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Module 3/Eigenvalues & Diagonalization

Invariant Subspaces

An enrichment look at subspaces that a matrix maps into themselves. We define AA-invariant subspaces, see why one-dimensional invariant subspaces are exactly the eigenlines, describe the invariant subspaces of a diagonalizable matrix, and read off the axis and perpendicular plane of a 3D rotation.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

Let A=(2003)A=\begin{pmatrix}2&0\\0&3\end{pmatrix}. Select ALL of the following subspaces that are AA-invariant.

How many one-dimensional invariant subspaces (invariant lines) does A=(1221)A=\begin{pmatrix}1&2\\2&1\end{pmatrix} have?

What you’ll be able to do

  • State Definition 7.7 of an AA-invariant subspace and check whether a given line or subspace is invariant.
  • Identify the invariant subspaces that exist for every matrix: the trivial ones {0}\{\mathbf{0}\} and Rn\mathbb{R}^n, together with ker⁡A\ker A and im⁡A\operatorname{im}A.
  • Explain, via Theorem 7.9, why the one-dimensional invariant subspaces of AA are exactly its eigenlines.
  • Describe the invariant subspaces of a diagonalizable matrix using Theorem 7.10.
  • Interpret the axis and the perpendicular plane of a 3×33\times3 rotation as its real invariant subspaces (Remark 7.6, Example 7.9).

In your course

· MATH2015 · Linear Algebra & Probability
§7.3 Optional: Invariant subspaces
  • Definition 7.7A-invariant subspace
    A subspace W⊆RnW\subseteq\mathbb{R}^n is AA-invariant if Aw∈WAw\in W whenever w∈Ww\in W.
  • Example 7.8Invariant lines of scalings, shears, and rotations
    Trivial subspaces {0},Rn\{\mathbf{0}\},\mathbb{R}^n and ker⁡A,im⁡A\ker A,\operatorname{im}A are always invariant; a scaling with distinct factors has the coordinate axes, a shear has one invariant line, a 90∘90^\circ rotation has none.
  • Theorem 7.9One-dimensional invariant subspaces
    A one-dimensional subspace WW is AA-invariant if and only if WW is an eigenline spanned by an eigenvector of AA.
  • Theorem 7.10Invariant subspaces of diagonalizable matrices
    If AA is diagonalizable, every kk-dimensional invariant subspace is spanned by kk linearly independent eigenvectors of AA.
  • Example 7.9Real invariant subspaces of a 3D rotation
    The rotation A=(010001100)A=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix} has axis span⁡{(1,1,1)}\operatorname{span}\{(1,1,1)\} (1-dim) and perpendicular plane x1+x2+x3=0x_1+x_2+x_3=0 (2-dim) as its nontrivial real invariant subspaces.
  • Remark 7.6Axis and plane of a 3x3 rotation
    For a 3×33\times3 rotation A≠IA\neq I, the unique 1-dim real invariant subspace is the axis of rotation (fixed: Av=vAv=v), and the perpendicular plane is a 2-dim invariant subspace.
Enrichment (the course marks §7.3 optional).
1

What is an invariant subspace?

A subspace W⊆RnW\subseteq\mathbb{R}^n is AA-invariant if AA never sends a vector of WW outside of WW: Aw∈WAw\in W for every w∈Ww\in W (Definition 7.7). Geometrically, the transformation AA may stretch, rotate, or shear vectors inside WW, but it keeps the whole action trapped in WW. To test a subspace it is enough to check a spanning set: if W=span⁡{w1,…,wk}W=\operatorname{span}\{w_1,\dots,w_k\}, then WW is invariant iff each Awi∈WAw_i\in W.

2

Invariant subspaces you always have

Several invariant subspaces come for free. The trivial ones are the zero subspace {0}\{\mathbf{0}\} and the whole space Rn\mathbb{R}^n (Example 7.8.1). In addition, ker⁡A\ker A and im⁡A\operatorname{im}A are always AA-invariant (Example 7.8.3): if w∈ker⁡Aw\in\ker A then Aw=0∈ker⁡AAw=\mathbf{0}\in\ker A, and AwAw lies in im⁡A\operatorname{im}A by definition of the image. At the extreme, if A=IA=I (or more generally A=λIA=\lambda I) then every subspace is invariant, since Aw=λwAw=\lambda w stays on the same line.

3

One-dimensional invariant subspaces are eigenlines

Suppose W=span⁡{w}W=\operatorname{span}\{w\} with w≠0w\neq\mathbf{0}. Then Aw∈WAw\in W means Aw=λwAw=\lambda w for some scalar λ\lambda -- which is exactly the statement that ww is an eigenvector. So a line is invariant precisely when it is an eigenline (Theorem 7.9). Consequences: a scaling with distinct factors has only the coordinate axes as invariant lines (Example 7.8.4); a shear has a single invariant line (Example 7.8.5); and a 90∘90^\circ rotation in R2\mathbb{R}^2 has none, because it has no real eigenvalues (Example 7.8.6).

4

Higher dimensions, diagonalizable matrices, and rotations

When AA is diagonalizable, every kk-dimensional invariant subspace is spanned by kk linearly independent eigenvectors of AA (Theorem 7.10). For a real matrix with a complex-conjugate pair of eigenvectors v±=x±iyv_\pm=x\pm iy, the smallest real invariant subspace is the 22-dimensional plane span⁡{x,y}\operatorname{span}\{x,y\}. The headline example is a 3×33\times3 rotation A≠IA\neq I: it has exactly one real eigenline, the axis of rotation (which it fixes, Av=vAv=v), and the perpendicular plane is a 22-dimensional invariant subspace on which AA acts as a planar rotation (Remark 7.6, Example 7.9).

Theorem 7.9 -- One-dimensional invariant subspaces

A one-dimensional subspace WW is invariant under the matrix AA if and only if WW is an eigenline, i.e. W=span⁡{v}W=\operatorname{span}\{v\} for some eigenvector vv of AA.

Intuition. On a line, staying on the line means the output is a scalar multiple of the input: Av=λvAv=\lambda v. That is the eigenvector equation, so invariant lines and eigenlines are the same thing. If every eigenvalue is simple there are only finitely many invariant lines; a repeated eigenvalue gives a whole eigenspace full of them.
Theorem 7.10 -- Invariant subspaces of diagonalizable matrices

If AA is diagonalizable, then every kk-dimensional AA-invariant subspace is spanned by kk linearly independent eigenvectors of AA.

Intuition. A diagonalizable matrix has an eigenbasis, so it acts independently along each eigen-direction. A subspace can only be preserved if it is built out of those directions -- you cannot keep just a little bit of two different eigenlines and stay invariant unless the whole eigenplane is included. This turns the hunt for invariant subspaces into choosing subsets of eigenvectors.
Remark 7.6 -- Axis and plane of a 3x3 rotation

For every 3×33\times3 rotation matrix A≠IA\neq I, the unique one-dimensional real invariant subspace is the axis of the rotation, and the plane perpendicular to it is a two-dimensional invariant subspace. The matrix fixes the axis (Av=vAv=v for every vv along it) and restricts to a planar rotation on the perpendicular plane.

Intuition. A 3D rotation leaves one direction untouched -- the axis -- and spins everything in the perpendicular plane. The axis is the real eigenline (eigenvalue 11); the spinning plane comes from the complex-conjugate pair of eigenvalues, whose eigenvectors' real and imaginary parts span it.

Worked examples

Example 1

Is the line W=span⁡{(1,2)}W=\operatorname{span}\{(1,2)\} invariant under A=(2003)A=\begin{pmatrix}2&0\\0&3\end{pmatrix}?

  1. 1

    By Definition 7.7, WW is AA-invariant iff Aw∈WAw\in W for every w∈Ww\in W. Since WW is a line, it suffices to test the spanning vector (1,2)(1,2).

  2. 2

    Compute A(12)=(2⋅13⋅2)=(26)A\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}2\cdot1\\3\cdot2\end{pmatrix}=\begin{pmatrix}2\\6\end{pmatrix}.

  3. 3

    Check whether (2,6)(2,6) lies on WW, i.e. whether (2,6)=c(1,2)(2,6)=c(1,2) for some scalar cc. The first coordinate forces c=2c=2, but then c(1,2)=(2,4)≠(2,6)c(1,2)=(2,4)\neq(2,6).

Answer. No. A(1,2)=(2,6)∉span⁡{(1,2)}A(1,2)=(2,6)\notin\operatorname{span}\{(1,2)\}, so WW is not AA-invariant. For this scaling the only invariant lines are the two coordinate axes.
Example 2

Find all one-dimensional invariant subspaces of A=(1221)A=\begin{pmatrix}1&2\\2&1\end{pmatrix}.

Example 3

Find the real invariant subspaces of the 3×33\times3 rotation A=(010001100)A=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix} (Example 7.9).