User:Egm6321.f12.team4.harris/Homework5 R5.3

Problem R5.3 – Exponential of a Diagonal Matrix
sec20-3

Statement
Show that
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$$ exp[\mathbf A] = \Phi Diag[e^{\lambda_1}, e^{\lambda_2},...,e^{\lambda_n}] \Phi ^{-1} $$     (5.3.1)
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when


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$$ [ \mathbf \Lambda] = Diag[\lambda_1, \lambda_2, ... \lambda_n] \in \mathbb C^{n \times n} $$ (5.3.2)
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Given
From the solution of R5.2 above, it can be shown that
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$$ exp[D] = Diag [e^{d_1}, e^{d_2}, ..., e^{d_n}] \in \mathbb C^{n \times n} $$ (5.3.3)
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when


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$$ [D]= Diag [d_1, d_2, ... d_n] $$     (5.3.4)
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Solution
To solve this problem, one should be aware of the matrix theory
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$$ (TAT^{-1})^n=TA^nT^{-1} $$     (5.3.5)
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This theory can be shown as follows:
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$$ (TAT^{-1})^n=(TAT^{-1})(TAT^{-1})...(TAT^{-1})=TA(T^{-1}T)A(T^{-1}T)...A(T^{-1}T)AT^{-1} $$
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Since $$ TT^{-1} = 1 $$
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$$ TA(T^{-1}T)A(T^{-1}T)...A(T^{-1}T)AT^{-1}=TAAA...AT^{-1}=TA^nT^{-1} $$
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The exponentiation of $$ \mathbf A $$ can be written as
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$$ exp[\mathbf A]=\sum_{n=0}^{\infty} \frac{1}{k!} A^k $$     (5.3.6)
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Since $$ \mathbf A $$ is diagonalizable, it can be decomposed as $$ [\mathbf A] = [\Phi \mathbf \Lambda \Phi^{-1}] $$
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$$ exp[\mathbf A]=exp[\Phi \mathbf \Lambda \Phi^{-1}]= \sum_{n=0}^{\infty} \frac{1}{k!} [\Phi \mathbf \Lambda \Phi^{-1}]^k $$     (5.3.7)
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From the matrix theory described above in equation 5.3.5, one can see that
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$$ \sum_{n=0}^{\infty} \frac{1}{k!} [\Phi \mathbf \Lambda \Phi^{-1}]^k=\sum_{n=0}^{\infty} \frac{1}{k!} [\Phi \mathbf \Lambda ^k \Phi^{-1}] $$     (5.3.8) Since $$ \Phi $$ and $$ \Phi ^{-1} $$ do not depend on k, they can be pulled outside of the summation as follows
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$$ \Phi \sum_{n=0}^{\infty} \frac{1}{k!} [\mathbf \Lambda ^k ] \Phi^{-1} $$     (5.3.9)
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The summation in equation 5.3.9 can be written in the form shown in equation 5.3.6:
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$$ exp[\mathbf A] = \Phi exp[\mathbf \Lambda] \Phi ^{-1} $$     (5.3.10)
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From equations 5.3.2-5.3.4, one can see that
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$$ exp[\mathbf A] = \Phi Diag[e^{\lambda_1}, e^{\lambda_2}, ..., e^{\lambda_n}] \Phi ^{-1} $$
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Thus equation 5.3.1 is valid.

Author and References

 * Solved by -- Kaitlin Harris and Seong Hyeon Hong
 * Typed by -- Kaitlin Harris