(a) $||A+B|| ≤ ||A|| + ||B||$
(b) $||AB|| ≤ ||A|| \cdot ||B||$
(c) $||αA|| = |α| \cdot ||A||$
41. Let $R$ be the radius of convergence of the real or complex power series $G(x)$.
(a) Prove that if $||A|| < R$ then $G(A)$ converges.
(b) Deduce that for all matrices $A$ the following power series converge:$$\text{sin}(A)=\sum_{k=0}^∞ (-1)^k \dfrac{A^{2k+1}}{(2k+1)!}$$$$\text{cos}(A)=\sum_{k=0}^∞(-1)^k \dfrac{A^{2k}}{(2k)!}$$$$\text{exp(A)}=\sum_{k=0}^∞\dfrac{A^k}{k!}$$ 42. Let $P$ be a nonsingular $n \times n$ matrix, and denote the variable $t$ by the matrix $tI$ (in light of the theory of differential equations).
(a) Prove $PG(At)P^{-1}=G(PAtP^{-1})=G(PAP^{-1}t)$, so it suffices to consider power series for matrices in canonical form.
(b) Prove that if $A$ is the direct sum of matrices $A_1,...,A_m$, then $G(At)$ is the direct sum of the matrices $G(A_1t),...,G(A_mt)$.
(c) Show that if $Z$ is the diagonal matrix with entries $z_1,...,z_n$ then $G(Zt)$ is the diagonal matrix with entries $G(z_1t),...,G(z_nt)$.
43. Letting $A$ and $B$ be commuting matrices, show $\text{exp}(A+B)=\text{exp}(A)\text{exp}(B)$.
44. Letting $λ ∈ K$, show$$\text{exp}(λIt+M)=e^{λt}\text{exp}(M)$$ 45. Let $N$ be the $r \times r$ matrix with $1$s on the first superdiagonal and zeros elsewhere. Show$$\text{exp}(Nt)=\begin{bmatrix}1 & t & \dfrac{t^2}{2!} & \cdots & \cdots & \dfrac{t^{r-1}}{(r-1)!} \\ ~ & 1 & t & \dfrac{t^2}{2!} &~ & \vdots \\ ~ & ~ & \ddots & \ddots & \ddots & \vdots \\ ~ & ~ & ~ & \ddots & t & \dfrac{t^2}{2!} \\ ~ & ~ & ~ & ~ & 1 & t \\ ~ & ~ & ~ & ~ & ~ & 1 \end{bmatrix}$$Deduce that if $J$ is the $r \times r$ elementary Jordan matrix with eigenvalue $λ$ then$$\text{exp}(Jt)=\begin{bmatrix}e^{λt} & te^{λt} & \dfrac{t^2}{2!}e^{λt} & \cdots & \cdots & \dfrac{t^{r-1}}{(r-1)!}e^{λt} \\ ~ & e^{λt} & te^{λt} & \dfrac{t^2}{2!}e^{λt} &~ & \vdots \\ ~ & ~ & \ddots & \ddots & \ddots & \vdots \\ ~ & ~ & ~ & \ddots & te^{λt} & \dfrac{t^2}{2!}e^{λt} \\ ~ & ~ & ~ & ~ & e^{λt} & te^{λt} \\ ~ & ~ & ~ & ~ & ~ & e^{λt} \end{bmatrix}$$ Proof: (40)(a) We have$$||A+B||=\sum_{i,j}|a_{ij}+b_{ij}| ≤ (\sum_{i,j}|a_{ij}|)+(\sum_{i,j}|b_{ij}|)=||A||+||B||$$(b) We have$$||AB||=\sum_{i,j}|\sum_{k=0}^n a_{ik}b_{kj}| ≤ \sum_{i,j} \sum_{k=0}^n |a_{ik}| \cdot |b_{kj}| ≤$$$$\sum_{i,j,i',j'} |a_{ij}| \cdot |b_{i'j'}| = (\sum_{i,j}|a_{ij}|)(\sum_{i,j}|b_{ij}|) = ||A|| \cdot ||B||$$(c) We have$$||αA|| = \sum_{i,j}|αa_{ij}| = |α| \sum_{i,j} |a_{ij}| = |α| \cdot ||A||$$(41)(a)(Method due to Project Crazy Project) For each entry $i,j$ entry $a_{(k)ij}$ of $A^k$, we note that $|a_(k){ij}| ≤ ||A^k|| ≤ ||A||^k < R^k$. Therefore, we note $\sum_{k=0}^N |α_ka_{(k)ij}| ≤ \sum_{k=0}^N |α_k| \cdot r^k$, where $r$ has been chosen $||A|| < r < R$. By the Cauchy-Hadamard theorem this last power series displays the same radius of convergence $R$ and thus converges for $r$ as $N$ approaches infinity. Thus the partial power series for $G(A)$ in the $i,j$ coordinate converges absolutely, and thus converges.
(b) Since these functions are shown to have radius of convergence $R = ∞$ over $K$, by the above they converge for matrices of arbitrary absolute value, i.e. for all matrices.
(42)(a) Lemma 1: Let $x_n → x$ and $y_n → y$ be convergent sequences of $m × m$ matrices over $K$. Then $x_ny_n → xy$. Proof: We see $x → x_n$ if and only if for any $ε > 0$ we have $||x-x_N|| < ε$ for sufficiently large $N$, since the forward is evident when the entries all converge within range of $ε/m^2$, and the converse forces all entries to converge within $ε$. For sufficiently large $N$ we have - for some asymptotically small matrices $||ε_{(N)1}||,||ε_{(N)2}||$ - the result$$||xy-x_Ny_N|| = ||xy-(x+ε_{(N)1})(y+ε_{(N)2})|| =$$$$||xε_{(N)2}+ε_{(N)1}y+ε_{(N)1}ε_{(N)2}|| ≤ ||x|| \cdot ||ε_{(N)2}|| + ||y|| \cdot ||ε_{(N)1}|| + ||ε_{(N)1}|| \cdot ||ε_{(N)2}||$$which vanishes to zero.$~\square$
Now we see$$PG(At)P^{-1}=(\text{lim }P)(\text{lim }G_N(At))(\text{lim }P)=$$$$\text{lim }PG_N(At)P^{-1} = \text{lim }G_N(PAtP^{-1}) = G(PAtP^{-1}) = G(PAP^{-1}t)$$ (b) By considering lemma 2 of 12.3.38, we see that the power series' summands may be computed in blocks, leading to independent convergences as blocks.
(c) This is simply a special case of (b).
(43) Lemma 2: $\lim_{n → ∞}\dfrac{x^n}{\lfloor n/2 \rfloor !}=0$ for any real $x$. Proof: When $n > x^2$$$\lim_{n → ∞}\dfrac{x^n}{\lfloor n/2 \rfloor !} ≤ \lim_{n → ∞}x\prod_{k=1}^{\lfloor n/2 \rfloor} \dfrac{x^2}{k} = \lim_{n → ∞}x \prod_{k = 0}^{\lfloor x^2 \rfloor}\dfrac{x^2}{k} \prod_{k = \lceil x^2 \rceil}^{\lfloor n/2 \rfloor}\dfrac{x^2}{k} =$$$$x \prod_{k = 0}^{\lfloor x^2 \rfloor}\dfrac{x^2}{k} \lim_{n → ∞} \prod_{k = \lceil x^2 \rceil}^{\lfloor n/2 \rfloor}\dfrac{x^2}{k} = 0~~\square$$ Lemma 3: Let $x_n → x$ and $y_n → y$ be convergent sequences of $m × m$ matrices over $K$ such that $|x_n-y_n| → 0$. Then $x=y$. Proof: Assume $x \neq y$ so $|x-y| > 0$. Let $|x-x_n| < |x-y|/2 - ε$ for $n > n_1$ and some chosen $0 < ε < |x-y|/2$, let $|y-y_n| < |x-y|/2$ for $n > n_2$, let $|x_n-y_n| < ε$ for $n > n_3$, and choose $N > \text{max}(n_1,n_2,n_3)$. We have$$|x-y| = |(x-x_N)+(x_N-y_N)+(y_N-y)| ≤$$$$|x-x_N|+|x_N-y_N|+|y-y_N| < |x-y|$$a contradiction.$~\square$
Utilizing lemma 1, we have $\lim_{N → ∞}\text{exp}_n(A)\lim_{N → ∞}\text{exp}_n(B)=\lim_{N → ∞}\text{exp}_n(A)\text{exp}_n(B)$, so we may compare terms of the two sequences$$\text{exp}_n(A)\text{exp}_n(B)=(\sum_{k=0}^n\dfrac{A^k}{k!})(\sum_{k=0}^n\dfrac{B^k}{k!})=\sum_{j=0}^n\sum_{k=0}^n\dfrac{A^jB^k}{j!~k!}=\sum_{j,k ≤ n} \dfrac{A^jB^k}{j!~k!}$$$$\text{exp}_n(A+B)=\sum_{k=0}^n \dfrac{(A+B)^k}{k!} = \sum_{k=0}^n \dfrac{\sum_{j=0}^k \dfrac{k!}{j!(k-j)!}A^jB^{k-j}}{k!} =$$$$\sum_{k=0}^n \sum_{j=0}^k \dfrac{A^jB^{k-j}}{j!(k-j)!} = \sum_{j+k ≤ n}\dfrac{A^jB^k}{j!~k!}$$ and then compare their differences (without loss of generality assume $||A|| ≤ ||B||$)$$|\text{exp}_n(A+B)-\text{exp}_n(A)\text{exp}_n(B)| = |\sum_{j,k ≤ n < j+k}\dfrac{A^jB^k}{j!~k!}| ≤ \sum_{j,k ≤ n < j+k} \dfrac{||A||^j||B||^k}{j!~k!} ≤$$$$\dfrac{n^2||B||^{2n}}{\lfloor n/2 \rfloor !} = \dfrac{||B||^2 \prod_{k=2}^n\dfrac{k^2||B||^{2k}}{(k-1)^2||B||^{2k-2}}}{\lfloor n/2 \rfloor !} ≤ ||B||^2\dfrac{(4||B||^2)^{n-1}}{\lfloor n/2 \rfloor !}$$When $4||B||^2 < 1$ we clearly have the above vanishing to zero as $n$ approaches infinity, so assume $4||B||^2 ≥ 1$$$||B||^2\dfrac{(4||B||^2)^{n-1}}{\lfloor n/2 \rfloor !} ≤ ||B||^2\dfrac{(4||B||^2)^n}{\lfloor n/2 \rfloor !}$$and by lemma 2, the term still tends toward $0$. Thus by lemma 3 we conclude $\text{exp}(A+B)=\text{exp}(A)\text{exp}(B)$.
(44) This is clear from the fact $\text{exp}(λt) = e^{λt}$, the ring isomorphism between $K$ and matrices $KI$, and the preceding exercise.
(45) The first part is clear from the fact that $N^k$ is the matrix with $1$s along the $k^{th}$ superdiagonal (observable most easily by induction from the linear transformation form of $N$), and since $N^r = 0$, we have $\text{exp}(Nt)=\text{exp}_r(Nt)$ is the matrix described above. The second part is clear from the observation in (44) with $M=Nt$ coupled with the first part.$~\square$
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