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3.19 Projector

Theorem

Suppose \(\dim V<\infty\) and \(f:V\to V\) be a k-linear map.

If \(f\) is diagonalizable, then for \(c_1,...,c_r\) the different eigenvalues, \(\exists\) projectors \(0\neq p_{i}:V\to V,1\leq i\leq r\) such that:

  1. \(f=f\circ p_{1}+\cdots+f\circ p_{r}=c_{1}\circ p_{1}+\cdots+c_{r}\circ p_{r}\)

  2. \(Id_V=p_1+...+p_r\)​

  3. \(p_i\circ p_j=0\) if \(i\neq j\)​

  4. \(p_i^2=p_i,\forall i\)​

  5. \(Imp_{i}=W_{i}=\{\text{eigenvectors of eigenvalue }c_{i}\}\cup\{0_{V}\}\)​

Conversely, if \(\exists c_{1},...,c_{r}\in k,c_{i}\neq c_{j}\) if \(i\neq j\) such that 1 2 3 hold, then \(f\) is diagonalizable where \(c_1,...,c_r\) are all the eigenvalue of \(f\). Moreover, 4 5 are also true.

Proof

\(\Rightarrow\)) Suppose \(f\) is diagonalizable and \(c_1,..,c_r\) are its different eigenvalues

Let \(W_i=\ker(f-c_iId),1\leq i\leq r\). Since \(f\) is diagonalizable, then \(V=W_1\oplus...\oplus W_r\)

Then we can have projectors \(p_1,...,p_r\) associated to this decomposition

We've already proved that 2 3 4 5 are true, then we are going to prove 1

Let \(v\in V\), then put \(f\) in both sides of 2, we get \(f(v)=f\circ p_{1}\left(v\right)+\cdots+f\circ p_{r}\left(v\right)\)

Since def of projectors, we get \(f(v)=c_{1}\circ p_{1}\left(v\right)+\cdots+c_{r}\circ p_{r}\left(v\right)=\left( c_{1}p_{1}+\cdots+c_{r}p_{r}\right)\left(v\right)\)

Thus \(f=c_1p_1+...+c_rp_r\)

\(\Leftarrow\)) Let \(c_{1},...,c_{r}\in k,c_{i}\neq c_{j}\) if \(i\neq j\) and \(f:V\to V\), \(k\)-linear map

Let \(p_i:V\to V,1\leq i\leq r\) be \(k\)-linear maps such that 1 2 3 hold

To prove 4 is true

By 2), \(Id_{V}=p_{1}+...+p_{r}\) (that is \(\forall v\in V,v=p_1(v)+...+p_r(v)\))

So given \(i\), \(1\leq i\leq r\), \(p_{i}=p_{i}\circ Id_{V}=p_i(p_1+...+p_r)=\sum_{j=1}^{r}p_i\circ p_j=p_i^2\), then 4) hold

To prove \(f\) is diagonalizable

Idea: 1. all of them are eigenvalue 2. there is no other eigenvalue 3. I will use 1

  1. all of them are eigenvalue

    By 1, we know that \(f=c_{1}\circ p_{1}+\cdots+c_{r}\circ p_{r}\), we get \(f\circ p_{i}=\left(\sum_{j=1}^{r}c_{j}p_{j}\right)\circ p_{i}=\sum_{j=1}^{r}c_{j} p_{j}\circ p_{i}=c_{i}p_{i}^{2}=c_{i}p_{i}\)

    So \(\forall v\in Imp_i\), \(\exists w:v=p_i(w)\), then \(f(v)=f(p_i(w))=c_ip_i(w)=c_iv\). Thus \(c_i\) is an eigenvalue of \(f\)

  2. there is no other eigenvalue

    Suppose \(\lambda\in k\) be an eigenvalue of \(f\) which different from \(c_1,...,c_r\)

    Consider \(f-\lambda Id_{V}=\sum_{j=1}^{r}c_{j}p_{j}-\lambda\sum_{j=1}^{r}p_{j}=\sum_{j=1}^{r} \left(c_{j}-\lambda\right)p_{j}\)​

    Let \(v\) be an eigenvector corresponding to \(\lambda\), then \(0_{V}=(f-\lambda Id)(v)=\underbrace{(c_{1}-\lambda)p_{1}(v)}_{\in Imp_1}+...+\underbrace{(c_{r}-\lambda)p_{r}(v)}_{\in Imp_r}\)​

    Use 2 3 4, we know that \(V=Imp_1\oplus ...\oplus Imp_r\), then \((c_i-\lambda)p_i(v)=0,\forall i,1\leq i\leq r\)​

    Since \(v\neq 0_V,\exists i,1\leq i\leq r:p_i(v)\neq 0_V\), For this \(i\), we have that \(c_i=\lambda\)​

To prove 5 hold

Since \(f\) is diagonalizable, then

\(\subseteq\)) Take eigenvalue \(c_i\) and \(v_{i}\in \text{Im}p_{i}\), then \(f(v_{i})=f\left(p\left(w\right)\right)=c_{i}p_{i}\left(w\right)=c_{i}v_{i}\), thus \(v_i\in W\)

\(\supseteq\)) Take \(v_i\in W\), then \(f(v_{i})=c_{i}v_{i}\) \(=c_{1}\circ p_{1}\left(v_{i}\right)+\cdots+c_{r}\circ p_{r}\left(v_{i}\right)=c_{i} p_{i}\left(v_{i}\right)\), then \(v_i=p_i(v_i)\)

Thus \(v_i\in Imp_i\)

Example

\(V\) is k-linear space, \(f:V\to V,k\)-linear map such that \(m_{f}(x)=(x-3)^{2}(x-5)^{3}\)

We know \(f\) is not diagonalizable, so \(\ker(f-3Id)\oplus \ker (f-5Id)\subsetneq V\)

Let \(W_{3}=\ker (f-3Id)^{2}\supseteq \ker(f-3Id)\) and \(W_{5}=\ker(f-5Id)^{3}\supseteq\ker(f-5Id)\)

Let \(\mathcal{B}_3\) be a basis of \(W_3\), \(\mathcal{B}_3=\{v_1,...,v_i\}\) with \(1\leq i\leq \dim V=m\)

Let \(\mathcal{B}_{5}\) be a basis of \(W_{5}\), \(\mathcal{B}_{5}=\{v_{i+1},...,v_{m}\}\), we will see that \(\mathcal{B}_{3}\cup \mathcal{B}_{5}=\mathcal{B}\) is a basis of \(V\)

Given \(v\in V\), we know that \(m_{f}(f)(v)=0_{V}=\left(f-3Id\right)^{2}\left(\left(f-5Id\right)^{3}\left(v\right )\right)\)

Consider \((x-3)^2\) and \((x-5)^{3}\), we know that they are coprime

Thus \(\exists p(x),q(x)\in k[x]:1=(x-3)^{2}p(x)+(x-5)^{3}q(x)\)

Then \(Id_{V}=(f-3Id_{V})^{2}\circ p(f)+(f-5Id_{V})^{3}\circ q(f)\), given \(v\in V\)

\(v=Id_{V}(v)=\underbrace{(f-3Id_{V})^{2}\circ p(f)\left(v\right)}_{w}+\underbrace{(f-5Id_{V})^{3}\circ q(f)\left (v\right)}_{w'}\)

Since \(m_f(f)=0\), we have that \((f-5Id_{V})^{3}(w)=\underbrace{\left(f-5Id_{V}\right)^3\left((f-3Id_{V})^2\circ p(f)\left(v\right)\right)}_{m_{f}(f)\left(v\right)}=0_{V}\), also \((f-3Id_{V})^{2}(w^{\prime})=0_{V}\)

Thus \(w\in W_5,w'\in W_3\), then \(V=W_5+W_3\)

To prove it's direct sum

Take \(z\in W_3\cap W_5\), \(z = Id_{V}(z) = \underbrace{p(f) \circ \left( (f - 3Id_{V})^2 (z) \right)}_{0 \text{ because } z \in W_3 = \ker (f - 3Id_{V})^2}+ \underbrace{q(f) \circ \left( (f - 5Id_{V})^3 (z) \right)}_{0 \text{ because } z \in W_5 }=0\)

So \(V=W_3\oplus W_5\) and \(V=\ker(f-3Id)^{2}\oplus\ker(f-5Id)^{3}\)​

Theorem(The primary decomposition theorem)

Let \(V\) be a finite dimensional \(k\)-vector space, \(f:V\to V\) be a \(k\)-linear map

Suppose \(m_{f}(x)=p_{1}(x)^{m_1}\cdot...\cdot p_{r}(x)^{m_{r}}=\prod_{i=1}^{r}p_{i}\left( x\right)^{m_i}\) where \(p_i(x)\neq p_j(x)\) if \(i\neq j\) and \(p_i(x)\) is irreducible and monic \(\forall i\) and \(m_i>0,\forall i\)

Let \(W_{i}=\ker p_{i}(f)^{m_i}\) \(\forall i,1\leq i\leq r\), then

  1. \(V=W_{1}\oplus...\oplus W_{r}\)​

  2. \(W_i\) is \(f\)-invariant \(\forall i\)​

  3. Let \(f_{i}=f|_{W_i}:W_{i}\to W_{i}\), \(\forall i\), then \(m_{f_i}(x)=p_i(x)^{m_i}\)​

image​

Proof

To prove 1, having such a decomposition means we have projectors \(E_i:V\to V\) \(\forall i,1\leq i\leq r\) such that \(E_i^2=E_i\), \(E_i\circ E_j=0,\) if \(i\neq j\) and \(Id=E_1+...+E_r\), then i know \(V=ImE_1\oplus ...\oplus ImE_r\) (since)and to prove \(ImE_i=W_i\)

Let's find the projectors \(E_i\), \(1\leq i\leq r\) such that \(E_{i}|_{W_i}=Id_{W_i}\) and given \(j\neq i\), \(E_i|_{W_j}=0\)

Given \(i\), \(1\leq i\leq r\), let \(\gamma_{i}(x)=\frac{m_{f}\left(x\right)}{p_{i}\left(x\right)^{m_{i}}}=\prod_{j=1,j\neq i}^{r}p_{j}\left(x\right)^{m_{j}}\in k[x]\)

Then \(\gamma_{1}(x),\gamma_{2}\left(x\right),\ldots,\gamma_{r}\left(x\right)\) have no common irreducible factors at same time

So their greatest common divisor is \(1\), thus there exists polynomials \(g_i(x),1\leq i\leq r\) such that \(1=g_{1}(x)\gamma_{1}(x)+\cdots+g_{r}(x)\gamma_{r}(x)=\sum_{i=1}^{r}\underbrace{g_{i}\left(x\right )\gamma_{i}\left(x\right)}_{h_i(x)}\), this will help me to find the projectors

Let's define \(E_{i}=h_i(f),\forall i,1\leq i\leq r\)

  1. \(Id=E_{1}+...+E_{r}\)​

    \(E_{i}=h_{i}(f)=g_{i}\left(f\right)\circ\underbrace{\frac{m_{f}}{p_{i}^{m_{i}}}\left(f\right)} _{\neq0}\). Since \(1=\sum^r_{i=1}h_i(x)\), we get \(Id_{V}=\sum_{i=1}^{r}h_{i}(f)=\sum_{i=1}^{r}E_{i}\)

  2. \(E_{i}\circ E_{j}=0\)

    Moreover, if \(i\neq j\), \(E_{i}\circ E_{j}=h_{i}(f)\circ h_{j}(f)=\underbrace{\left(h_{i}\left(x\right)\circ h_{j}\left (x\right)\right)}_{\text{is a multiple of }m_f(x)}\left(f\right)=0\)​

  3. \(E_i^2=E_i\)​

    Moreover, \(E_{i}=E_{i}\circ Id_{V}=E_{i}\circ\sum_{j=1}^{r}E_{j}=\sum_{j=1}^{r}E_{i}\circ E _{j}=E_{i}^{2}\), so \(E_i=E_i^2\)

So \(V=\operatorname{\mathrm{Im}}E_{1}\oplus...\oplus\operatorname{\mathrm{Im}}E_{r}\)

  1. \(\text{Im}E_{i}=\left(W_{i}=\ker p_{i}(f\right)^{m_{i}})\)

    (\(\subseteq\)) Given \(v \in \text{Im} E_i\) , let \(v = E_i(v)\), so \(p_{i}(f)^{m_{i}}v=p_{i}(f)^{m_{i}}(E_{i}(v))=p_{i}(f)^{m_{i}}\circ h_{i}(f)(v)\) \(=p_{i}(f)^{m_{i}}\circ g_{i}(f)\circ\frac{m_{f}(f)}{p_{i}(f)^{m_{i}}}(v)=g_{i}(f )\circ p_{i}(f)^{m_{i}}\circ\frac{m_{f}(f)}{p_{i}(f)^{m_{i}}}(v)=q_{i}\left(f\right )\circ m_{f}(f)(v)=0\)

    Thus \(v\in\ker p_{i}(f)^{m_{i}}\)

    (\(\supseteq\)) Suppose \(v \in \ker p_i(f)^{m_i} = W_i\), NTP \(v\in\operatorname{\mathrm{Im}}h_{i}(f)\)

    \(v = Id_V(v) = (E_1 + \dots + E_r)(v)\) \(= h_1(f)(v) + h_2(f)(v) + \dots + h_r(f)(v)\)

    If \(j \neq i\) , \(1 \leq j \leq r\) , then: \(h_j(f)(v) = 0\), then \(g_{j}(f)\frac{m_{f}}{{p_j}^{m_{j}}}(v)=0_{V}\) since \(\frac{m_{f}}{{p_j}^{m_{j}}}\) is divisible by \(p_{i}^{m_i}\) and \(v\in\ker p_{i}^{m_i}\)

    Thus \(v=h_{i}(f)(v)=E_{i}(v)\Rightarrow v\in\text{Im}E_{i}\)

Thus \(V=W_{1}\oplus...\oplus W_{r}\)

To prove 2, we can prove it directly, Let \(v\in W_{i}=\ker p_{i}(f)^{m_{i}}\), NTP \(f(v)\in W_{i}\iff p_{i}(f)^{m_i}(f(v))=0\)​

\(p_{i}(f)^{m_{i}}(f(v))=p_i(f)^{m_i}\circ f(v)=f\circ p_i(f)^{m_i}(v)=f(\underbrace{p_i(f)^{m_i}(v)}_{0_v})=0_V\)

To prove 3, given a restriction \(f_i\) of \(f\), \(m_{f_i}(x)\mid m_f(x)\), NTP \(m_{f_{i}}(x)=p_{i}(x)^{m_1}\iff\)They divides each other

  1. We have \(W_i = \text{Im} E_i = \ker p_i(f)^{m_i}\), so \(p_{i}(f)^{m_{i}}(v)=0,\forall v\in W_{i}\), thus \(m_{f_i}(x)\mid p_i(x)^{m_i}\)

  2. Conversely, suppose \(\phi(x)\in k[X]:\phi(f_{i})=0\) , then \(\phi(f_i): W_i \rightarrow W_i\)

    \(\left(\phi(x\right)\cdot\frac{m_{f}(x)}{p_{i}(x)^{m_{i}}})(f):V\to V=0\). Why is this equal to zero?

    Because given \(v \in V\) : \(v = E_1(v) + \dots + E_r(v)\) where \(E_i(v) \in W_i\) .

    Of course, \(\left( \phi(x) \cdot \frac{m_{f}(x)}{p_{i}(x)^{m_i}}\right) (f) \cdot E_{i}(v) =\) \(\frac{m_{f}(x)}{p_{i}(x)^{m_{i}}}(f)\cdot\phi(f)\cdot E_{i}(v)\)

    Since \(E_i(v)\in W_i\), then \(\phi(f)\cdot E_{i}(v)=\phi(f_{i})\cdot E_{i}(v)=0\)

    Thus \(\left(\phi(x)\cdot\frac{m_{f}(x)}{p_{i}(x)^{m_{i}}}\right)(f)=0\)

    Given \(j, 1 \leq j \leq r, j \neq i\) , we want to prove that \(\left(\phi(x)\cdot\frac{m_{f}(x)}{p_{i}(x)^{m_{i}}}\right)(f)\cdot E_{j}(v)=0\)???OO because \(E_{j}(v) \in \ker p_{j}(f)^{m_j}\) and \(p_{j}(x)^{m_{j}}\mid\frac{m_{f}(x)}{p_{i}(x)^{m_{i}}}\)

    Then \(m_{f}(x)\mid\phi(x)\cdot\frac{m_{f}(x)}{p_{i}(x)^{m_{i}}}\), this means that \(\exists q(x) \in k[x] \text{ such that }\)​\(\phi(x) \cdot \frac{m_f(x)}{p_i(x)^{m_i}} = m_f(x) \cdot q(x)\)

    \(\Leftrightarrow \phi(x) m_f(x) = m_f(x) q(x) p_i(x)^{m_i}\). So \(\phi(x) = q(x) \cdot p_i(x)^{m_i}\).

    We can fix \(\phi(x)=m_{f_i}(x)\), then we get that \(p_{i}(x)^{m_i}\mid m_{f_i}(x)\)

    Thus \(m_{f_{i}}(x)=p_{i}(x)^{m_{i}}\)

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