Reducing subspace

In linear algebra, a reducing subspace $$W$$ of a linear map $$T:V\to V$$ from a Hilbert space $$V$$ to itself is an invariant subspace of $$T$$ whose orthogonal complement $$W^\perp$$ is also an invariant subspace of $$T.$$ That is, $$T(W) \subseteq W$$ and $$T(W^\perp) \subseteq W^\perp.$$ One says that the subspace $$W$$ reduces the map $$T.$$

One says that a linear map is reducible if it has a nontrivial reducing subspace. Otherwise one says it is irreducible.

If $$V$$ is of finite dimension $$r$$ and $$W$$ is a reducing subspace of the map $$T:V\to V$$ represented under basis $$B$$ by matrix $$M \in\R^{r\times r}$$ then $$M$$ can be expressed as the sum

$$ M = P_W M P_W + P_{W^\perp} M P_{W^\perp}$$

where $$P_W \in\R^{r\times r}$$ is the matrix of the orthogonal projection from $$V$$ to $$W$$ and $$P_{W^\perp} = I - P_{W}$$ is the matrix of the projection onto $$W^\perp.$$ (Here $$I \in \R^{r\times r}$$ is the identity matrix.)

Furthermore, $$V$$ has an orthonormal basis $$B'$$ with a subset that is an orthonormal basis of $$W$$. If $$Q \in \R^{r\times r}$$ is the transition matrix from $$B$$ to $$B'$$ then with respect to $$B'$$ the matrix $$Q^{-1}MQ$$ representing $$T$$ is a block-diagonal matrix

$$Q^{-1}MQ = \left[ \begin{array}{cc} A & 0 \\ 0 & B \end{array} \right] $$

with $$ A\in\R^{d\times d},$$ where $$ d= \dim W$$, and $$ B\in\R^{(r-d)\times(r-d)}.$$