# Chapter 43. Invariant Subspaces # Chapter 43. Invariant Subspaces An invariant subspace is a subspace that is preserved by a linear operator. Let \(V\) be a vector space over a field \(F\), and let $$ T:V\to V $$ be a linear operator. A subspace \(U\subseteq V\) is called invariant under \(T\), or \(T\)-invariant, if $$ T(U)\subseteq U. $$ Equivalently, $$ u\in U \quad \Longrightarrow \quad T(u)\in U. $$ The operator may move vectors inside \(U\), but it does not move them outside \(U\). This is the standard definition of an invariant subspace. ## 43.1 Basic Meaning An invariant subspace is a smaller space on which the operator acts by itself. If $$ U\subseteq V $$ is \(T\)-invariant, then the restriction $$ T|_U:U\to U $$ is a well-defined linear operator on \(U\). This is the main reason invariant subspaces are important. They allow us to study a large operator by studying its action on smaller subspaces. For example, if \(T:\mathbb{R}^3\to\mathbb{R}^3\) preserves a plane \(U\), then the behavior of \(T\) on that plane can be analyzed separately from the rest of the space. ## 43.2 First Examples Every operator has at least two invariant subspaces: $$ \{0\} $$ and $$ V. $$ The zero subspace is invariant because $$ T(0)=0. $$ The whole space is invariant because $$ T(V)\subseteq V. $$ These are called trivial invariant subspaces. The interesting question is whether an operator has nontrivial invariant subspaces, meaning subspaces other than \(\{0\}\) and \(V\). ## 43.3 Kernel and Image The kernel and image of a linear operator are invariant subspaces. Let $$ T:V\to V $$ be linear. The kernel is $$ \ker(T)=\{v\in V:T(v)=0\}. $$ If \(v\in\ker(T)\), then $$ T(v)=0. $$ Since $$ 0\in\ker(T), $$ we have $$ T(v)\in\ker(T). $$ Thus $$ \ker(T) $$ is \(T\)-invariant. The image is $$ \operatorname{im}(T)=\{T(v):v\in V\}. $$ If \(w\in\operatorname{im}(T)\), then $$ w=T(v) $$ for some \(v\in V\). Applying \(T\) again gives $$ T(w)=T(T(v))=T^2(v). $$ Since \(T^2(v)\) is still an output of \(T\), it lies in \(\operatorname{im}(T)\). Therefore $$ \operatorname{im}(T) $$ is \(T\)-invariant. Kernel, range, and eigenspaces are standard examples of invariant subspaces. ## 43.4 Eigenspaces Eigenspaces are invariant subspaces. Let \(\lambda\in F\) be an eigenvalue of \(T\). The eigenspace associated with \(\lambda\) is $$ E_\lambda=\ker(T-\lambda I). $$ Thus $$ E_\lambda=\{v\in V:T(v)=\lambda v\}. $$ If \(v\in E_\lambda\), then $$ T(v)=\lambda v. $$ Since \(E_\lambda\) is a subspace, and \(v\in E_\lambda\), we have $$ \lambda v\in E_\lambda. $$ Therefore $$ T(v)\in E_\lambda. $$ So $$ E_\lambda $$ is invariant under \(T\). This means that an eigenvector does more than give a special direction. The whole eigenspace is preserved by the operator. ## 43.5 Example: A Diagonal Matrix Let $$ A= \begin{bmatrix} 2&0&0\\ 0&3&0\\ 0&0&5 \end{bmatrix}. $$ The coordinate axes are invariant under \(A\). For example, $$ U=\operatorname{span}\{e_1\} $$ is invariant because $$ Ae_1=2e_1. $$ Similarly, $$ \operatorname{span}\{e_2\} $$ and $$ \operatorname{span}\{e_3\} $$ are invariant. The coordinate planes are also invariant. For example, $$ W=\operatorname{span}\{e_1,e_2\}. $$ If $$ w=ae_1+be_2, $$ then $$ Aw=2ae_1+3be_2. $$ This remains in \(W\). Hence \(W\) is invariant. Diagonal matrices make invariant subspaces visible because each coordinate direction is scaled independently. ## 43.6 Example: A Shear Let $$ A= \begin{bmatrix} 1&1\\ 0&1 \end{bmatrix}. $$ Then $$ A \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} x+y\\ y \end{bmatrix}. $$ The \(x\)-axis $$ U=\operatorname{span} \left\{ \begin{bmatrix} 1\\ 0 \end{bmatrix} \right\} $$ is invariant, because $$ A \begin{bmatrix} x\\ 0 \end{bmatrix} = \begin{bmatrix} x\\ 0 \end{bmatrix}. $$ The \(y\)-axis is not invariant. Indeed, $$ A \begin{bmatrix} 0\\ 1 \end{bmatrix} = \begin{bmatrix} 1\\ 1 \end{bmatrix}, $$ which does not lie on the \(y\)-axis. Thus a subspace may look natural geometrically but fail to be invariant under a given operator. ## 43.7 Restriction to an Invariant Subspace If \(U\) is invariant under \(T\), then the restriction $$ T|_U:U\to U $$ is defined by $$ T|_U(u)=T(u). $$ The invariant condition ensures that the output lies in \(U\). The restriction is linear because \(T\) is linear. For \(u_1,u_2\in U\), $$ T|_U(u_1+u_2)=T(u_1+u_2)=T(u_1)+T(u_2). $$ Since $$ T(u_1),T(u_2)\in U, $$ this is addition inside \(U\). Similarly, $$ T|_U(cu)=T(cu)=cT(u)=cT|_U(u). $$ Thus invariant subspaces allow us to form smaller linear operators from larger ones. ## 43.8 Matrix Form Let \(T:V\to V\) be linear, and suppose \(U\subseteq V\) is \(T\)-invariant. Choose a basis $$ (u_1,\ldots,u_k) $$ of \(U\), and extend it to a basis of \(V\): $$ (u_1,\ldots,u_k,w_1,\ldots,w_{n-k}). $$ Since \(U\) is invariant, each $$ T(u_j) $$ is a linear combination only of $$ u_1,\ldots,u_k. $$ Therefore the matrix of \(T\) in this basis has block form $$ \begin{bmatrix} A&B\\ 0&C \end{bmatrix}. $$ The zero block appears because vectors in \(U\) have no component outside \(U\) after applying \(T\). This block upper triangular form is one of the main uses of invariant subspaces. ## 43.9 Reducing Subspaces A stronger condition occurs when both a subspace and its complement are invariant. Suppose $$ V=U\oplus W, $$ and both \(U\) and \(W\) are invariant under \(T\). Then \(T\) decomposes into two independent operators: $$ T|_U:U\to U $$ and $$ T|_W:W\to W. $$ In a basis adapted to the decomposition, the matrix of \(T\) has block diagonal form $$ \begin{bmatrix} A&0\\ 0&C \end{bmatrix}. $$ This is stronger than block upper triangular form. It means the operator does not mix the two subspaces in either direction. Such a decomposition allows the study of \(T\) to split into the study of smaller operators. ## 43.10 Invariant Lines A one-dimensional invariant subspace is an invariant line through the origin. Let $$ U=\operatorname{span}\{v\}, \qquad v\neq 0. $$ Then \(U\) is invariant under \(T\) if and only if $$ T(v)\in \operatorname{span}\{v\}. $$ This means there exists a scalar \(\lambda\) such that $$ T(v)=\lambda v. $$ Therefore invariant lines are exactly eigenspaces generated by eigenvectors. Thus finding one-dimensional invariant subspaces is the same as finding eigenvectors. ## 43.11 Invariant Planes A two-dimensional invariant subspace is an invariant plane through the origin. Let $$ U=\operatorname{span}\{u_1,u_2\}. $$ The subspace \(U\) is invariant under \(T\) if and only if both $$ T(u_1) $$ and $$ T(u_2) $$ belong to \(U\). This condition is enough because every vector in \(U\) has the form $$ au_1+bu_2. $$ Then $$ T(au_1+bu_2)=aT(u_1)+bT(u_2), $$ which lies in \(U\) if both basis images lie in \(U\). Thus to check invariance of a finite-dimensional subspace, it is enough to check a basis. ## 43.12 Cyclic Subspaces Given a vector \(v\in V\), the cyclic subspace generated by \(v\) under \(T\) is $$ \mathcal{K}(T,v)=\operatorname{span}\{v,Tv,T^2v,T^3v,\ldots\}. $$ This subspace is invariant under \(T\). Indeed, if $$ w=a_0v+a_1Tv+\cdots+a_kT^kv, $$ then $$ T(w)=a_0Tv+a_1T^2v+\cdots+a_kT^{k+1}v. $$ This again lies in $$ \mathcal{K}(T,v). $$ Cyclic subspaces appear in Krylov methods, rational canonical form, and the study of minimal polynomials. ## 43.13 Invariant Subspaces and Polynomials If \(U\) is invariant under \(T\), then \(U\) is invariant under every polynomial in \(T\). Let $$ p(t)=a_0+a_1t+\cdots+a_mt^m. $$ Then $$ p(T)=a_0I+a_1T+\cdots+a_mT^m. $$ If \(u\in U\), then $$ T(u)\in U. $$ Repeated application gives $$ T^k(u)\in U $$ for every \(k\geq 0\). Since \(U\) is closed under linear combinations, $$ p(T)u\in U. $$ Therefore \(U\) is invariant under \(p(T)\). This fact connects invariant subspaces with minimal polynomials and canonical forms. ## 43.14 Generalized Eigenspaces Let \(T:V\to V\) be linear, and let \(\lambda\) be an eigenvalue. The generalized eigenspace associated with \(\lambda\) is $$ G_\lambda=\ker((T-\lambda I)^m) $$ for a sufficiently large positive integer \(m\). This subspace is invariant under \(T\). To see this, let $$ N=T-\lambda I. $$ Since \(N\) commutes with \(T\), we have $$ N^mT=TN^m. $$ If $$ v\in G_\lambda, $$ then $$ N^m v=0. $$ Therefore $$ N^mT(v)=TN^m(v)=T(0)=0. $$ So $$ T(v)\in G_\lambda. $$ Generalized eigenspaces are the invariant pieces used in Jordan canonical form. ## 43.15 Direct Sums of Invariant Subspaces Suppose $$ V=U_1\oplus U_2\oplus\cdots\oplus U_k, $$ and each \(U_i\) is invariant under \(T\). Then \(T\) splits into operators $$ T|_{U_i}:U_i\to U_i. $$ In a basis formed by concatenating bases of the \(U_i\), the matrix of \(T\) is block diagonal: $$ \begin{bmatrix} A_1&0&\cdots&0\\ 0&A_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&A_k \end{bmatrix}. $$ This is the algebraic reason direct sum decompositions are powerful. They convert one large operator into several smaller operators. ## 43.16 Invariant Subspaces and Triangular Form A chain of invariant subspaces gives a triangular matrix. Suppose \(V\) has a basis $$ (v_1,\ldots,v_n) $$ such that $$ U_k=\operatorname{span}\{v_1,\ldots,v_k\} $$ is invariant under \(T\) for each \(k=1,\ldots,n\). Then the matrix of \(T\) in this basis is upper triangular. Indeed, invariance of \(U_k\) means $$ T(v_k)\in U_k. $$ Therefore the \(k\)-th column of the matrix has zero entries below row \(k\). Thus nested invariant subspaces correspond to triangular representations. ## 43.17 Invariant Subspaces and Diagonalization Diagonalization is the best-case invariant subspace decomposition. If \(T\) has a basis of eigenvectors $$ (v_1,\ldots,v_n), $$ then each line $$ U_i=\operatorname{span}\{v_i\} $$ is invariant under \(T\). Moreover, $$ V=U_1\oplus\cdots\oplus U_n. $$ In this basis, the matrix of \(T\) is diagonal: $$ \begin{bmatrix} \lambda_1&0&\cdots&0\\ 0&\lambda_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&\lambda_n \end{bmatrix}. $$ Thus diagonalization means decomposing the vector space into one-dimensional invariant subspaces. ## 43.18 Invariant Subspaces Over Complex Fields Over an algebraically closed field, every finite-dimensional operator on a nonzero vector space has at least one eigenvalue. Therefore it has at least one one-dimensional invariant subspace. For complex vector spaces, this means every operator $$ T:V\to V $$ with $$ \dim V\geq 1 $$ has an invariant line. This follows from the characteristic polynomial: over \(\mathbb{C}\), it has a root \(\lambda\). Then $$ \ker(T-\lambda I) $$ contains a nonzero vector and is invariant. This fact is one reason complex linear algebra has especially clean spectral theory. ## 43.19 Invariant Subspaces Over Real Fields Over \(\mathbb{R}\), a finite-dimensional operator may have no invariant line. For example, the rotation $$ R_{\pi/2}= \begin{bmatrix} 0&-1\\ 1&0 \end{bmatrix} $$ has no real eigenvectors. Therefore it has no one-dimensional real invariant subspace. However, the whole plane is invariant. More generally, real operators may have invariant planes corresponding to pairs of complex conjugate eigenvalues. Thus real linear algebra often replaces invariant lines by invariant planes. ## 43.20 Nontrivial Invariant Subspaces In finite-dimensional complex vector spaces, nontrivial invariant subspaces are common. If $$ \dim V>1, $$ then an eigenline gives a nontrivial invariant subspace. In infinite-dimensional spaces, the situation is much more delicate. The invariant subspace problem asks, in one classical form, whether every bounded operator on a complex separable Hilbert space of dimension greater than one has a nontrivial closed invariant subspace. This problem remains unsolved in that setting. This infinite-dimensional problem belongs to functional analysis, but it shows how central the idea of invariance is beyond finite-dimensional linear algebra. ## 43.21 How to Test Invariance To test whether a subspace \(U\) is invariant under \(T\), use a basis of \(U\). Let $$ U=\operatorname{span}\{u_1,\ldots,u_k\}. $$ Then \(U\) is invariant under \(T\) if and only if $$ T(u_j)\in U $$ for every $$ j=1,\ldots,k. $$ For a matrix \(A\), this means checking whether each \(Au_j\) is a linear combination of the basis vectors of \(U\). Equivalently, if \(Q\) is the matrix with columns \(u_1,\ldots,u_k\), then \(U\) is invariant under \(A\) if there exists a \(k\times k\) matrix \(B\) such that $$ AQ=QB. $$ The matrix \(B\) is the matrix of the restricted operator \(T|_U\) in the chosen basis of \(U\). ## 43.22 Summary An invariant subspace of a linear operator $$ T:V\to V $$ is a subspace \(U\subseteq V\) satisfying $$ T(U)\subseteq U. $$ Equivalently, every vector in \(U\) is sent back into \(U\). The restriction $$ T|_U:U\to U $$ is then a well-defined linear operator. Important examples include $$ \{0\},\quad V,\quad \ker(T),\quad \operatorname{im}(T), $$ eigenspaces, generalized eigenspaces, and cyclic subspaces. Invariant subspaces explain block triangular form, block diagonal form, diagonalization, Jordan form, and rational canonical form. They are the subspaces on which an operator can be studied separately. The central idea is that an invariant subspace is a part of the vector space that the operator does not leave.