--- title: "Ackermann's formula" chunk: 2/3 source: "https://en.wikipedia.org/wiki/Ackermann's_formula" category: "reference" tags: "science, encyclopedia" date_saved: "2026-05-05T13:34:54.532302+00:00" instance: "kb-cron" --- ( A C L ) 0 = ( A − B k T ) 0 = I ( A C L ) 1 = ( A − B k T ) 1 = A − B k T ( A C L ) 2 = ( A − B k T ) 2 = A 2 − A B k T − B k T A + ( B k T ) 2 = A 2 − A B k T − ( B k T ) [ A − B k T ] = A 2 − A B k T − B k T A C L ⋮ ( A C L ) n = ( A − B k T ) n = A n − A n − 1 B k T − A n − 2 B k T A C L − … − B k T A C L n − 1 {\displaystyle {\begin{aligned}(\mathbf {A} _{\rm {CL}})^{0}=&\ (\mathbf {A} -\mathbf {Bk} ^{\rm {T}})^{0}=\mathbf {I} \\[4pt](\mathbf {A} _{\rm {CL}})^{1}=&\ (\mathbf {A} -\mathbf {Bk} ^{\rm {T}})^{1}=\mathbf {A} -\mathbf {Bk} ^{\rm {T}}\\[4pt](\mathbf {A} _{\rm {CL}})^{2}=&\ (\mathbf {A} -\mathbf {Bk} ^{\rm {T}})^{2}\\[2pt]&=\mathbf {A} ^{2}-\mathbf {ABk} ^{\rm {T}}-\mathbf {Bk} ^{\rm {T}}\mathbf {A} +(\mathbf {Bk} ^{\rm {T}})^{2}\\[2pt]&=\mathbf {A} ^{2}-\mathbf {ABk} ^{\rm {T}}-(\mathbf {Bk} ^{\rm {T}})[\mathbf {A} -\mathbf {Bk} ^{\rm {T}}]\\[2pt]&=\mathbf {A} ^{2}-\mathbf {ABk} ^{\rm {T}}-\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}\\[4pt]\vdots \ &\\[4pt](\mathbf {A} _{\rm {CL}})^{n}=&\ (\mathbf {A} -\mathbf {Bk} ^{\rm {T}})^{n}\\[2pt]&=\mathbf {A} ^{n}-\mathbf {A} ^{n-1}\mathbf {Bk} ^{\rm {T}}-\mathbf {A} ^{n-2}\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}-\ldots -\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}^{n-1}\end{aligned}}} Replacing the previous equations into Δ(ACL) yields Δ ( A C L ) = ( A n − A n − 1 B k T − A n − 2 B k T A C L − … − B k T A C L n − 1 ) ⏞ ( A C L ) n + … + α 2 ( A 2 − A B k T − B k T A C L ) + α 1 ( A − B k T ) + α 0 I ⏞ ∑ k = 0 n − 1 α k A C L k = ( A n + α n − 1 A n − 1 + … + α 2 A 2 + α 1 A + α 0 I ) − ( A n − 1 B k T + A n − 2 B k T A C L + … + B k T A C L n − 1 ) + … − α 2 ( A B k T + B k T A C L ) − α 1 ( B k T ) = Δ ( A ) − ( A n − 1 B k T + A n − 2 B k T A C L + … + B k T A C L n − 1 ) − … − α 2 ( A B k T + B k T A C L ) − α 1 ( B k T ) {\displaystyle {\begin{aligned}\Delta (\mathbf {A} _{\rm {CL}})&=\overbrace {(\mathbf {A} ^{n}-\mathbf {A} ^{n-1}\mathbf {Bk} ^{\rm {T}}-\mathbf {A} ^{n-2}\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}-\ldots -\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}^{n-1})} ^{(\mathbf {A} _{\rm {CL}})^{n}}+\overbrace {\ldots +\alpha _{2}(\mathbf {A} ^{2}-\mathbf {ABk} ^{\rm {T}}-\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}})+\alpha _{1}(\mathbf {A} -\mathbf {Bk} ^{\rm {T}})+\alpha _{0}\mathbf {I} } ^{\sum _{k=0}^{n-1}\alpha _{k}\mathbf {A} _{\rm {CL}}^{k}}\\[4pt]&=(\mathbf {A} ^{n}+\alpha _{n-1}\mathbf {A} ^{n-1}+\ldots +\alpha _{2}\mathbf {A} ^{2}+\alpha _{1}\mathbf {A} +\alpha _{0}\mathbf {I} )-(\mathbf {A} ^{n-1}\mathbf {Bk} ^{\rm {T}}+\mathbf {A} ^{n-2}\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}+\ldots +\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}^{n-1})+\ldots -\alpha _{2}(\mathbf {ABk} ^{\rm {T}}+\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}})-\alpha _{1}(\mathbf {Bk} ^{\rm {T}})\\[4pt]&=\Delta (\mathbf {A} )-(\mathbf {A} ^{n-1}\mathbf {Bk} ^{\rm {T}}+\mathbf {A} ^{n-2}\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}+\ldots +\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}}^{n-1})-\ldots -\alpha _{2}(\mathbf {ABk} ^{\rm {T}}+\mathbf {Bk} ^{\rm {T}}\mathbf {A} _{\rm {CL}})-\alpha _{1}(\mathbf {Bk} ^{\rm {T}})\end{aligned}}} Rewriting the above equation as a matrix product and omitting terms that kT does not appear isolated yields Δ ( A C L ) = Δ ( A ) − [ B A B ⋯ A n − 1 B ] [ ⋆ ⋮ k T ] {\displaystyle \Delta (\mathbf {A} _{\rm {CL}})=\Delta (\mathbf {A} )-{\begin{bmatrix}\mathbf {B} &\mathbf {AB} &\cdots &\mathbf {A} ^{n-1}\mathbf {B} \end{bmatrix}}{\begin{bmatrix}\star \\\vdots \\\mathbf {k} ^{\rm {T}}\end{bmatrix}}} From the Cayley–Hamilton theorem, Δ(ACL) = 0, thus [ B A B ⋯ A n − 1 B ] [ ⋆ ⋮ k T ] = Δ ( A ) {\displaystyle {\begin{bmatrix}\mathbf {B} &\mathbf {AB} &\cdots &\mathbf {A} ^{n-1}\mathbf {B} \end{bmatrix}}{\begin{bmatrix}\star \\\vdots \\\mathbf {k} ^{\rm {T}}\end{bmatrix}}=\Delta (\mathbf {A} )} Note that C = [ B A B ⋯ A n − 1 B ] {\displaystyle {\mathcal {C}}={\begin{bmatrix}\mathbf {B} &\mathbf {AB} &\cdots &\mathbf {A} ^{n-1}\mathbf {B} \end{bmatrix}}} is the controllability matrix of the system. Since the system is controllable, C {\displaystyle {\mathcal {C}}} is invertible. Thus, [ ⋆ ⋮ k T ] = C − 1 Δ ( A ) {\displaystyle {\begin{bmatrix}\star \\\vdots \\\mathbf {k} ^{\rm {T}}\end{bmatrix}}={\mathcal {C}}^{-1}\Delta (\mathbf {A} )} Both sides can then be multiplied by the vector [ 0 ⋯ 0 1 ] {\displaystyle {\begin{bmatrix}0&\cdots &0&1\end{bmatrix}}} giving [ 0 ⋯ 0 1 ] [ ⋆ ⋮ k T ] = [ 0 ⋯ 0 1 ] C − 1 Δ ( A ) {\displaystyle {\begin{bmatrix}0&\cdots &0&1\end{bmatrix}}{\begin{bmatrix}\star \\\vdots \\\mathbf {k} ^{\rm {T}}\end{bmatrix}}={\begin{bmatrix}0&\cdots &0&1\end{bmatrix}}\,{\mathcal {C}}^{-1}\Delta (\mathbf {A} )} and thus k T {\displaystyle \mathbf {k} ^{\rm {T}}} has necessarily the following form: k T = [ 0 ⋯ 0 1 ] C − 1 Δ ( A ) {\displaystyle \mathbf {k} ^{\rm {T}}={\begin{bmatrix}0&\cdots &0&1\end{bmatrix}}\,{\mathcal {C}}^{-1}\Delta (\mathbf {A} )} Since the system is assumed to be controllable, this necessary condition is also sufficient. == Example == Consider