Reducing subspace

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

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

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

$${\displaystyle M=P_{W}MP_{W}+P_{W^{\perp }}MP_{W^{\perp }}}$$

where ${\displaystyle P_{W}\in \mathbb {R} ^{r\times r}}$ is the matrix of the orthogonal projection from ${\displaystyle V}$ to ${\displaystyle W}$ and ${\displaystyle P_{W^{\perp }}=I-P_{W}}$ is the matrix of the projection onto ${\displaystyle W^{\perp }.}$¹ (Here ${\displaystyle I\in \mathbb {R} ^{r\times r}}$ is the identity matrix.)

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

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

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