# Powers of a matrix

• December 8th 2009, 10:54 AM
Volcanicrain
Powers of a matrix
Hi!

If I have the following matrix, call it B, (3x3 with entries separated by commas)
8, 12, 0
0, 8, 12
0, 0, 8

How can I find a matrix A such that A^3 = B?
Since B is not diagonalizable I have no idea how to approach this problem.
• December 8th 2009, 01:18 PM
qmech
Trial and Error gives:

2,1,-0.5
0,2,1
0,0,2

whose cube is the matrix you want.
• December 9th 2009, 07:39 AM
HallsofIvy
Quote:

Originally Posted by Volcanicrain
Hi!

If I have the following matrix, call it B, (3x3 with entries separated by commas)
8, 12, 0
0, 8, 12
0, 0, 8

How can I find a matrix A such that A^3 = B?
Since B is not diagonalizable I have no idea how to approach this problem.

"Brute strength" works. Because B is "upper triangular", you know A will also be "upper triangular" and look for a solution of the form $\begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}$.

Now, it's not too hard to calculate that $A^2= \begin{bmatrix}a^2 & ab+ be & ac+be+ ef \\ 0 & d^2 & de+ef \\ 0 & 0 & f^2\end{bmatrix}$ and then that
$A^3= \begin{bmatrix}a^3 & a^2b+abd+ bd^2 & a^2c+ abe+ acf+ bde+ e^f+ cf^2 \\ 0 & d^3 & de^2+ e^2f \\ 0 & 0 & f^3\end{bmatrix}$
Set that equal to B and you have 6 equations to solve for a, b, c, d, e, and f.
• December 9th 2009, 07:47 AM
Volcanicrain
Thanks!
I had considered that before in my mind, but I wasn't sure if there wasn't an easier way.
But it really isn't so bad!
Thanks a lot!