each \(i\), the vector \(T(\mathbf{b}_i)\) can be expressed as a linear combination of \(\mathbf{b}_1, \ldots, \mathbf{b}_i\) only. This shows that 1 implies 2.

Show 2 ⇒ 3

Assume that a basis \(\{\mathbf{b}_1, \ldots, \mathbf{b}_n\}\) exists such that \(T(\mathbf{b}_i)\) is a linear combination of \(\mathbf{b}_1, \ldots, \mathbf{b}_i\). Define subspaces \(V_i = \text{span}\{\mathbf{b}_1, \ldots, \mathbf{b}_i\}\). Each subspace \(V_i\) is \(T\)-invariant by construction (since applying \(T\) to any vector \(\mathbf{b}_i\) keeps it within \(V_i\)). Hence, condition 2 implies condition 3.

Show 3 ⇒ 1

Assume the existence of \(T\)-invariant subspaces \(V_1 \subseteq V_2 \subseteq \cdots \subseteq V_n = V\) where \(\text{dim}(V_i) = i\). Construct a basis for each \(V_i\) such that \(\{\mathbf{b}_1, \ldots, \mathbf{b}_i\}\) is a basis for \(V_i\). The matrix representation of \(T\) in this basis is upper triangular because each \(V_i\) is \(T\)-invariant. Therefore, condition 3 implies condition 1.

Conclusion

We have shown that 1 implies 2, 2 implies 3, and 3 implies 1. By transitivity of implication, these conditions are equivalent. Thus, any of these conditions being true ensures the others are also true.

Key concepts

Upper Triangular Matrix

An upper triangular matrix is a special kind of square matrix where all the elements below the diagonal are zeros. This type of matrix is very important in linear algebra, especially when dealing with linear operators on vector spaces. Let’s break down its characteristics and significance for better understanding:

When a matrix is upper triangular it simplifies many calculations and solutions of linear systems. For instance, if a matrix is of size 3x3, it will look like this:

\[ \begin{bmatrix} a & b & c \ 0 & d & e \ 0 & 0 & f \end{bmatrix} \]

This structure ensures that each row can be used to easily solve for the corresponding variable in systems of equations by back-substitution. It's especially useful in understanding eigenvalues since the eigenvalues of an upper triangular matrix are the entries on its main diagonal.

Additionally, when a linear transformation's matrix with respect to some basis is upper triangular, there is a special kind of ordering and simplicity in how it acts on those basis vectors. This directly connects to one of the given exercise conditions: each vector within a set is mapped to a linear combination restricted to itself and preceding vectors.

Invariant Subspaces

Invariant subspaces are subspaces of a vector space that remain unchanged when a linear transformation is applied. This means if you have a vector subspace, whenever you apply a linear operator to any vector within this subspace, the resulting vector still lies within the same subspace.

To imagine it, consider a linear operator as a kind of machine, and inputs as vectors from a subspace. When the machine processes these vectors, if all the output vectors belong to the same subspace, this subspace is invariant under the operator.

For deeper understanding, let's use this connection with the exercise. The exercise describes a sequence of invariant subspaces \( V_1 \subseteq V_2 \subseteq \ldots \subseteq V_n = V \). The dimensions grow one by one, with each subspace adding a new dimension, and all are preserved under the action of the operator \( T \).

This concept is essential in understanding how a transformation can structure a vector space, influencing how complicated or simple computations can become.

Basis of Vector Space

A basis of a vector space is a set of vectors that are both linearly independent and span the entire vector space. This means that every vector in the space can be expressed uniquely as a combination of these basis vectors.

Understanding a basis is like finding the essential building blocks of the space. Each vector in the basis adds a new dimension, which allows you to fill the entire space by linear combination of these vectors.

The significance becomes clearer when paired with the given exercise condition. For example, the condition requiring a matrix to be upper triangular showcases how a specific order of basis vectors allows the matrix to have entries only up to the diagonal. Such an ordering ensures each vector's transformation (under the linear operator \( T \)) can be built up from the previous vectors in the basis.

Thus, working with a well-chosen basis can greatly simplify many problems in linear algebra, from matrix representation to solving linear systems efficiently.