(b) Prove $x,y,z∉ℚ$ but $x^n+y^n+z^n∈ℚ$ for all $n∈ℕ$.
24. Prove that an $n×n$ matrix $A$ over a field $F$ of characteristic $0$ is nilpotent iff $\text{Tr}(A^k)=0$ for all $1≤k≤n$.
25. Prove than two $n×n$ matrices $A$ and $B$ over a field $F$ of characteristic $0$ have the same characteristic polynomial iff $\text{Tr}(A^k)=\text{Tr}(B^k)$ for all $1≤k≤n$.
26. When $A$ and $B$ are two $n×n$ matrices over a field $F$ of characteristic $0$, show the characteristic polynomials of $AB$ and $BA$ are the same.
Proof: (22) When $I_n=\{1,2,...,n\}$, let $A_i=\{A⊆I_n~|~|A|=i\}$ so that we may say $$s_i=\sum_{A∈A_i}\prod_{a∈A}α_a$$ Now, define $$q_1=s_{j-1}p_1-js_j$$$$q_i=s_{j-i}p_i-q_{i-1}$$ By rearranging Newton's $j^\text{th}$ Formula we see that we must prove $$p_j=s_1p_{j-1}-(s_2p_{j-2}-(...-(s_{j-1}p_1-js_j)...)=q_{j-1}$$ To this end, we shall inductively prove, $$q_i=\sum_{A∈A_{j-i}} \sum_{a∈A} α_a^{i+1} \prod_{\substack{b∈A \\ b≠a}} α_b$$ which will establish the formula when $i=j-1$. For the base case $i=1$, $q_1=s_{j-1}p_1-js_j$ is seen to be the sum of all combinations of roots multiplied $j-1$ at a time with a squaring of one of those roots, which agrees with the inductive pattern stated above.
Now, for the inductive step, we see $q_i=s_{j-i}p_i-q_{i-1}$. Note $s_{j-i}p_i$ is the sum of all roots multiplied $j-i$ at a time with one root to the $(i+1)^\text{th}$ power, plus the sum of all roots multiplied $j-i+1=j-(i-1)$ at a time with one root to the $i^\text{th}=((i-1)+1)^{th}$ power. In fact, that second sum mentioned is inductively $q_{i-1}$, so that in fact $q_i$ accords with the pattern above and the induction is complete.
(23) (a) Letting $f(X)=(X-x)(X-y)(X-z)$, we see $p_1=1$, $p_2=2$, and $p_3=3$. Newton's Formulas provides a sufficient system: $$p_1-s_1=0⇒s_1=1$$ $$p_2-s_1p_1+2s_2=0⇒s_2=-\dfrac{1}{2}$$ $$p_3-s_1p_2+s_2p_1-3s_3=0⇒s_3=\dfrac{1}{6}$$ Since $s_4=0$, we may proceed to calculate $p_4$: $$p_4-s_1p_3+s_2p_2-s_3p_1=0⇒p_4=\dfrac{25}{6}$$ (b) Since we have $s_1$,$s_2$, and $s_3$, we may explicitly observe the polynomial $f(X)=x^3-x^2-\dfrac{1}{2}x-\dfrac{1}{6}$. Multiplying by $6$ one may prove the polynomial $6x^3-6x^2-3x-1$ has no rational roots by checking it against the rational root theorem. Meanwhile, $p_n$ may be deduced for all $n∈ℕ$ by the inductive process shown at the end of (a) which makes all of its calculations in $ℚ$, hence $p_n∈ℚ$.
(24) ($⇒$) We've seen the negative of the trace of a matrix manifests as the coefficient of $x^{n-1}$ in its characteristic polynomial. Hence, since a nilpotent matrix has a minimal polynomial of the form $x^m$, we see the characteristic polynomial must be of the form $x^{m'}$ so that $\text{Tr}(A)=0$. Since $A^k$ is also nilpotent for all $k$, the same argument applies. ($⇐$) This shall follow from the more general argument in (25).
(25) ($⇒$) Put $A$ and $B$ in Jordan form over the algebraic closure $K$ of $F$. We see that their primary diagonals' sum is the same given that their characteristic polynomials are the same and hence have the same roots in the same multiplicities. In general, given an upper triangular matrix $C$ with primary diagonal elements $c_{ii}=γ_i$, it is simple to show by induction that $C^k$ is an upper triangular matrix with primary diagonal elements $γ_i^k$, i.e. the strict upper triangular elements have no effect on the primary diagonal, ultimately implying through the Jordan forms of $A$ and $B$ that $\text{Tr}(A^k)=\text{Tr}(B^k)$. ($⇐$) Once again given the Jordan forms $A'$ and $B'$ of $A$ and $B$ over $K$, we may deduce from the Newton's Formula system implied by the diagonal sums entailed by $\text{Tr}(A'^k)=\text{Tr}(B'^k)$ for $k∈I_n$ that the elementary symmetric functions in the roots of the characteristic polynomials for $A$ and $B$ are identical (here characteristic $0$ or at least $> n$ is needed so that the term $(-1)^jjs_j$ doesn't vanish), hence their characteristic polynomials are identical.
(26) For any two matrices $A$ and $B$ with $AB=C$ we see $$\text{Tr}(AB)=\sum_{m=1}^n c_{mm}=\sum_{m=1}^n \sum_{k=1}^n a_{mk}b_{km} = \sum_{k=1}^n \sum_{m=1}^n b_{km}a_{mk} = \text{Tr}(BA)$$ Thus for all $k$ we have $\text{Tr}((AB)^k)=\text{Tr}(A(BA)^{k-1}B)=\text{Tr}((BA)^{k-1}BA)=\text{Tr}((BA)^k)$ hence by (25) $AB$ and $BA$ have the same characteristic polynomial.$~\square$
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