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This is problem 6.7.2 from Hoffman and Kunze.

Let $T$ be a linear operator on $\mathbb{R}^2$, which in the standard basis is given by, $$\begin{bmatrix}2 & 1 \\ 0 & 2\end{bmatrix}.$$ Let $W_1$ be the subspace spanned by the vector $\epsilon_1=(1,0)$. First, prove that $W_1$ is invariant under $T$. Second, prove that there is no subspace $W_2$ which is invariant under $T$ and complementary to $W_1$, such that $W_1\oplus W_2=\mathbb{R}^2$.

The first part is trivial. Any vector in $W_1$ must be of the form $(a,0)$. We note that $T(a,0)=(2a,0)$, which is still in the desired form. So, $W_1$ is invariant under $T$.

Here is my attempt at the second part:

First, we note that $T$ is not diagonalizable.

Next, we have from Theorem 9 that if $V=W_1\oplus W_2$ then there exist projection operators, $E_1,E_2$ satisfying the following properties:

  • $E_1E_2=0$.
  • $I=E_1+E_2$.
  • The range of $E_i$ is $W_i$.

We note then that because the two subspaces are invariant under $T$, that their direct sum must also be invariant under $T$. That is, for $v\in\mathbb{R}^2$, we have that $Tv\in\mathbb{R}^2$. Any $v$ can be written in the form $c_1\epsilon_1+c_2\epsilon_2$. So we also have that $Tv=c_1'\epsilon_1+c_2'\epsilon_2$. From this we have, $T=c_1'E_1+c_2'E_2$. By Theorem 11 then, $T$ is diagonalizable, from where the contradiction arises.

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    $\begingroup$ "$T$ is not diagonalizable" is the key element here. The rest is just expanding on what "diagonalizable" means (and is correct, though it could be tightened-up). But I think you need to do more that just say that $T$ is not diagonalizable. $\endgroup$ – Paul Sinclair May 23 '18 at 23:49
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Suppose $\mathbb R^2 = W_1 \oplus W_2$. Then to span the second coordinate of $\mathbb R^2$, vectors in $W_2$ must be multiples of $(b, c)$ where $c \ne 0$. The coordinate matrix of $T(b, c)$ is $$\begin{bmatrix}2 & 1\\ 0 & 2\end{bmatrix} \begin{bmatrix}b\\ c\end{bmatrix} = \begin{bmatrix}2b + c\\ 2c\end{bmatrix} = 2\begin{bmatrix} b\\ c\end{bmatrix} + c\begin{bmatrix} 1\\ 0\end{bmatrix},$$ so $T(b, c) = 2(b, c) + c(1, 0)$ which cannot be in $W_2$ because $c \ne 0$ prevents $ 2(b, c) + c(1, 0)$ from being a multiple of $(b, c)$. Thus, there is no such subspace $W_2$ that is invariant under $T$.

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