Invariant Subspaces and First Examples

What It Means for a Subspace to Be Invariant

A linear map may send some parts of a vector space back into themselves while sending other vectors elsewhere. Those preserved parts are useful because the map can be studied on them separately. The central question is whether applying the map to every vector in a chosen subspace keeps the result inside that same subspace.

Let \(V\) be a vector space over \(\mathbb{R}\), and let \(T:V\to V\) be linear. A subspace \(W\) of \(V\) is a subset containing \(\mathbf{0}\) that is closed under addition and scalar multiplication. The image of \(W\) under \(T\) is the set \(T(W)=\{T(\mathbf{w}):\mathbf{w}\in W\}\). Invariance is a containment condition: the image need not equal \(W\).

The quantifier “every” matters. Checking one vector in \(W\) is not enough; a successful test must establish that the image of an arbitrary vector in \(W\) remains in \(W\). The zero subspace \(\{\mathbf{0}\}\) and the whole space \(V\) are always invariant: linearity gives \(T(\mathbf{0})=\mathbf{0}\), and \(T\) maps into \(V\) by definition.

A First Example: An Eigenspace

An eigenvector of \(T\) is a nonzero vector \(\mathbf{v}\) for which \(T(\mathbf{v})=\lambda\mathbf{v}\) for some scalar \(\lambda\), called its eigenvalue. The line spanned by an eigenvector is a natural candidate for an invariant subspace: multiplying that vector by a scalar still gives a vector on the same line.

This calculation illustrates the definition directly: start with an arbitrary vector in the proposed subspace, apply the map, and check that the result still belongs to it. In this example, the map scales every vector on the eigenspace by \(2\).

Kernels and Images

The kernel and image of a linear map are subspaces. For a linear map \(S:V\to V\), the kernel is \(\ker S=\{\mathbf{v}\in V:S(\mathbf{v})=\mathbf{0}\}\), and the image is \(\operatorname{im}S=\{S(\mathbf{v}):\mathbf{v}\in V\}\). These spaces are sometimes preserved by a second map \(T\); commutation of the maps gives a useful condition.

Proof. Suppose \(\mathbf{w}\in\ker S\). Then \(S(\mathbf{w})=\mathbf{0}\), and commutation gives \(S(T(\mathbf{w}))=T(S(\mathbf{w}))=T(\mathbf{0})=\mathbf{0}\). Hence \(T(\mathbf{w})\in\ker S\), proving \(T(\ker S)\subseteq\ker S\). Next, suppose \(\mathbf{y}\in\operatorname{im}S\). By the definition of image, there is some \(\mathbf{x}\in V\) with \(\mathbf{y}=S(\mathbf{x})\). Therefore \(T(\mathbf{y})=T(S(\mathbf{x}))=S(T(\mathbf{x}))\), which belongs to \(\operatorname{im}S\). Thus \(T(\operatorname{im}S)\subseteq\operatorname{im}S\), as required. \(\square\)

Taking \(S=T\) meets the commutation condition because \(TT=TT\). Therefore, the kernel and image of \(T\) itself are invariant under \(T\). The general statement also explains why kernels and images of a different map need not be invariant unless there is a suitable relationship between the maps.

Coordinate Subspaces

A coordinate subspace is formed by requiring selected coordinates to be zero. Such a subspace is not automatically invariant: the matrix may move a vector into a coordinate that was required to vanish. The test is to apply the matrix to a general vector satisfying the coordinate restrictions.

For the plane \(W=\{(x,y,0)^{\top}\}\) and a general \(3\times3\) matrix \(\mathbf{C}=[c_{ij}]\), the third coordinate of \(\mathbf{C}(x,y,0)^{\top}\) is \(c_{31}x+c_{32}y\). This vanishes for every real \(x,y\) exactly when \(c_{31}=c_{32}=0\): necessity follows by taking \((x,y)=(1,0)\) and \((0,1)\), and sufficiency follows by substitution. This criterion is specific to this coordinate plane and makes the “every vector” requirement explicit.

Polynomial Subspaces Under Differentiation

Invariance is not limited to matrices acting on coordinate vectors. Let \(V=\mathbb{R}[x]\), the vector space of real polynomials, and let \(D:V\to V\) be differentiation, defined by \(D(p)=p'\). For a nonnegative integer \(k\), let \(P_{\leq k}\) be the subspace of polynomials of degree at most \(k\), including the zero polynomial.

In fact, \(P_{\leq k}\) is invariant under differentiation for every nonnegative integer \(k\). If \(p(x)=\sum_{j=0}^{k}a_jx^j\), then \(p'(x)=\sum_{j=1}^{k}ja_jx^{j-1}\), whose degree is at most \(k-1\) when the derivative is nonzero; if the derivative is zero, it is also in \(P_{\leq k}\). Thus every element of \(P_{\leq k}\) maps back into that space.

A subspace of polynomials can fail the test. For example, \(U=\operatorname{span}\{1+x\}\) contains \(1+x\), but \(D(1+x)=1\). The polynomial \(1\) is not a scalar multiple of \(1+x\), so \(D(1+x)\notin U\); consequently \(U\) is not invariant. The relevant question is not whether differentiation produces another polynomial, but whether it produces one in the particular subspace \(U\).

Common Pitfalls

Confusing containment with equality. Invariance requires \(T(W)\subseteq W\); it does not require \(T(W)=W\). For instance, the zero map on \(\mathbb{R}^2\) sends every subspace to \(\{\mathbf{0}\}\). That image is contained in each subspace, even when it is strictly smaller.

Checking too few vectors. A calculation on one vector generally cannot establish invariance. For the coordinate plane in the example, checking only \(\mathbf{e}_1\) and \(\mathbf{e}_2\) would not be a complete argument unless the effect on their linear combinations were also justified. The direct calculation with arbitrary \(x,y\) does that.

Using an output that leaves the space as if it passed. A mistaken calculation might claim that \(W=\{(x,y,0)^{\top}\}\) is invariant under \(\mathbf{B}\) because \(\mathbf{B}\mathbf{e}_1=(1,0,1)^{\top}\) is still a vector in \(\mathbb{R}^3\). But membership in the ambient space is not enough: its third coordinate is \(1\), so it is not in \(W\). The correction is to check membership in the proposed subspace itself.