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While self studying Linear Algebra from Hoffman Kunze I am unable to understand some deductions in Theorem-6 on Page 204 .

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$(1)$ In 7th last line from below I am not able to deduce how $q- q(c_{j})$ equals $( x-c_{j} ) h$ despite the definition of $q$ being clear to me ( It's given in line above.

$(2)$ How did in 3rd last line from below $p(T)\alpha=0$? And how does that implies the belonging of $q(T)\alpha$ in $W$? The equality of $p(T)\alpha = (T-c_{j}I ) q(T)\alpha$ is clear but I don't know how they became equal to $0$.

Kindly help.

user264745
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  • I had the exact same question as your first one. I missed $c_j$ is a root of $q-q(c_j)$ polynomial. Following is easy to show: if $s$ is a constant polynomial over $F$ (i.e. $s=c\cdot x^0$, for some $c\in F$), then $s(\alpha)=c\cdot (\alpha )^0=c\cdot 1_V$. I suppose you have seen this thing, if not here is a recap. Since $q(c_j)$ is a constant polynomial (it is technically an element of $F$, we abuse notation), we have $q(c_j)=q(c_j)\cdot 1_F=q(c_j)$. So $q-q(c_j)=q(c_j)-q(c_j)=0$. – user264745 Jan 18 '23 at 19:09

1 Answers1

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(1) Of course, $c_j$ is a root of $q-q(c_j)$. Hence there is some polynomial $h$ such that $q-q(c_j) = (x-c_j)\cdot h$.

(2) Since $p$ is the minimal polynomial of $T$, one has $p(T) = 0$ by definition.

Moreover, $W$ was defined to be the span of all characteristic vectors of $T$. Here, one has $0 = (T-c_j I)q(T)\alpha$. In other words, $q(T)\alpha \in \mathrm{ker}(T-c_jI)$, which means that $q(T)\alpha$ is a characteristic vector of $T$ to the eigenvalue $c_j$ (or $q(T)\alpha = 0$).

Marktmeister
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