Section 7.1 Invariant subspaces
Remark 7.1.2.
- The whole space \(V\) and the trivial subspace \(\{0\}\) are always invariant under such a map \(T\text{.}\)
- If \(W\leq V\) is invariant under \(T\) then, by restricting \(T\) to \(W\text{,}\) \(T\) defines an \(F\)-linear map from \(W\) to \(W\text{.}\) We denote this map by \(T|_W\) and it is called the restriction of \(T\) to \(W\).
- In the Definition 7.1.1 above we only require \(T(W)\subseteq W\text{.}\) We do not require \(T(w)=w\) for every \(w\in W\text{.}\) In other words, \(T\) maps every vector of \(W\) into \(W\text{,}\) however \(T\) need not fix vectors of \(W\) pointwise.
Example 7.1.3.
\begin{equation*}
e_1\mapsto e_2,\quad e_2\mapsto e_1,\quad\text{and}\quad e_3\mapsto e_1+e_2+e_3.
\end{equation*}
The subspace \(W\) of \(\R^3\) spanned by \(e_1,e_2\) is invariant under \(T\text{.}\) Note that \(T(e_1)\neq e_1\) but \(T(e_1)\in W\text{.}\) The matrix of \(T\) with respect to \(\mathfrak{B}\) is the following.
\begin{equation*}
[T]_{\mathfrak{B}}=\begin{pmatrix}0\amp 1\amp 1\\1\amp 0\amp 1\\0\amp 0\amp 1\end{pmatrix}
\end{equation*}
Example 7.1.4. (Subspaces invariant under projection of \(\R^2\) onto \(X\)-axis).
Let \(T\colon\R^2\to\R^2\) be a projection onto \(X\)-axis, i.e., \(T(x,y)=(x,0)\text{.}\) Then the \(X\)-axis, \(W=\{(x,0)\in\R^2:x\in\R\}\) is invariant under \(T\text{.}\) In fact, \(T(w)=w\) for every \(w\in W\text{.}\)
Consider the standard ordered basis \(\mathfrak{B}=(e_1,e_2)\) of \(\R^2\text{.}\) The matrix of \(T\) with respect to \(\mathfrak{B}\) is the following block diagonal matrix.
\begin{equation*}
[T]_{\mathfrak{B}}=\begin{pmatrix}1\amp 0\\0\amp 0\end{pmatrix}
\end{equation*}
Consider \(S\) to be a proper nonzero subspace of \(\R^2\) which is invariant under \(T\text{.}\) We must have \(\dim_FS=1\text{.}\) Suppose that \(S=\langle ae_1+be_2\rangle\) for some \(a,b\in \R\text{.}\) We have
\begin{equation*}
T\left(\alpha\cdot(ae_1+be_2)\right)=T\left(\alpha ae_1+\alpha be_2\right)=\alpha a e_1.
\end{equation*}
Therefore, \(S\) must be spanned by \(e_1\text{,}\) i.e., \(S\) is the \(X\)-axis.
In other words, only proper nonzero subspace of \(\R^2\) invariant under the projection \(T\) is the \(X\)-axis.
Proposition 7.1.5.
Let \(V\) be a finite-dimensional vector space over a field \(F\) and \(T\colon V\to V\) be an \(F\)-linear map. Suppose that \(W\) is an invariant subspace of \(T\text{.}\) Then there exists a basis \(\mathfrak{B}\) of \(V\) such that with respect to \(\mathfrak{B}\) the matrix of \(T\) has the following block form.
\begin{equation*}
[T]_{\mathfrak{B}}=\begin{pmatrix}A\amp B\\\mathbf{0}\amp D\end{pmatrix}
\end{equation*}
Proof.
\begin{equation*}
\mathfrak{B}=\{w_1,\ldots,w_r,v_1,\ldots,v_s\}
\end{equation*}
of \(V\text{.}\) Thus \(T(w_i)=\sum_j\alpha_{ji}w_j\) because \(W\) is invariant under \(T\text{.}\) On the other hand \(T(v_k)=\sum_j\beta_{jk}w_j+\sum_\ell\gamma_{\ell k}v_\ell\text{.}\) If we consider
\begin{equation*}
A=(\alpha_{ji})_{1\leq i,j\leq r},\quad B=(\beta_{jk})_{1\leq k\leq s}^{1\leq j\leq r},\quad\text{and}\quad D=(\gamma_{\ell k})_{1\leq k\leq s}^{1\leq\ell\leq s}
\end{equation*}
then we get the required result.
