# Abstract Nonsense

## Direct Sum of Linear Transformations and Direct Sum of Matrices (Pt. I)

Point of post: In this post we discuss the concepts of the direct sum of linear transformations, the direct sum of matrices, and their connection. This is all in preparation to take a broader view of section 40 in Halmos.

Motivation

It’s a leitmotif in mathematics to attempt to break up, when possible, objects into smaller subobjects which are simpler but still capture the essence of the entire object. In this post we show one way to do this for linear transformations. Namely, we show how given endomorphisms on spaces $\mathscr{U}_1,\cdots,\mathscr{U}_m$ which are naturally isomorphic to $\mathscr{W}_1,\cdots,\mathscr{W}_m\leqslant \mathscr{V}$ and endomorphisms $T_k\in\text{End}\left(\mathscr{U}_k\right),\text{ }k\in[m]$ there is a natural way to ‘put them together’ to get an endomorphism on $\displaystyle \mathscr{W}_1\oplus\cdots\oplus\mathscr{W}_m$. We will then see how the structures of the individual endomorphisms affect the structure of the ‘put together one’.

Direct Sum of Linear Transformations and Matrices

In essence, the direct sum of a linear transformation or matrix is breaking it down into it’s smaller components. Those of you who have taken matrix analysis or some similar course will have undoubtedly crossed paths with this notation/concept when dealing with the Jordan decomposition theorem.

Let $\mathscr{V}$ be an $F$-space and $\mathscr{W}_1,\cdots,\mathscr{W}_m\leqslant\mathscr{V}$  such that $\displaystyle \mathscr{V}=\bigoplus_{k=1}^{m}\mathscr{W}_k$ and let $\mathscr{U}_1,\cdots,\mathscr{U}_m$ be $F$-spaces such that there exists isomorphisms $\phi_k:\mathscr{U}_k\to\mathscr{W}_k,\text{ }k\in[m]$ . Then, if $T_k\in\text{End}\left(\mathscr{U}_k\right),\text{ }k\in[m]$ then we define the direct sum of $T_1,\cdots,T_m$ on $\mathscr{V}$ to be the endomorphism $T\in\text{End}\left(\mathscr{V}\right)$ by the rule

$\displaystyle T(v)=T\left(\sum_{k=1}^{m}w_k\right)=\sum_{k=1}^{m}T_k\left(\phi^{-1}_k(w_k)\right)$

where evidently $\displaystyle \sum_{k=1}^{m}w_k$ is the unique representation of $v$ with $w_k\in\mathscr{W}_k,\text{ }k\in[m]$.

The two special cases to consider is when $\mathscr{U}_k=\mathscr{W}_k$ and where $\phi_k=\text{id}_{\mathscr{W}_k}$ in which case the above reduces to

$\displaystyle T(v)=T\left(\sum_{k=1}^{m}w_k\right)=\sum_{k=1}^{m}T_k(w_k)$

and the case when we consider $\mathscr{W}_1\boxplus\cdots\boxplus\mathscr{W}_k$ and $\phi_k$ is the ismomorphism

$\phi_k:\{\bold{0}\}\boxplus\cdots\boxplus\mathscr{W}_k\boxplus\cdots\boxplus\{\bold{0}\}\to\mathscr{W}_k:(\bold{0},\cdots,w_k,\cdots,\bold{0})\mapsto w_k$

When $\mathscr{U}_k$ and $\phi_k$ is not specified it will be assumed that we are dealing the the former case.

We now show some simple consequences of the above definition

Theorem: Let $\mathscr{V}$, $\mathscr{W}_k,\mathscr{U}_k, \phi_k, T_k$ and $T$ be as in the definition of direct sum. Then,

$\displaystyle T\left(\mathscr{V}\right)=\bigoplus_{k=1}^{m}T_k\left(\phi^{-1}_k\left(\mathscr{W}_k\right)\right)=\bigoplus_{k=1}^{m}T_k\left(\mathscr{U}_k\right)$

Proof: We first prove that

$\displaystyle T\left(\mathscr{V}\right)=\sum_{k=1}^{m}T_k\left(\mathscr{U}_k\right)$

To see this we note that

\displaystyle \begin{aligned}T\left(\mathscr{V}\right) &= \left\{T(v):v\in\mathscr{V}\right\}\\ &= \left\{T\left(\sum_{k=1}^{m}w_k\right):w_k\in\mathscr{W}_k,\text{ }\text{ }k\in[m]\right\}\\ &= \left\{\sum_{k=1}^{m}T_k\left(\phi^{-1}_k\left(w_k \right)\right): w_k\in\mathscr{W}_k\text{ }\text{ }k\in[m]\right\}\\ &= \left\{\sum_{k=1}^{m}x_k : x_k\in T_k\left(\phi^{-1}_k\left(\mathscr{W}_k\right)\right)\right\}\\ &= \sum_{k=1}^{m}T_k\left(\phi^{-1}_k\left(\mathscr{W}_k\right)\right)\\ &= \sum_{k=1}^{m}T_k\left(\mathscr{U}_k\right)\end{aligned}

Thus, it suffices to show that the set $\left\{T\left(\mathscr{W}_1\right),\cdots,T\left(\mathscr{W}_m\right)\right\}$ is independent. But, this follows directly from the fact that for each $k\in[m]$ we have that

$\displaystyle \{\bold{0}\}\subseteq T\left(\mathscr{U}_k\right)\cap\sum_{\substack{j\in[m]\\j\ne k}}T\left(\mathscr{U}_j\right)\subseteq \mathscr{W}_k\cap\sum_{\substack{j\in[m]\\j\ne k}}\mathscr{W}_j=\{\bold{0}\}$

The conclusion follows. $\blacksquare$

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

December 22, 2010 -

## 3 Comments »

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