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This question already has an answer here:

I am being asked to find a matrix $B$ where $B^5 = A$

$$A = \begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}$$

In the first part of the question I was asked to find the eigenvalues & eigenvectors for the matrix which I found successfully. If someone could help me finish this then that would be great.

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marked as duplicate by Math Lover, amd, Jean-Claude Arbaut, Chinnapparaj R, KReiser Dec 6 '18 at 3:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ So what did you find? For the eigenvalues +vectors? $\endgroup$ – Mason Dec 5 '18 at 21:47
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    $\begingroup$ I edited your post to more properly $\LaTeX$ify it. Cheers! $\endgroup$ – Robert Lewis Dec 5 '18 at 22:18
  • $\begingroup$ @Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ and $$A = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ $\endgroup$ – Ray Fitzgerald Dec 13 '18 at 18:52
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HINT: If you have found the eigenvalues and eigenvectors, then you should easily be able to diagonalize this matrix as $A=PDP^{-1}$, where $D$ is diagonal. Then use the fact that $A^n=PD^n P^{-1}$, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries.

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    $\begingroup$ Maybe, use $D$ instead of $B$ to avoid the confusion as the OP is asking to find $B$ such that $B^5 = A$. Also, this problem could be considered as a duplicate, I think. $\endgroup$ – Math Lover Dec 5 '18 at 21:39
  • $\begingroup$ @MathLover Oh, sorry. I will change it to avoid confusion. $\endgroup$ – Frpzzd Dec 5 '18 at 21:41
  • $\begingroup$ @Frpzzd any chance you could expand on the latter part of that suggestion. I know how to do matrix diagonalisation but what do you mean by "Then use the fact that An=PDnP−1, and the fact that the powers of a diagonal matrix are the matrices consisting of the powers of its entries." Thanks $\endgroup$ – Ray Fitzgerald Dec 13 '18 at 18:56
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We can also use projector decomposition. The left and right eigenvectors are equal, because of the symmetry. So the projector decomposition is $$A=\lambda_1 \frac{v_1 \circ v_1}{v_1 \cdot v_1}+\lambda_2 \frac{v_2 \circ v_2}{v_2 \cdot v_2}$$ And $$f(A)=f(\lambda_1) \frac{v_1 \circ v_1}{v_1 \cdot v_1}+f(\lambda_2) \frac{v_2 \circ v_2}{v_2 \cdot v_2}$$ Where $(a \circ b)_{ij}=a_i b_j$.

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