16.-Linear-Algebra-by-Kenneth-Hoffman · Problem 6.7.2

Elementary Canonical Forms

Mathematics — Linear Algebra · invariant-subspace-no-complement

Problem

Let TT be the linear operator on R2R^2, the matrix of which in the standard ordered basis is [2102].\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}. Let W1W_1 be the subspace of R2R^2 spanned by the vector ϵ1=(1,0)\epsilon_1 = (1, 0). (a) Prove that W1W_1 is invariant under TT. (b) Prove that there is no subspace W2W_2 which is invariant under TT and which is complementary to W1W_1: R2=W1⊕W2.R^2 = W_1 \oplus W_2. (Compare with Exercise 1 of Section 6.5.)
Intuitive explanation

The key insight is recognizing that invariant subspaces are "closed" under the transformation—applying TT keeps you inside the subspace. For part (a), you just need to check where TT sends the basis vector of W1W_1. Part (b) is trickier: if a complementary invariant subspace W2W_2 existed, it would have to be 1-dimensional, so you'd need to find a vector that TT maps back into its own span. The catch is that this matrix is not diagonalizable—it's a Jordan block—which forces a structural obstruction that prevents such a decomposition from existing.

Intuitive explanation

The key insight is recognizing that invariant subspaces are "closed" under the transformation—applying TT keeps you inside the subspace. For part (a), you just need to check where TT sends the basis vector of W1W_1. Part (b) is trickier: if a complementary invariant subspace W2W_2 existed, it would have to be 1-dimensional, so you'd need to find a vector that TT maps back into its own span. The catch is that this matrix is not diagonalizable—it's a Jordan block—which forces a structural obstruction that prevents such a decomposition from existing.

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