A question on "Linear Algebra" by Kenneth Hoffman

Question

I'm reading "Linear Algebra" by Kenneth Hoffman and Ray Kunze.

I'm now lost at $\S$6.4 Theorem 6: the proof looks OK, but when I pick an example, somehow it does not tally. Please find below the theorem and proof, and my example in $\color{blue}{\texttt{blue}}$, and question in $\color{red}{\texttt{red}}$ in step 8. Please kindly help me: where did I mistake?

Theorem: Let $V$ be a finite-dimensional vector space over the field $F$ and let $T$ be a linear operator on $V$. Then $T$ is diagonalizable if and only if the minimal polynomial for $T$ has the form $p = (x - c_1) \dots (x - c_k)$ where $c_1, \dots , c_k$ are distinct elements of $F$. $$\color{blue}{\texttt{Example: Choose } V=\mathbb R^3, F=\mathbb R, T=\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 1\\ 0 & 0 & 1 \end{bmatrix}, p=(x-1)(x-2), c_1=1, c_2=2 }$$

The proof is: (the (1)(2).. numbers are added by me) Proof

(1) We have noted earlier that, if $T$ is diagonalizable, its minimal polynomial is a product of distinct linear factors (see the discussion prior to Example 4).

(2)To prove the converse, let $W$ be the subspace spanned by all of the characteristic vectors of $T$, and suppose $W \ne V$. $$\color{blue}{\texttt{characteristic vectors:} v_1=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \texttt{for } c_1=1, v_2=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \texttt{for } c_2=2.\\ W=\langle v1, v2 \rangle, W\ne V=\mathbb R^3. }$$ (3)By the lemma used in the proof of Theorem 5, there is a vector $\alpha$ not in $W$ and a characteristic value $c_j$ of $T$ such that the vector $\beta= (T - c_jI)\alpha$ lies in W. $$\color{blue}{\texttt{Choose: } \alpha=\begin{bmatrix} 0 \\ 0.5 \\ 0.5 \end{bmatrix}, c_j=c_1=1, \\ \beta=\begin{bmatrix} 0&0&0 \\ 0&1&1 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 0 \\ 0.5 \\ 0.5 \end{bmatrix} =\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \beta=v_2 \in \langle v_1, v_2 \rangle = W. }$$ (4)Since $\beta$ is in $W$, $\beta = \beta_1+\dots\beta_k$ where $T\beta_i = c_i\beta_i$, $1\le i\le k$, and therefore the vector $h(T)\beta = h(c_1)\beta_1+\dots+h(c_k)\beta_k$ is in $W$, for every polynomial $h$.

(5)Now $p = (x-c_j)q$, for some polynomial $q$. $$\color{blue}{ p=(x-1)(x-2)=(x-1)q \Rightarrow q=(x-2) }$$ (6)Also $q- q(c_j) = (x - c_j)h$. $$\color{blue}{ q-q(c_1)=q-q(1)=(x-2)-(1-2)=(x-1)=(x-1)h \Rightarrow h=1 }$$ (7)We have $q(T)\alpha - q(c_j)\alpha = h(T)(T - c_jI)\alpha = h(T)\beta$. $$\color{blue}{ q(T)=(T-2I)=\begin{bmatrix} -1&0&0 \\ 0&0&1 \\ 0&0&-1 \end{bmatrix}, \\ q(T)\alpha=\begin{bmatrix} -1&0&0 \\ 0&0&1 \\ 0&0&-1 \end{bmatrix} \begin{bmatrix} 0 \\ 0.5 \\ 0.5 \end{bmatrix} =\begin{bmatrix} 0 \\ 0.5 \\ -0.5 \end{bmatrix}\\ q(c_j)\alpha=q(c_1)\alpha=q(1)\alpha=(1-2)\alpha=-\alpha=\begin{bmatrix} 0 \\ - 0.5 \\ - 0.5 \end{bmatrix}, \\ q(T)\alpha-q(c_j)\alpha=\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = h(T)\beta = 1\beta\\ }$$ (8)But $h(T)\beta$ is in $W$ and, since $0 = p(T)\alpha = (T - c_jI)q(T)\alpha$, $$\color{blue}{ p(T)=0\\ q(T)\alpha = \begin{bmatrix} -1&0&0 \\ 0&0&1 \\ 0&0&-1 \end{bmatrix} \begin{bmatrix} 0 \\ 0.5 \\ 0.5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0.5 \\ -0.5 \end{bmatrix} \\ 0=p(T)\alpha = (T - c_jI)q(T)\alpha = \begin{bmatrix} 0 & 0&0\\ 0& 1&1 \\ 0&0&0 \end{bmatrix} \begin{bmatrix} 0 \\ 0.5 \\ -0.5 \end{bmatrix} }$$ the vector $q(T)\alpha$ is in $W$. $$\color{red}{ q(T)\alpha = \begin{bmatrix} 0 \\ 0.5 \\ -0.5 \end{bmatrix} \notin W= \langle v_1, v_2 \rangle = \left \langle \begin{bmatrix} 1\\ 0 \\ 0 \end{bmatrix} , \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right \rangle\\ \texttt{--What happen? why this step does not tally?} }$$ (9)Therefore, $q(c_j)\alpha$ is in $W$.

(10)Since $\alpha$ is not in $W$, we have $q(c_j) = 0$.

(11)That contradicts the fact that $p$ has distinct roots. QED.

Answer 1

This is a proof by contradiction. In your concrete example, the contradiction comes earlier than in the proof. Because the proof has to follow all possible paths to absurdity, to make the contradiction argument work. The chosen example may already fail at an earlier stage.

The example you cooked up was build such that your $W$ was not the full space spanned by all eigenvectors (or characteristic vectors). Thus, the contradiction strikes at exactly the place in the proof, where the fact '$W$=spanned by all eigenvectors' is going to be used.

If you would start with a non-diagonalizable matrix (say $\pmatrix{0&1\\0&0}$) then the contradiction should happen at the very latest point in the proof at (10)/(11).