# Find a matrix B such that $B^5 = A$ [duplicate]

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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.

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

• So what did you find? For the eigenvalues +vectors? – Mason Dec 5 '18 at 21:47
• I edited your post to more properly $\LaTeX$ify it. Cheers! – Robert Lewis Dec 5 '18 at 22:18
• @Mason i got λ=4,λ=-2 for eigenvals. Corresponding to $$A = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ and $$A = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ – Ray Fitzgerald Dec 13 '18 at 18:52

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.
• 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. – Math Lover Dec 5 '18 at 21:39
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 $$(a \circ b)_{ij}=a_i b_j$$.