Find a matrix B such that $B^5 = A$

Question

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.

Answer 1

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.

Answer 2

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 $(a \circ b)_{ij}=a_i b_j$.