19.3 Matrix exponential[2D8]
[1MF]We define the exponential of matrices as
where we agree that \(A^ 0={\mathbb {I}}\), the identity matrix.
- E408
[1MG] Prerequisites:Section . 12.4,2, 1, 4, 2.
We equip the space of the matrices \(ℂ^{n× n}\) with one of the norms seen in Section 12.5.
Show that the series \(∑_{k=0}^∞{A^ k}/{k!}\) converges.
Show that
\begin{equation} \exp (A)=\lim _{N→∞}\Big({\mathbb {I}}+A/N\Big)^ N \label{eq:exp(A)=lim(I+A/N)N} \end{equation}409where \({\mathbb {I}}\) is the identity matrix in \(ℝ^{n× n}\); and that convergence is uniform in every compact neighborhood of \(A\). (Hint: make good use of the similar result 2.)
Hidden solution: [UNACCESSIBLE UUID ’1MH’]
- E408
[1MJ] If \(A\) is invertible then
\[ A\exp (B)A^{-1}=\exp (A BA ^{-1})\quad . \]- E408
[1MK] The derivative of
\[ t∈ℝ↦ \exp (tA) \]is \(A\exp (tA)\). Hidden solution: [UNACCESSIBLE UUID ’1MM’]
- E408
[1MN] If \(A,B\) commute, then
\[ A\exp (B)=\exp (B)A~ ~ ~ ,~ ~ ~ \exp (A+B)=\exp (A)\exp (B)~ . \]In particular \(\exp (A)\) is always invertible and its inverse is \(\exp (-A)\).
Hidden solution: [UNACCESSIBLE UUID ’1MP’]
- E408
[1MQ] Let
\[ A= \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\quad ,\quad B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\quad :\quad \]compute
\[ \exp (A) \exp (B)\quad ,\quad \exp (B) \exp (A)\quad ,\quad \exp (A+B)\quad ; \]You will get that they are all different from each other. Hidden solution: [UNACCESSIBLE UUID ’1MR’]
- E408
[1MS]If \(A,B\) then the directional derivative of \(\exp \) at the point \(A\) in the direction \(B\) is \(B\exp (A)\), i.e.
\[ \frac{d\hskip5.5pt}{d{t}} \exp (A+tB)|_{t=0}=B\exp (A)~ ~ . \]- E408
[1MT]Difficulty:*.Show that
\[ \det (\exp (A))=\exp ({\operatorname {tr}}(A))\quad . \]Hint: use Jacobi’s formula 14 to calculate the derivative of \(\det (\exp (tA))\). Use the previous result 3 — see also 3. Another proof can be obtained by switching to Jordan’s form (using 2).
Hidden solution: [UNACCESSIBLE UUID ’1MV’]
- E408
[1MW] Difficulty:*.In the general case (when we do not know if \(A,B\) commute) we proceed as follows. Let’s define \([A,B]=AB-BA\).
Setting \(B_ 0=B\) and \(B_{n+1}=[A,B_ n]\) you have
\begin{eqnarray*} B_ n& =& A^ nB - n A^{n-1}BA + \frac{n(n-1)} 2 A^{n-2}BA^ 2 + \cdots + (-1)^ n\, BA^ n =\\ & =& ∑_{k=0}^ n (-1)^ k\binom {n}{k} A^{n-k} B A^{k}~ ~ ; \end{eqnarray*}let’s define now \(Z=Z(A,B)\)
\begin{equation} Z{\stackrel{.}{=}}∑_{n=0}^∞\frac{B_ n}{n!}~ ~ ,\label{eq:Z(A,B)} \end{equation}410(note that \(Z\) is linear in \(B\)): prove that the above series converges, and that
\begin{equation} \exp (A)B\exp (-A)=Z~ ~ ;\label{eq:exp_ A_ B_-A_ Z} \end{equation}411from this finally it is shown that
\[ \exp (A)\exp (B)\exp (-A)=\exp (Z)~ ~ . \]
(These formulas can be seen as consequences of the Baker–Campbell–Hausdorff formula [ 48 ] ). Hidden solution: [UNACCESSIBLE UUID ’1MX’]
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[1MY]Prerequisites:1.In general (even when \(A,B\) do not commute)
\[ \exp (A+B)=\lim _{N\to \infty }\Big(\exp (A/N) \exp (B/N) \Big)^ N \]Hidden solution: [UNACCESSIBLE UUID ’1MZ’]
- E408
[1N0]Prerequisites:2.Difficulty:**.In the general case (even when we do not know if \(A,B\) commute), we can express \(\exp (A+sB)\) using a power series. Define
\[ C(t)=\exp (-tA)B \exp (tA) \]and (recursively) set \(Q_ 0={\mathbb {I}}\) (the identity matrix) and then
\[ Q_{n+1}(t)=∫_ 0^ t C(𝜏) Q_ n(𝜏)\, {\mathbb {d}}𝜏 \]then
\begin{equation} \exp (-A)\exp (A+sB)=∑_{n=0}^∞ s^ n Q_ n(1)~ ~ ;\label{eq:exp_ A+sB} \end{equation}412this series converges for every \(s\).
In particular, the directional derivative of \(\exp \) at the point \(A\) in the direction \(B\) is
\[ \frac{d\hskip5.5pt}{d{s}} \exp (A+sB)|_{s=0}=\exp (A) Q_ 1(1) = ∫_ 0^ 1 \exp ((1-𝜏) A)B\exp (𝜏 A) \, {\mathbb {d}}𝜏~ ~ . \]( Hint: Use the exercise 2 with \(Y(t,s) = \exp (-tA)\exp (t(A+sB))\) and then set \(t=1\). )
Hidden solution: [UNACCESSIBLE UUID ’1N1’]
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[1N2]Prerequisites:8.Difficulty:*.
Prove the relations
\[ \frac{d\hskip5.5pt}{d{t}} \exp (A+tB)|_{t=0}=\int _ 0^ 1 \exp (s A)B\exp ((1-s) A) \, {\mathbb {d}}s~ ~ . \]