Problem 9 - Algebra

Question

Let \(E\) be a vector space over \(k\), of dimension \(n\). Let \(T: E \rightarrow E\) be a linear map such that \(T\) is nilpotent, that is \(T^{m}=0\) for some positive integer \(m\). Show that there exists a basis of \(E\) over \(k\) such that the matrix of \(T\) with respect to this basis is strictly upper triangular.

Step-by-Step Solution

Answer

To show that there exists a basis of \(E\) over \(k\) such that the matrix of \(T\) with respect to this basis is strictly upper triangular: 1. Find a nontrivial subspace \(W\) of \(E\) invariant under \(T\) (e.g., kernel of \(T^m\)). 2. Build a chain of invariant subspaces called the flag. 3. Choose a basis for each subspace in the flag and concatenate them to form a basis for \(E\). 4. Show that the matrix representation of \(T\) with respect to the concatenated basis is strictly upper triangular. This is because the action of \(T\) on a basis vector belonging to an invariant subspace results in a linear combination of basis vectors from the same invariant subspace only.

Step 1: Find an Invariant Subspace

Since \(T\) is nilpotent, there exists a positive integer \(m\) such that \(T^m = 0\). So, for any vector \(v \in E\), we have \(T^m(v) = 0\). In particular, this means that the kernel of \(T^m\) (which is the space of vectors \(v\) such that \(T^m(v) = 0\)) is a nontrivial subspace of \(E\) invariant under \(T\). Let \(W_1 = \text{ker}(T^m)\), and notice that \(T(W_1) \subseteq W_1\). Thus, \(W_1\) is an invariant subspace of \(E\).

Step 2: Build the Flag

Now we build a chain of subspace known as a flag. Let \(W_0 = \{0\}\) and \(W_1\) as previously defined. Proceed inductively as follows: Suppose we have found an invariant subspace \(W_i \subset E\) for some positive integer \(i\). Then, since \(T(W_i) \subseteq W_i\), it follows that there exists a vector space quotient \(W_{i+1}/W_i\) and an induced linear map \(T_i: W_{i+1}/W_i \rightarrow W_{i+1}/W_i\) such that \(T_i(x + W_i) = T(x) + W_i\) for all \(x \in W_{i+1}\). Moreover, since \(T\) is nilpotent, so is \(T_i\), which implies that there exists an invariant subspace \(V_{i+1}/W_i\) of \(W_{i+1}/W_i\). Let \(W_{i+1}\) be the preimage of \(V_{i+1}/W_i\) in \(W_{i+1}\) under the quotient map. Then \(W_{i+1}\) contains \(W_i\) and is invariant under \(T\). Continue this process until the entire vector space \(E\) is spanned by the constructed subspaces.

Step 3: Choose a Basis for Each Invariant Subspace

For each invariant subspace \(W_i\), choose a basis \(B_i\) such that \(B_i\) is also a basis for \(W_{i+1} / W_i\). The concatenated basis formed by joining all bases \(B_i\) together is a basis for the entire vector space \(E\). We will denote this basis by \(B = \{v_1, v_2, \ldots, v_n\}\). Notice that each \(v_i\) belongs to one of the invariant subspaces, \(W_i\).

Step 4: Show the Matrix Representation of T is Strictly Upper Triangular

Let \(A = [a_{ij}]\) be the matrix representation of \(T\) with respect to the basis \(B\). Our goal is to show that \(a_{ij} = 0\) for all \(i \leq j\). Consider the action of \(T\) on a basis vector \(v_i \in B\). Since \(v_i\) belongs to \(W_i\), we have \(T(v_i) \in W_i\). Moreover, because the basis vectors in \(W_i\) are linearly independent of the ones in \(W_j\) for \(j > i\), it follows that \(T(v_i)\) must be a linear combination of basis vectors from \(W_i\) only. In other words, the matrix entries \(a_{ij}\) must be zero for all \(j > i\). Thus, the matrix \(A\) is strictly upper triangular.

Key Concepts

Invariant Subspace

Understanding an invariant subspace is key when analyzing the behavior of linear maps. An invariant subspace with respect to a linear map is a subset of a vector space that, when acted upon by the map, remains within the subset. In simpler terms, if you pick any vector from this special area of our space and apply our transformation—boom—it will not leap out of this area. It's like a safe zone where all vectors abide by the rules of the transformation.

For the nilpotent linear map problem, these subspaces allow us to neatly organize the vector space into smaller, more manageable pieces. Each piece, or invariant subspace, will be used as a stepping stone to piece together the final basis that satisfies our strictly upper triangular matrix condition. It's like finding a set of smaller boxes inside a large box, wherein each small box can be moved independently without affecting the other boxes. These boxes, or invariant subspaces, make it easier to understand the whole space by examining one piece at a time.

Strictly Upper Triangular Matrix

A strictly upper triangular matrix is the secret clubhouse of matrices—an exclusive club where all the action happens above the diagonal, and the diagonal itself and everything below remain strictly zero. These matrices are important because they reveal a lot to us about the linear map they represent, like telling us the map is nilpotent in our current scenario.

For a nilpotent linear map, finding a basis that leads to a strictly upper triangular matrix representation isn't just about aesthetics—it tells us about how the transformation scales down whenever it's applied repeatedly. It's essentially a game of 'hot potato' where the vector components get passed along up the ladder with each application of the linear map, until they're eventually thrown out of the game (become zero).

Basis of Vector Space

A basis of a vector space is your ultimate toolkit for building anything within that space. Just like having a hammer, screwdriver, and nails, the basis gives you the fundamental building blocks to construct any vector. It's a set of vectors that are both linearly independent—they don't rely on each other like a self-sufficient remote team—and they span the entire space, meaning you can reach any point in the space using them.

In our case, we're looking to find a special basis where our nilpotent linear map creates a magical spectacle—a strictly upper triangular matrix. Imagine choosing the perfect set of lego pieces such that when you construct your lego model (apply the linear map), the model gets taller rather than wider with every additional layer (application). Or simply put, each basis vector, when transformed, doesn't interfere with the 'lower' vectors of our basis.

Linear Map Matrix Representation

The matrix representation of a linear map is like a passport for the map, giving us all the deets about its identity in the land of linear algebra. It converts the abstract concept of a linear map into a concrete array of numbers we can touch and feel (mathematically speaking). With respect to a chosen basis, this matrix tells you exactly how the linear map will juggle the components of any vector around.

For a nilpotent map to have a strictly upper triangular matrix representation means matching the map's 'passport' with the 'perfect photo'—the right basis where all the non-zero entries of the transformation are captured above the diagonal. It's all about finding the right angle (basis) to snap its picture (matrix representation) so that it reveals the true nature of the linear map—its power to eventually send vectors to the land of zeros after enough applications.