3.20 Projector

Assume $A\in\mathbb{C}^{4\times 4}$ is diagonalizable. Find all the diagonal matrix $B$ in $\mathbb{C}^{4\times 4}$ that satisfy the following conditions
1. $A$ is similar to $B$
2. $m_A(x)\mid x(x^2-1)(x^2+2)$
3. $\chi_A(x)\in \R[x]$
4. $x^{2} \nmid \chi_{A}(x)$
5. $A$ is not invertible
Since $A$ is diagonalizable, then $A=PDP^{-1}\Rightarrow \det A=\det(PDP^{-1})=\det (P)\det (D)\det (P^{-1})=\det(PP^{-1})\det D=\det D $

If $A$ is not invertible, then $0$ is a root of $\chi_{A}(x)$, that is $x\mid \chi_A(x)$

But $x^2\nmid \chi_A(x)$, then $0$ is a simple root(multiplicity is $1$), in particular $\chi_A(x)=x\cdot p(x)$ and $m_{A}(x)=xq(x)$ and $q(x)\mid p(x)$ and $q$ has a single root

If $\chi_A(x)$ has no roots in $\mathbb{C}$, then the roots of $q(x)$ are in $\R$, then $m_{A}(x)\mid x(x^{2}-1)$

Then $m_{A}(x)=x(x-1),x(x+1),x(x^{2}-1)$

Then $\chi_A(x)=x(x-1)^3,x(x+1)^3,x(x+1)^2(x-1),x(x+1)(x-1)^2$

If $\chi_A(x)$ has root in $\mathbb{C}$, then $m_{A}(x)=x(x+\sqrt{2}i)(x-\sqrt{2}i)r(x)$ with $\deg r(x)\leq 1$

If $\deg r(x)=0$, then $\chi_{A}(x)=kx(x+\sqrt{2}i)^{2}(x-\sqrt{2}i)$ or $kx(x+\sqrt2i)(x-\sqrt2i)^2$

But it's contradiction to 3

If $\deg r(x)=1$, then since 2, $r(x)$ is a factor of $x^2-1$, then it can be $(x+1)$ or $(x-1)$

Therefore, \(B\in\left\lbrace \begin{pmatrix} 0 & & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{pmatrix}, \begin{pmatrix} 0 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{pmatrix}, \begin{pmatrix} 0 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & -1 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & -1 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 \end{pmatrix}, \begin{pmatrix} 0 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & -1 & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & 1 & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 \end{pmatrix} \begin{pmatrix} 0 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & -\sqrt{2i} & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \sqrt{2i} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & 1 \end{pmatrix}, \begin{pmatrix} 0 & \placeholder{} & \placeholder{} & \placeholder{} \\ \placeholder{} & -\sqrt{2i} & \placeholder{} & \placeholder{} \\ \placeholder{} & \placeholder{} & \sqrt{2i} & \placeholder{} \\ \placeholder{} & \placeholder{} & \placeholder{} & -1 \end{pmatrix}\right\rbrace\)

Definition: $V$ is a vector space over $\mathbb{K}$, $f: V \to V$ is a linear operator, then $f$ is a projection if $f^2 = f$.

Remark: $f$ is a projection iff $\forall x \in \text{Im} f, f(x) = x$ . If $f$ is a projection, then $V = \text{Ker} f \oplus \text{Im} f$ .

Proposition: $V$ is a vector space over $\mathbb{K}$ and $S, T \leq V$ such that $V = S \oplus T$ . Then there exists a unique projection $f: V \to V$ such that $S = \text{Ker} f$ and $T = \text{Im} f$ .

Find a projection $f: \mathbb{R}^3 \to \mathbb{R}$ such that $\text{Ker} f = \text{span} \{ (1, 2, 3) \}$ and $\text{Im} f = \text{span} \{ (0, 1, 0), (3, 2, 1) \}$.

Solution

$V, W$ are vector spaces, $\mathcal{B} = \{v_1, \ldots, v_n\}$ is a basis for $V$. $\{w_1, \ldots, w_n\}$ is a set in $W$. There exists a unique $f: V \to W$ such that $f(v_i) = w_i$ for each $i = 1, \ldots, n$.

$B = \{ (1, 2, 3), (0, 1, 0), (3, 2, 1) \}$ a basis for $\mathbb{R}^3$

Since $f(1,2,3) = (0,0,0)$, $f(0,1,0) = (0,1,0)$, $f(3,2,1) = (3,2,1)$

Then $[f]_{B}= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Since $f \text{ is a projection iff } f(x) = x \quad \forall x \in \text{Im}(f).$

Then $[f]_{e} = C(B, E) [f]_{b} C(E,B)^{-1}$, $[f]_e = \begin{bmatrix} 1 & 0 & 3 \\ 2 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 3 \\ 2 & 1 & 2 \\ 3 & 0 & 1 \end{bmatrix}^{-1}$$= \begin{bmatrix} 9/8 & 0 & -3/8 \\ 1/4 & 1 & -3/4 \\ 3/8 & 0 & -1/8 \end{bmatrix}$

Then $f(x,y,z)=\left(\frac{9x-3z}{8},\quad\frac{x+4y-3z}{4},\quad\frac{3x-z}{8}\right)$
Let $V$ be a v.s. over K and let $W_1$ and $W_2$ be subspace of $V$ such that $V=W_1\oplus W_2$, then $V^*=W_1^\circ\oplus W^\circ _2$

Remark: $V^{*}=L\left(V,\mathbb{K}\right)$ and $W^{\circ}=\left\lbrace f:V\to K:f\left(w\right)=0,\forall w\in W\right\rbrace$

Theorem

Consider $f_i:V\to V$ such that $f_i$ is a projection on $W_i$ along $W_j$

We want $F_{i}:V^{*}\to V^*$, define the map is $(g:V\to K)\mapsto (F_{i}(g):V\to K)$

$v\mapsto F_i(g)(v)=g(f_i(v))$
1. $(F_i)^2=F_i$
  
  $F_{i}^{2}(g)(v)=F_{i}(F_{i}(g))(v)=F_i(g)(f_i(v))=g(f_i(f_i(v)))=g(f_i(v))=F_i(g)(v)$ 2. $F_iF_j=0$ if $i\neq j$
  
  $(F_{i}F_{j})(g)(v)=F_{i}(F_{j}(g))(v)=F_{j}(g)(f_{i}(v))=g(f_{j}f_{i}(v))=g(0)=0$ 3. $id_{V^*}=F_{1}+F_{2}$
  
  $(F_{1}+F_{2})(g)(v)=F_{1}(g)(v)+F_{2}(g)(v)=g\left(f_{1}\left(v\right)\right)+g\left (f_{2}\left(v\right)\right)=g\left(\left(f_{1}+f_{2}\right)\left(v\right)\right)= g\left(id_{v}\left(v\right)\right)=g\left(v\right)$ 4. $ImF_i=W_j^\circ$
  
  $\subseteq$) Assume $h\in ImF_{i}$, then $\exists g\in V^{*}:F_{i}\left(g\right)=h$
  
  Let $v\in W_{j},h(v)=F_{i}(g)(v)=g\left(f_{i}\left(v\right)\right)=g\left(0\right)=0$, thus $h(v)=0,\forall v\in W_j$
  
  Thus $h\in W_j^\circ$
  
  $\supseteq$)Let $g\in W_j^\circ$, then $g(w_j)=0,\forall w_j\in W_j$. Now if $v\in V$, $\exists!w_1\in W_1$ and $w_2\in W_2$ such that $v=w_1+w_2$, $g(v)=g(w_{i}+w_{j})=g\left(w_{i}\right)$
  
  $F_{i}(g)(v)=g(f_{i}(w_{1}+w_{2}))=g(w_{i})=g\left(v\right)$, thus $F_i(g)(v)=g(v),\forall v\in V$, then $F_i(g)=g$
Thus $V^{*}=W_{1}^{\circ}\oplus W_{2}^{\circ}$
Definition: Let $V$ be a finite-dimensional vector space over $K$. Let $f: V \to V$ be a linear operator.

We say that $f$ is nilpotent of index $r$ if $m_{r}\left(x\right)=x^{r}$. That is, $f^r = 0$ but $f^{i}\neq0$ if $i < r$.

Prove that if $f$ is nilpotent of index $r$ and $\dim V = n$, then $\{0\}\subsetneq\text{Ker}\,f\subsetneq\text{Ker}\,f^{2}\subsetneq\text{Ker}\,f^{3} \subsetneq\ldots\subsetneq\text{Ker}\,f^{r-1}\subsetneq\text{Ker}\,f^{r}=V$

$\text{Ker} f^r = V$ since $f^r = 0 \implies f^r(v) = 0 \ \forall v$

$\text{Ker } f^{i-1} \subsetneq \text{Ker } f^i$

$\subseteq$) $v \in \text{Ker }f^{i-1}\Rightarrow f^{i}(v)=0 \Rightarrow f(0) = f(f^{i-1}(v)) = f^{i}(v)$
$\therefore v\in \text{Ker }f^{i}$

$\supseteq$) By def $m_{f}(x)=x^{r}\Rightarrow f^{r}=0$ and $f^{r-1} \neq 0$ $\Rightarrow \exists v \in \text{Ker }f^{r} \setminus \text{Ker }f^{r-1}$

Now, $f(v) \in \text{Ker }f^{r-1}$ and $f(v) \notin \text{Ker }f^{r-2}$

$f(v) \in \text{Ker }f^{r-i}\setminus \text{Ker }f^{r-i-1} \text{ for }i = 0, \ldots, r-1$

$\dim V = n$
If $f$ is nilpotent of index $n$, there exist $f$-inv. subspaces $S_i \subseteq V$ with dim$S_i = i$ $\forall i = 0, \ldots, n$

$\{0\}\subsetneq\text{Ker }f\subsetneq\text{Ker }f^{2}\subsetneq\ldots\subsetneq\text{Ker }f^{n-2}\subsetneq\text{Ker }f^{n-1}\subsetneq V=\text{Ker }f^{n}$

$S_n = V = \text{span} \{ v, fv, \ldots, f^{n-1}(v) \}$

$S_{n-1}=\text{span}\{fv,f^{2}(v),\ldots,f^{n-1}(v)\}$

$S_{n-2}=\text{span}\{f^{2}(v),f^{3}(v),\ldots,f^{n-1}(v)\}$

$\vdots$

$S_{2} = \text{span}\{ f^{n-2}(v), f^{n-1}(v) \}$

$S_1 = \text{span} \{ f^{n-1}(v) \}$

$S_0 = \{ 0 \}$

$f(S_n) = S_{n-1} \subseteq S_n$

$f(S_{n-1}) = S_{n-2} \subseteq S_{n-1}$

$\vdots$

$f(S_i) = S_{i-1} \subseteq S_i \quad \text{\checkmark}$