# Abstract Nonsense

## Invariant Subspaces

Point of Post: In this post we’ll discuss the equivalent of section 39 in Halmos, but in more detail.

Motivation

Often in mathematics it’s fruitful to take a look at a class of commonly occurring, easily dealt with set of examples to enrich the overall tapestry of one’s knowledge. Linear transformations are no-different. We’d like to study the idea now of invariant subspaces. The idea being that if $T$ is a linear homomorphism with a non-trivial invariant subspace then $T$ is ‘nicer’ in some sense. We make this more rigorous below.

Invariant Subspaces

In essence, an invariant subspace are precisely those subspaces of a vector space for which the restriction of an endomorphism to that subspace is an endomorphism. But, one may wonder “Why isn’t this true for any subspace? Why do we need to define a special kind of subspace?” The answer is, in a word, silly. Indeed, the ‘shortcoming’ of some subspace with respect to some endomorphism is not that the endomorphism is no longer linear, but merely that there’s no need for the endomorphism to map the subspace back into itself! I’m sure the reader can  easily come up with an endomorphism on $\mathbb{C}^2$ and a subspace $\mathscr{M}$ for which $\mathscr{M}$ is not mapped into itself by this endomorphism! Thus, for the sake of wanting to restrict endomorphisms to subspaces and still have a new endomorphism we’d like only to consider subspaces for which the endomorphism maps that subspace back into itself. We now put this more rigorously.

Let $\mathscr{V}$ be an $F$-space and $T\in\text{End}\left(\mathscr{V}\right)$.  We call $\mathscr{M}\leqslant\mathscr{V}$ an invariant subspace of $T$ if $T\left(\mathscr{M}\right)\subseteq\mathscr{M}$. We now prove what we hinted at above, namely that invariant subspaces of $\mathscr{V}$ are precisely those for which the restriction of $T$ is an endomorphism:

Theorem: Let $\mathscr{V}$ be an $F$-space and $T\in\text{End}\left(\mathscr{V}\right)$. Then, for $\mathscr{M}\leqslant\mathscr{V}$; $\mathscr{M}$ is invariant under $T$ if and only if $T_{\mid \mathscr{M}}\in\text{End}\left(\mathscr{M}\right)$.

Proof: Suppose first that $\mathscr{M}$ is invariant under $T$. Then, we know that $T_{\mid\mathscr{M}}:\mathscr{M}\to\mathscr{M}$ and since trivially

$T(\alpha x+\beta y)=\alpha T(x)+\beta T(y)$

for all $x,y\in\mathscr{M}$ and $\alpha,\beta\in F$ We have that $T_{\mid\mathscr{M}} \in \text{End}\left(\mathscr{M}\right)$.

Conversely, suppose that $T_{\mid\mathscr{M}}\in\text{End}\left(\mathscr{M}\right)$, then by definition we have that $T_{\mid\mathscr{M}}:\mathscr{M}\to\mathscr{M}$ so that $T\left(\mathscr{M}\right)\subseteq\mathscr{M}$ from where the conclusion follows. $\blacksquare$

It turns out though that this isn’t the only interesting characterization of invariant subspaces for finite dimensional spaces. Indeed, there is a characterization via the matrix representation of the endomorphism by a suitable ordered basis. More specifically:

Theorem: Let $\mathscr{V}$ be an $n$-dimensional $F$-space, $\mathscr{M}\leqslant \mathscr{V}$, and $T\in\text{End}\left(\mathscr{V}\right)$. Then, if $\mathcal{B}=(x_1,\cdots,x_m,x_{m+1},\cdots,x_n)$ is an ordered basis for $\mathscr{V}$ such that $\{x_1,\cdots,x_m\}$ is a basis for $\mathscr{M}$ ;then $\mathscr{M}$ is invariant under $T$ if and only if

$\left[T\right]_{\mathcal{B}}=\left(\begin{array}{c|c}A & B\\ \hline 0 & C\end{array}\right)$

where the size of the zero matrix is $(n-m)\times m$.

Proof: Suppose first that $\mathscr{M}$ is invariant under $T$. We note then that

$\displaystyle T(x_j)=\sum_{i=1}^{m}\alpha_{i,j}x_i$ for $j\in[m]$. In particular, $\alpha_{i,j}=0$ for $i\in\{m+1,\cdots,n\}$ and $j\in[m]$. Thus, we see that

$\displaystyle [T]_{\mathcal{B}}=\left(\begin{array}{ccc|ccc}\alpha_{1,1} & \cdots & \alpha_{1,m} & \alpha_{1,m+1} & \cdots & \alpha_{1,n}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ \alpha_{m,1} & \cdots & \alpha_{m,m} & \alpha_{m,m+1}& \cdots & \alpha_{m,n}\\ \hline 0 & \cdots & 0 & \alpha_{m+1,m+1} & \cdots & \alpha_{m+1,n}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ 0 & \cdots & 0 & \alpha_{n,m+1} & \cdots & \alpha_{n,n}\end{array}\right)$

from where the conclusion follows by noticing that the lower left-hand block of zeros is of size $(n-m)\times m$.

Conversely, suppose that

$\displaystyle \left[T\right]_{\mathcal{B}}=\left(\begin{array}{c|c}A & B\\ \hline 0 & C\end{array}\right)$

where the size of the zero block is $(n-m)\times m$. We see then that for $j\in[m]$ we have that $\alpha_{i,j}=0,\text{ }i=m+1,\cdots,n$ in other words for each $x_j,\text{ }j\in[m]$ we have that

$\displaystyle T(x_j)=\sum_{i=1}^{m}\alpha_{i,j}x_i$

namely, $T(x_j)\in\mathscr{M}$. It follows $T(\{x_1,\cdots,x_m\})\subseteq\mathscr{M}$. But since $\mathscr{M}$ this evidently implies that

$\displaystyle \mathscr{M}\supseteq \text{span }T(\{x_1,\cdots,x_m\})=T\left(\text{span }\{x_1,\cdots,x_m\}\right)=T\left(\mathscr{M}\right)$

from where it follows that $\mathscr{M}$ is invariant under $T$. $\blacksquare$

Remark: The second part of the above theorem should really be stated as $\alpha_{i,j}=0$ for $i=m+1,\cdots,n$ and $j\in[m]$. But, the matrix representation really drives it home.

References:

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

December 20, 2010 -