Exercise 5.A.10

Proof. (a) The eigenvalues are $1,\dots ,n$ and their corresponding eigenvectors are the vectors in the standard basis of ${𝔽}^n$.

(b) We claim that all subspaces invariant under $T$ are spans of vectors in the standard basis.

Suppose $U$ is an invariant subspace of $T$ and $v={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{k=1}^m{a}_{i_k}{e}_{i_k}\in U$, where $i_1<\cdots <i_m$ and all $a_{i_k}$’s are nonzero. We will show that $e_{i_k}\in U$ for each $k$. First, all is clear if $m=1$, so suppose $m>1$. Note that

Also, $\mathit{Tv}-i_{1}v\in U$ because $v\in U$ and $U$ is a subspace. Moreover, the coefficients in the sum above are all nonzero. Repeating this $m-1$ times (subtracting $i_2(\mathit{Tv}-i_{1}v)$ from $\mathit{Tv}-i_{1}v$, ...), we’re left with a nonzero vector in $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(e_{i_m})$ which is also in $U$. In particular, $e_{i_m}\in U$, so $v-a_{i_m}{e}_{i_m}\in U$. Repeating again with this last vector, we find that $e_{i_{m-1}}\in U$. Continuing like this, we see that $e_{i_k}\in U$ for all $k$.

Now, we have $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(e_{i_1},\dots <!--\; FUNCTION\; APPLICATION\; -->,e_{i_m})\subseteq U$. If this inclusion is not an equality, we can find another vector $w\in U$ which is not in the span of the $e_{i_k}$’s. Writing $w$ as a linear combination of vectors in the standard basis, we can find an integer $j$, distinct from all the $i_k$, such that the coefficient of $e_j$ is nonzero. Thus, if we repeat the previous procedure with $w$ instead of $v$, we see that $e_j\in U$, so $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->(e_{i_1},\dots <!--\; FUNCTION\; APPLICATION\; -->,e_{i_m},e_j)\subseteq U$. If this inclusion is still not an equality, we can repeat this last procedure again and add another basis vector to the $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->$. Each time the dimension increases by $1$. Since $U$ is finite-dimensional, we must have equality at the some point. □