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 $M\in \mathbb {R} ^{r\times r}$ then $M$ can be expressed as the sum

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

where $P_{W}\in \mathbb {R} ^{r\times r}$ 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 $I\in \mathbb {R} ^{r\times r}$ is the identity matrix.)

Furthermore, $V$ has an orthonormal basis $B'$ with a subset that is an orthonormal basis of $W$. If $Q\in \mathbb {R} ^{r\times r}$ 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 $A\in \mathbb {R} ^{d\times d},$ where $d=\dim W$, and $B\in \mathbb {R} ^{(r-d)\times (r-d)}.$