# using the spectral decomposition of a matrix

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?

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what do you mean $A$ is similar to its characteristic polynomial by the way? – Vishal Gupta Jun 15 '13 at 8:59

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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I 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 $$2^{A} = 2E_{1}+2^{2}E_{2}$$ – sarah jamal Oct 1 '12 at 10:49

Hint

If $A=VDV^{-1}$ is an eigenvalue decomposition of $A$, then $f(A) = Vf(D)V^{-1}$.

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