# Matrices in power diagonal question

1. Nov 2, 2011

### nhrock3

find B matrices so $$B^{3}=A=\left(\begin{array}{cc}14 & 13\\13 & 14\end{array}\right)$$

,the diagonal form of A is $$D=\left(\begin{array}{cc}a & 0\\0 & b\end{array}\right)$$

i got weird numbers so for convinience the eigenvalues are a,b

so there is U for which

$$U^{-1}AU=\left(\begin{array}{cc}a^{\frac{1}{3}} & 0\\0 & b^{\frac{1}{3}}\end{array}\right)^{3}$$

i can find U
because A is simetric so U is consists of the orthonormal eigenvectors of A
what to do next,how to find B

?

2. Nov 2, 2011

### Staff: Mentor

You have U-1AU = D, where D is a diagonal matrix whose entries are the eigenvalues of A.

It turns out that U-1A1/3U is a cube root of D (or D1/3), where D1/3 is also a diagonal matrix whose entries are the cube roots of the entries of D. In short, U-1A1/3U = D1/3.

To verify that U-1A1/3U is a cube root of D,
notice that (U-1A1/3U)3
= (U-1A1/3U)(U-1A1/3U)(U-1A1/3U)
= U-1A1/3UU-1A1/3UU-1A1/3U
= U-1A1/3A1/3A1/3U
= U-1AU = D.

In other words, if we cube U-1A1/3U, we get D, so U-1A1/3U is a cube root of D.

To get B, which is A1/3, multiply both sides of the equation U-1A1/3U = D1/3 on the left by U, and multiply both sides on the right by U-1.

3. Nov 2, 2011

### Dick

Why do you say the eigenvalues of A are 'weird'? They don't look weird to me. In fact they are both integer cubes. What did you get? I think you can do this whole exercise with easy numbers.

Last edited: Nov 2, 2011
4. Nov 3, 2011

### nhrock3

$$\left(\begin{array}{cc}14 & 13\\13 & 14\end{array}\right)$$

the caracterstic polinomial

$$P(t)=(t-14)^{2}-169=t^{2}-28t+27$$

t=1 t=27

for t=1 i get (-1,1) and for t=27 i got the same (-1,1)

so the transformation matrices is not invertible which is wrong

where is my mistake?

5. Nov 3, 2011

### Dick

Your mistake is in what you didn't show. (-1,1) can't have an eigenvalue of both 1 and 27. Which is it?