(Homework) Prove that there is no T-invariant subspace $W_2$ such that $R^2$ = $W_1 \oplus W_2$

Question

Let T be a linear operator on $R^2$ defined by T(x,y) = (2x+y, 2y) and $W_1 = span{(1,0)}$

Prove that there is no T-invariant subspace $W_2$ such that $R^2 = W_1 \oplus W_2$

At first I showed that $W_1$ is an eigenspace and T is not diagonalizable so $R^2$ cannot be decomposed into the direct sum of the eigenspaces of T. But this is not enough right since eigenspace is not the only T-invariant subspace.

So I tried this way instead

Suppose $R^2 = W_1 \oplus W_2$ . Then dim$W_2$ = 1 that is $W_2$ is spanned by a single vector (a,b) and b is nonzero. But T[(a,b)] = (2a+b, 2b) which is not in $W_2$. Therefore $W_2$ is not T-invariant.

Is my argument correct. We're using Friedberg in class and we've just covered eigenspace, invariant subspace and minimal polynomial.

Thanks

Answer 1

Let $V=\mathbf R^2$ with the standard basis, and let $M=\begin{pmatrix}2&1\\0&2\end{pmatrix}$, so that $M$ is the matrix of $T$ in the standard basis of $V$. Note that $M-2I$ is nonzero but $(M-2I)^2=0$, so the minimum polyonomial of $T$ is $(X-2)^2$. A decomposition $V=W_1\oplus W_2$ would imply $M$ is diagonalizable. Claim: diagonalizable transformations have minimum polynomials with no repeated roots, which contradicts the fact that $m_T=(X-1)^2$ has a double root.

Proof of the claim It is enough to recall that if $T:V\to V$ is a linear transformation the minimal polynomial of $T$ equals the greatest common divisor of the minimal polynomials of a basis of $V$. If we pick a basis $(v_i)$ of eigenvectors with eigenvalues $\lambda_i$ the corresponding minimal polynomial is $X-\lambda_i$, and hence the result.