Proposition. Let $\mathcal{V}$ be a nontrivial finite-dimensional vector space over a scalar field $\mathbb{F}$ with $n = \dim \mathcal{V}$. Let $T \in\mathcal{L}(\mathcal{V})$ be such that $T\neq 0$ and $T$ is not an injection. Then there exists unique $q\in\{1,\ldots,n\}$ such that \[ \{0_{\mathcal{V}}\} = \operatorname{nul}(T^0) \subsetneq \operatorname{nul}(T) \subsetneq \cdots \subsetneq \operatorname{nul}(T^{q-1}) \subsetneq \operatorname{nul}(T^q) = \operatorname{nul}(T^{q+j}) \quad \text{for all} \quad j \in \mathbb{N}. \]
Proof. Since the statement is trivial if $n=1$, we assume that $n \geq 2.$ We also assume that $T\neq 0.$ Let $k \in \mathbb{N}$ be such that $k \geq 2.$ First we prove that \[ \operatorname{nul}(T^{k-2}) \subsetneq \operatorname{nul}(T^{k-1}) \quad \Leftarrow \quad \operatorname{nul}(T^{k-1}) \subsetneq \operatorname{nul}(T^{k}). \] Assume \[ \operatorname{nul}(T^{k-1}) \subsetneq \operatorname{nul}(T^{k}). \] Then there exists $v\in\mathcal{V}$ such that $T^{k-1}v \neq 0$ and $T^{k}v = 0.$ Consequently, for $w= Tv \in \mathcal{V}$ we have $T^{k-2}w \neq 0$ and $T^{k-1}w = 0.$ Therefore, \[ \operatorname{nul}(T^{k-2}) \subsetneq \operatorname{nul}(T^{k-1}). \] Repeating this reasoning total of $k-1$ times we deduce that the assumption \[ \operatorname{nul}(T^{k-1}) \subsetneq \operatorname{nul}(T^{k}). \] implies \[ \{0_{\mathcal{V}}\} = \operatorname{nul}(T^0) \subsetneq \operatorname{nul}(T) \subsetneq \cdots \subsetneq \operatorname{nul}(T^{k-1}) \subsetneq \operatorname{nul}(T^k). \] The strict inclusions of the above subspaces imply \[ n \geq \dim \operatorname{nul}(T^{k}) \geq k. \] Thus, specifically, \[ \operatorname{nul}(T^{k-1}) \subsetneq \operatorname{nul}(T^{k}) \quad \Rightarrow \quad k \leq n. \] The contrapositive of this implication is \[ k \gt n \quad \Rightarrow \quad \operatorname{nul}(T^{k-1}) = \operatorname{nul}(T^{k}). \] Now, let us set \[ q = \max \Bigl\{ k \in \{1,\ldots,n\} \, : \, \operatorname{nul}(T^{k-1}) \subsetneq \operatorname{nul}(T^{k}) \Bigr\}. \] Then \[ \operatorname{nul}(T^{k-1}) \subsetneq \operatorname{nul}(T^{k}) \quad \text{for all} \quad k \in \{1,\ldots,q\} \] and \[ \operatorname{nul}(T^{k-1}) = \operatorname{nul}(T^{k}) \quad \text{for all} \quad k \in \mathbb{N}\setminus\{1,\ldots,q\}. \] QED
Example 1.
Consider a five-dimensional vector space $\mathcal{V}$ and a nilpotent operator $N$ on $\mathcal{V}$ such that
\[
N \neq 0, \quad N^2 = 0 \quad \text{and} \quad \dim \operatorname{nul} N = 3.
\]
Since $N^2 = 0$ and by the nullity-rank theorem we have
\[
\dim \operatorname{ran} N = 2 \quad \text{and} \quad \operatorname{ran} N \subset \operatorname{nul} N.
\]
Let us construct a special basis $\mathcal{B}$ for $\mathcal{V}$ for which $\mathsf{M}^{\mathcal{B}}_{\mathcal{B}}(N)$ is particularly simple.
To get Jordan chains in a basis, we have to start with vectors in $\operatorname{ran} N.$ This is a convenient way to state this: Let $v_1,$ $v_2,$ and $v_3$ be vectors in $\mathcal{V}$ such that the vectors
\[
Nv_1, \ Nv_2, \ v_3 \quad \text{form a basis for} \quad \operatorname{nul} N.
\]
Such vectors exist since we can choose a basis $w_1, w_2$ of $\operatorname{ran} N$ and, since $\operatorname{ran} N \subset \operatorname{nul} N$ and $\dim \operatorname{nul} N = 3$, there exists a vector $v_3 \in \operatorname{nul} N$ such that
\[
w_1, \ w_2, \ v_3 \quad \text{form a basis for} \quad \operatorname{nul} N.
\]
Now choose $v_1$ and $v_2$ such that $w_1 = Nv_1,$ and $Nv_2 = w_2.$
We are almost done. Just prove that
\[
Nv_1, \ v_1 \ Nv_2, \ v_2, \ v_3 \quad \text{are linearly independent}.
\]
Let $\alpha_{1,1}, \alpha_{1,0}, \alpha_{2,1}, \alpha_{2,0}, \alpha_{3,0}$ be such that
\[
\alpha_{1,1} Nv_1 + \alpha_{1,0} v_1+\alpha_{2,1} Nv_2 + \alpha_{2,0} v_2 + \alpha_{3,0} v_3 = 0.
\]
Apply $N$ to both sides of the previous equality and use that $N^2 = 0$ and $Nv_3 = 0$ to conclude:
\[
\alpha_{1,0} N v_1 + \alpha_{2,0} N v_2 = 0.
\]
Since $Nv_1$ and $Nv_2$ are linearly independent we conclude that $\alpha_{1,0} =0$ and $\alpha_{2,0} = 0.$ Consequently
\[
\alpha_{1,1} Nv_1 + \alpha_{2,1} Nv_2 + \alpha_{3,0} v_3 = 0.
\]
Since the vectors in the last equality forma basis for $\operatorname{nul} N$ we conclude that $\alpha_{1,1} =0$, $\alpha_{2,1} = 0$ and $\alpha_{3,0} = 0,$ proving the linear independence of vectors
\[
\mathcal{B} = \bigl\{ Nv_1, \ v_1 \ Nv_2, \ v_2, \ v_3 \bigr\}.
\]
Since $\mathcal{V}$ is five-dimensional we have a basis for $\mathcal{V}.$ We have
\[
\mathsf{M}^{\mathcal{B}}_{\mathcal{B}}(N) =
\left[ \begin{array}{cc|cc|c}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\\hline
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 \\\hline
0 & 0 & 0 & 0 & 0 \\
\end{array} \right]
\]
Example 2.
Consider a five-dimensional vector space $\mathcal{V}$ and a nilpotent operator $N$ on $\mathcal{V}$ such that
\[
\mathsf{M}^{\mathcal{B}}_{\mathcal{B}}(N) =
\left[ \begin{array}{ccc|cc}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\\hline
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 \\
\end{array} \right]
\]
relative to some basis $\mathcal{B}$ of $\mathcal{V}.$
Determine the degree of nilpotency of $N$, $\dim \operatorname{nul} N,$
$\dim \operatorname{nul}(N^2),$ $\dim \operatorname{nul}(N^3),$
$\dim \operatorname{nul}(N^4).$
Theorem. Let $\mathcal{V}$ be a vector space over a scalar field $\mathbb{F}.$ Let $\langle\,\cdot\,, \cdot\,\rangle$ be a positive definite inner product on $\mathcal{V}$ and denote by $\|\cdot\|$ the corresponding norm. Let $\mathcal U$ be a subspace of $\mathcal V$, let $v \in \mathcal{V}$ and let $u_0 \in \mathcal{U}.$ The following equivalence holds \[ \forall u \in \mathcal U \quad \| v - u_0 \| \leq \| v - u \| \quad \Leftrightarrow \quad v - u_0 \perp \mathcal U. \]
Proof. Assume that $\mathcal U$ is a subspace of $\mathcal V$, $v \in \mathcal{V}$ and $u_0 \in \mathcal{U}.$ We first prove the "if" part ($\Leftarrow$) of the equivalence. Assume that $v - u_0 \perp \mathcal U$ and let $u \in \mathcal U$ be arbitrary. Then $u_0 - u \in \mathcal U$ and therefore $v - u_0$ and $u_0 - u$ are orthogonal. Consequently, by the Pythagorean Theorem we have \[ \|v-u\|^2 = \|v - u_0 + u_0 - u \|^2 = \|v - u_0\|^2 + \|u_0 - u \|^2 \geq \|v - u_0\|^2. \] This proves the "if" part. Assume that $v - u_0 \perp \mathcal U$ is not true. That is there exists We prove the "only if" part ($\Rightarrow$) of the equivalence by proving its contrapositive. Assume that $v - u_0 \perp \mathcal U$ is not true. That is assume that there exists $u_1 \in \mathcal U$ such that \[ \langle v - u_0, u_1 \rangle \neq 0. \] This inequality implies that $u_1 \neq 0_{\mathcal V}.$ Set \[ \alpha = \langle v - u_0, u_1 \rangle \neq 0 \] and \[ u = u_0 + \frac{\alpha}{\| u_1\|^2} u_1 \in \mathcal U. \] Now calculate \begin{align*} \| v - u \|^2 & = \left\| v - u_0 - \frac{\alpha}{\| u_1\|^2} u_1 \right\|^2 \\ & = \left\langle v - u_0 - \frac{\alpha}{\| u_1\|^2} u_1 , v - u_0 - \frac{\alpha}{\| u_1\|^2} u_1 \right\rangle \\ & = \left\langle v - u_0, v - u_0 \right\rangle - \frac{\alpha}{\| u_1\|^2} \left\langle u_1, v - u_0 \right\rangle \\ & \phantom{xxxxxxxxxx} - \frac{\overline{\alpha}}{\| u_1\|^2} \left\langle v - u_0, u_1 \right\rangle + \frac{|\alpha|^2}{\| u_1\|^4} \left\langle u_1,u_2 \right\rangle \\ & = \left\langle v - u_0, v - u_0 \right\rangle - \frac{|\alpha|^2}{\| u_1\|^2} - \frac{|\alpha|^2}{\| u_1\|^2} + \frac{|\alpha|^2}{\| u_1\|^2} \\ & = \left\langle v - u_0, v - u_0 \right\rangle - \frac{|\alpha|^2}{\| u_1\|^2} \\ & \lt \| v - u_0 \|^2. \end{align*} The strict inequality in the preceding sequence of inequalities follows from the fact that $|\alpha|/\| u_1 \| \gt 0.$ Thus, we proved that \[ \exists u \in \mathcal U \quad \| v - u \| \lt \| v - u_0\|. \] This completes a proof of the contrapositive of the "only if" part of the theorem. QED.
Remark. Recall that in class I presented a more natural proof by setting \[ u = u_0 + t \alpha u_1 \in \mathcal U \] with $t \in \mathbb{F}\cap\mathbb{R}.$ Then I proceeded to calculate the quadratic expression \[ \left\langle v - u_0 - t \alpha u_1 , v - u_0 - t \alpha u_1 \right\rangle = \| v - u_0 \|^2 - 2 t |\alpha|^2 + t^2 |\alpha|^2 \|u_1\|^2 \] which is clearly less than $ \| v - u_0 \|^2$ for all $t \in \mathbb{F}\cap\mathbb{R}$ such that \[ 0 \lt t \lt \frac{2}{\|u_1\|^2}. \] Since our proof does not require this level of detail, I just selected the midpoint of the last interval for $t$ and proceeded with this specific $u.$
I came, I thought, I wrote, I taught, I thought more, I rewrote.
Verba volant, scripta manent