User:J D Bowen/Math725 HW12

1) Let $$p(x)\in\mathbb{C}[x] \ $$ and define $$(p) \ $$ as $$(p):=\left\{{pq:q\in\mathbb{C}[x]}\right\} \ $$. Observe that $$0\in\mathbb{C}[x] \implies 0=0\cdot p \in (p) \ $$.  We also have $$r,s \in\mathbb{C}[x] \implies \exists u,v : \ r=up, \ s=vp \ $$.  Then $$r+s=up+vp=(u+v)p \in (p) \ $$.  Finally, observe $$rs = (up)(vp) = (upv)p \in (p) \ $$.  So, $$(p) \ $$ is an ideal.

2) Let $$a,b \in \left\{{p:p(0)=1 }\right\} \ $$. Then $$(a+b)(0) = a(0)+b(0) = 1+1=2 \neq 1 \implies a+b\not\in\left\{{p:p(0)=1 }\right\} \ $$ which means this is not an ideal.

3) Define the set $$V=\left\{{-1,1}\right\} \ $$, and consider the set $$I=\left\{{p:p(V)=0 }\right\} \ $$. Obviously the zero polynomial $$0 \in I \ $$, since $$0(V)=0 \ $$.  Observe that $$a,b \in I \implies (a+b)(V) = a(V)+b(V) = 0+0=0, \ (ab)(V) = a(V)b(V) = 0\cdot 0 = 0 \ $$.  So $$a+b, \ ab \in I \ $$, and so $$I \ $$ is an ideal.

Observe that since $$x^2-1\in I \implies (x^2-1)\subseteq I \ $$. Further observe that any polynomial $$p\in I \ $$ vanishes on $$-1,1 \ $$, and so $$(x-1)|p, \ (x+1)|p \implies (x^2-1)|p \ $$, and so $$I\subseteq (x^2-1) \ $$. Hence $$I=(x^2-1) \ $$.

4) Suppose $$P:V\to V \ $$ is a projection $$P^2=P \ $$. This implies $$P^2-P = 0 \ $$.  If there was a smaller degree monic polynomial such that $$p(P) =0 \ $$, then we would have $$p|(x^2-x) \ $$, but $$x(x-1)=x^2-x \ $$, and $$x, \ x-1 \ $$ are prime.  Since $$\mathbb{C}[x] \ $$ is a Euclidean domain, it is a unique factorization domain, and hence the only possible factors of $$x^2-x \ $$ are $$x, \ x-1 \ $$.  But $$P\neq 0, I \ $$ and so $$x^2-x \ $$ is the minimal polynomial.

Therefore, the possible characteristic polynomials are elements of $$(x^2-x) \ $$, and so $$\lambda=-1,1 \ $$ are among the eigenvalues of $$P \ $$.

Define $$Q=I-P \ $$. Then $$(I-Q)^2-(I-Q)=0 \ $$, and so $$0 = I-2Q+Q^2-I+Q \implies Q^2-Q=0 \ $$, and so $$Q \ $$ is also a projection. The minimal polynomial is as given.

Suppose $$Ch_P(\lambda)=\lambda^n+a_{n-1}\lambda^{n-1}+\dots +a_0 \ $$ is the characteristic polynomial of $$P \ $$. Then

$$Ch_Q(\lambda)=|Q-\lambda I| = |I-P-\lambda I| = |-P+(1-\lambda)I| = (-1)^{\text{dim}(P)} |P-(1-\lambda)I| = (-1)^{\text{dim}(P)} Ch_P(1-\lambda) \ $$.

5) Suppose a transformation $$P \ $$ is invertible. Then it has inverse $$P^{-1} \ $$ such that $$P^{-1}P=PP^{-1}=I \ $$.  Now suppose $$f \ $$ is the minimal polynomial for $$P \ $$.

$$f(P)=0 \implies f(P)P^{-1}=0\cdot P^{-1} =0 \ $$. Suppose $$f \ $$ has no constant term: $$f(x)=x^n +a_{n-1}x^{n-1}+ \dots +a_1 x \ $$. Then $$f(P)P^{-1} = P^{n-1}+\dots +a_1 =0 \ $$. Then $$g(x)=x^{n-1}+a_{n-1}x^{n-2}+\dots+a_1 \ $$ is a monic polynomial that vanishes for $$P \ $$ with degree less than $$f \ $$, and so $$f \ $$ cannot be the minimal polynomial for $$P \ $$. Since it is, $$f \ $$ must have a constant term.

Hence, $$P \ $$ invertible implies $$f \ $$ has a constant term.

Now we show that $$f \ $$ having a constant term implies $$P \ $$ is invertible by showing that if $$P \ $$ is not invertible, then $$f \ $$ cannot have a constant term.

Suppose $$P \ $$ is not invertible. Then $$\text{ker}(P)\neq \varnothing \implies \exists \vec{v}: P\vec{v}=\vec{0} \ $$. This means that $$0 \ $$ is an eigenvector of $$P \ $$. Then $$0 \ $$ is a root of the characteristic polynomial, which means that it must be a root of the minimal polynomial: $$f(0)=0 \ $$. But this is just $$0^n + \dots+a_0 = 0 \ $$, which of course implies $$f \ $$ has no constant term.

6) Let $$T:V\to V \ $$ have eigenspaces $$U_{\lambda,T}=\text{ker}(T-\lambda I) \ $$. Then $$x\in U_\lambda \implies Tx=\lambda x \ $$.  Observe that $$T^{-1}x =T^{-1}(\lambda^{-1}\lambda)x =\lambda^{-1} T^{-1} \lambda x = \lambda^{-1} T^{-1} Tx = \lambda^{-1}x \ $$.

So $$\lambda^{-1} \ $$ is an eigenvalue of $$T^{-1} \ $$, and $$U_{\lambda^{-1},T^{-1}}=U_{\lambda,T} \ $$.

Observe that this implies the minimal polynomial of $$T \ $$ is $$(x-\lambda_1)\dots(x-\lambda_n) \ $$ and the minimal polynomial of $$T^{-1} \ $$ is $$(x-\lambda_1^{-1})\dots(x-\lambda_n^{-1}) \ $$.

The number of eigenvalues of $$T \ $$ are necessarily the same as $$T^{-1} \ $$.

7) Let $$T $$ be a linear map with eigenvalues $$\lambda_1, \dots, \lambda_m \ $$ and eigenvectors $$x_1, \dots, x_m \ $$.  Then we have $$\langle x_j, T^H x_j \rangle = \langle Tx_j, x_j \rangle =  \langle \lambda_j x_j, x_j \rangle = \lambda_j \langle x_j, x_j \rangle = \langle x_j, \overline{\lambda_j}x_j \rangle $$.

But $$T^H x_j = \overline{\lambda_j}x_j \ $$ implies that $$x_j \ $$ is an eigenvector of $$T^H \ $$ with eigenvalue $$\overline{\lambda_j} \ $$. Hence if the minimal polynomial for $$T \ $$ is $$p(x) \ $$, the minimal polynomial of $$T^H \ $$ is $$q(x):=p(\overline{x}) \ $$.