17 Suppose $\mathbf{F}=\mathbf{C}$ and $P \in \mathcal{L}(V)$ is such that $P^{2}=P$. Prove that the characteristic polynomial of $P$ is $z^{m}(z-1)^{n}$, where $m=\operatorname{dim}$ null $P$ and $n=\operatorname{dim} \operatorname{range} P$.

SOLUTION First we will prove that
$$
V=\operatorname{null} P \oplus \operatorname{range} P .
$$

To do this, first suppose $u \in$ null $P \cap$ range $P$. Then $P u=0$, and there exists $w \in V$ such that $u=P w$. Applying $P$ to both sides of the last equation, we have $P u=P^{2} w=P w$. But $P u=0$, so this implies that $P w=0$. Because $u=P w$, this implies that $u=0$. Because $u$ was an arbitrary vector in null $P \cap$ range $P$, this implies that null $P \cap$ range $P=\{0\}$.
Now suppose $v \in V$. Then obviously
$$
v=(v-P v)+P v .
$$

Note that $P(v-P v)=P v-P^{2} v=0$, so $(v-P v) \in$ null $P$. Clearly $P v \in \operatorname{range} P$. Thus the equation above shows that $v \in \operatorname{null} P+\operatorname{range} P$. Because $v$ was an arbitrary vector in $V$, this implies that $V=\operatorname{null} P+\operatorname{range} P$.

We have shown that null $P \cap \operatorname{range} P=\{0\}$ and $V=$ null $P+\operatorname{range} P$. Thus $V=$ null $P \oplus \operatorname{range} P$ (by 1.46 ).
Note that
$$
G(0, P) \supseteq E(0, P)=\operatorname{null} P .
$$

If $v \in \operatorname{range} P$, then there exists $u \in V$ such that $v=P u$. Applying $P$ to both sides shows that $P v=P^{2} u=P u=v$, which implies that $v \in E(1, P)$. Hence
$$
G(1, P) \supseteq E(1, P) \supseteq \text { range } P \text {. }
$$

Thus $m+n=\operatorname{dim} V$. Because the sum of the multiplicities of the eigenvalues of $P$ equal $\operatorname{dim} V$ (by 8.25 ), the inclusions above must be equalities and there are no other eigenvalues of $P$. Thus the characteristic polynomial of $P$ is $z^{m}(1-z)^{n}$.

Question

17 Suppose $\mathbf{F}=\mathbf{C}$ and $P \in \mathcal{L}(V)$ is such that $P^{2}=P$. Prove that the characteristic polynomial of $P$ is $z^{m}(z-1)^{n}$, where $m=\operatorname{dim}$ null $P$ and $n=\operatorname{dim} \operatorname{range} P$.

SOLUTION First we will prove that
$$
V=\operatorname{null} P \oplus \operatorname{range} P .
$$

To do this, first suppose $u \in$ null $P \cap$ range $P$. Then $P u=0$, and there exists $w \in V$ such that $u=P w$. Applying $P$ to both sides of the last equation, we have $P u=P^{2} w=P w$. But $P u=0$, so this implies that $P w=0$. Because $u=P w$, this implies that $u=0$. Because $u$ was an arbitrary vector in null $P \cap$ range $P$, this implies that null $P \cap$ range $P=\{0\}$.
Now suppose $v \in V$. Then obviously
$$
v=(v-P v)+P v .
$$

Note that $P(v-P v)=P v-P^{2} v=0$, so $(v-P v) \in$ null $P$. Clearly $P v \in \operatorname{range} P$. Thus the equation above shows that $v \in \operatorname{null} P+\operatorname{range} P$. Because $v$ was an arbitrary vector in $V$, this implies that $V=\operatorname{null} P+\operatorname{range} P$.

We have shown that null $P \cap \operatorname{range} P=\{0\}$ and $V=$ null $P+\operatorname{range} P$. Thus $V=$ null $P \oplus \operatorname{range} P$ (by 1.46 ).
Note that
$$
G(0, P) \supseteq E(0, P)=\operatorname{null} P .
$$

If $v \in \operatorname{range} P$, then there exists $u \in V$ such that $v=P u$. Applying $P$ to both sides shows that $P v=P^{2} u=P u=v$, which implies that $v \in E(1, P)$. Hence
$$
G(1, P) \supseteq E(1, P) \supseteq \text { range } P \text {. }
$$

Thus $m+n=\operatorname{dim} V$. Because the sum of the multiplicities of the eigenvalues of $P$ equal $\operatorname{dim} V$ (by 8.25 ), the inclusions above must be equalities and there are no other eigenvalues of $P$. Thus the characteristic polynomial of $P$ is $z^{m}(1-z)^{n}$.

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To prove the characteristic polynomial of $ P $, we first show that $ V $ can be decomposed into a direct sum of the null space and range of $ P $. ‖ ‖ step 2 ⋮ If $ u \in \text{null } P \cap \text{range } P $, then $ P u = 0 $ and $ u = P w $ for some $ w \in V $. Applying $ P $ gives $ P u = P^2 w = P w $, hence $ u = 0 $. This implies $ \text{null } P \cap \text{range } P = \{0\} $. ‖ ‖ step 3 ⋮ For any $ v \in V $, we can write $ v = (v - P v) + P v $, where $ (v - P v) \in \text{null } P $ and $ P v \in \text{range } P $, showing $ V = \text{null } P + \text{range } P $. ‖ ‖ step 4 ⋮ Since $ \text{null } P \cap \text{range } P = \{0\} $ and $ V = \text{null } P + \text{range } P $, we have $ V = \text{null } P \oplus \text{range } P $. ‖ ‖ step 5 ⋮ The generalized eigenspace $ G(0, P) $ contains the eigenspace $ E(0, P) = \text{null } P $, and similarly $ G(1, P) $ contains $ E(1, P) $ which contains the range of $ P $. ‖ ‖ step 6 ⋮ Since $ m + n = \text{dim } V $ and the sum of the multiplicities of the eigenvalues equals $ \text{dim } V $, the characteristic polynomial of $ P $ is $ z^m(1 - z)^n $, where $ m = \text{dim null } P $ and $ n = \text{dim range } P $. ‖ ∻Answer∻ ⚹ The characteristic polynomial of $ P $ is $ z^m(1 - z)^n $, where $ m = \text{dim null } P $ and $ n = \text{dim range } P $. ⚹ ∻Key Concept∻ ⚹Characteristic Polynomial of a Projection Operator⚹ ∻Explanation∻ ⚹The characteristic polynomial of a projection operator $ P $ that satisfies $ P^2 = P $ is given by $ z^m(1 - z)^n $, where $ m $ is the dimension of the null space of $ P $, and $ n $ is the dimension of the range of $ P $. This is because $ P $ has only 0 and 1 as its eigenvalues, with their multiplicities adding up to the dimension of the vector space $ V $.⚹◊What is the significance of $P^{2} = P$ in relation to the characteristic polynomial of $P$?⍭ Generate me a similar question◊

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