Exercise 8.B.11

Proof. Suppose $T$ has an upper-triangular matrix with respect to the basis $v_1,\dots ,v_n$. Suppose also that $\lambda$ appears on the $j$-th diagonal entry of $M(T)$. Then

for some $a_1,\dots ,a_{j-1}\in 𝔽$, and so

We have

for some $c_1,\dots ,c_{j-1}\in 𝔽$. If $c_{j-1}-\lambda =0$, then ${(T-\mathit{\lambda I})}^2{v}_j\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_{j-2})$. If $c_{j-1}-\lambda \ne 0$, then

We go on, either squaring by squaring $(T-\mathit{\lambda I})$ or subtracting a vector $u\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_{j-1})$ from $v_j$ in the argument of $(T-\mathit{\lambda I})$, and we will have ${(T-\mathit{\lambda I})}^2(v_j-u)$ in the span of the first $j-3$ vectors of the basis, then in span of the first $n-4$, an so on until it will be in the span of an empty list, that is, $\{0\}$. This means that $v_j-u\in G(\lambda ,T)$ for some $u\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_{j-1})$.

Let ${\nu }_1,\dots ,{\nu }_d$ denote the vectors of the chosen basis of $V$ that correspond to the columns of $M(T)$ in which $\lambda$ appears and in the order that they appear on the basis. We can repeat the previous process and find $u_1,\dots ,u_d$ such

where each $u_k$ is in the span of the basis vectors that come before ${\nu }_k$. We claim this list is linearly independent. To see this, fix $k\in \{1,\dots ,d\}$. Suppose ${\nu }_k$ is the $j$-th basis vector, i.e. ${\nu }_k=v_j$. This means that

Therefore, we can’t have ${\nu }_k-u_k\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->({\nu }_1-u_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\nu }_{k-1}-u_{k-1})$, because that would imply that

since $u_k\in \mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(v_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,v_{j-1})$. This argument can be repeated for each $k$. Therefore no vector in $(6)$ is in the span of the previous ones. It follows that the list in $(6)$ is linearly independent.

Hence $\dim <!--\; FUNCTION\; APPLICATION\; -->G(\lambda ,T)\ge d$. By 8.26, the dimension of $G(\lambda ,T)$ cannot be greater $d$, because we have $\dim <!--\; FUNCTION\; APPLICATION\; -->V$ diagonal entries and each one adds at least $1$ to the dimension of some generalized eigenspace. Thus $\dim <!--\; FUNCTION\; APPLICATION\; -->G(\lambda ,T)=d$, completing the proof. □