if I have $$ A= \begin{bmatrix} \hphantom{-}1 & 0 & 0\\ -1 & 1 & 1 \\ -1 & 0 & 2\\ \end{bmatrix} $$ I am given the characteristic values are $$ \lambda_{1}=1 \text{ and } \lambda_{2}=2 \text{ and A is similar to a characteristic polynomial} $$ I have found the spectral decomposition of A. I now need to use this to find $$2^{A}$$ how do I do this?
using the spectral decomposition of a matrix
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matrices
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1what do you mean $A$ is similar to its characteristic polynomial by the way? – 2013-06-15
3 Answers
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If $A = SDS^{-1}$ where $D$ is the diagonal matrix containing the eigenvalues of $A$, say,
$$D = \left(\begin{matrix} \lambda_{1} & 0 & 0 \\ 0 & \lambda_{2} & 0 \\ 0 & 0 & \lambda_{3} \\ \end{matrix}\right) $$
Then
$$ 2^{A} = S\left(\begin{matrix} 2^{\lambda_{1}} & 0 & 0 \\ 0 & 2^{\lambda_{2}} & 0 \\ 0 & 0 & 2^{\lambda_{3}} \\ \end{matrix}\right)S^{-1} $$
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Multiply out the decomposition, substituting $\lambda_i \to 2^{\lambda_i}$.
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0I worked out the spectral decomposition to be $$A=E_{1} + 2E_{2} $$ now if I am going to subsitute using the function $$2^{A}$$ do I say $$ 2^{A} = 2E_{1}+2^{2}E_{2}$$ – 2012-10-01
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Hint
If $A=VDV^{-1}$ is an eigenvalue decomposition of $A$, then $f(A) = Vf(D)V^{-1}$.