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 (p) \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=0,1 \ $ 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}) \ $.