Matrix Math Proof help!

[erndoglai]

Allright so i've been stuck on this problem for a while and i'm not exactly sure how to go about this

let A =
| .5 .5 0|
| .25 .25 .5 |
| .25 .25 .5 |

show that A^n =
(1/3) |1 1 1| + (1/[3* 4^n]) | 2 2 -4|
|1 1 1| |-1 -1 2|
|1 1 1| |-1 -1 2|
for n greater than or equal to 1.

I'm not 100% sure of the properties of matrices and I tried A^n as |.5^n .5^n 0| (raising all the elements inside of A ^n) and then trying to work it out... i've also tried expanding the bottom portion and combining both matrices.. so any help would be great.

 

[mmm4444bot]

… I tried A^n as … raising [each of] the elements inside … A [to the power of] n …

Hi erndoglai:

You're not multiplying the matrices properly. Determining A^n is not as simple as raising each element to the nth power.

For example, when matrix A is multiplied by itself once, we get A^2, which is the following.

\(\displaystyle \left[ \begin{array}{ccc} 3/8 & 3/8 & 1/4 & 5/16 & 5/16 & 3/8 & 5/16 & 5/16 & 3/8 \end{array} \right]\)

This multiplication involves nine repeats of: "multiply elements in column by corresponding elements in row, and add the three resulting products".

Look up matrix multiplication, in the index of your textbook, if you need to refresh your memory. Or, try searching at Google on keywords: matrix multiplication to find sites like THIS ONE.

Also, did you notice the jumbled display, in your post? If you want to "draw" matrices by manually spacing the alignments, then you'll need to use BBCode. (You can click the

button to see the coding for the display below.) Use the [Preview] button to see what your post will look like, before clicking [Submit].

                 |1 1 1|                     | 2 2 -4|
         (1/3) * |1 1 1|   +   (1/[3*4^n]) * |-1 -1 2|
                 |1 1 1|                     |-1 -1 2|

Cheers,

~ Mark

MY EDITS: Added [Preview] button info, and deleted suggestion to practice multiplying matrices by hand

 

[DrMike]

If n is an integer, you can define

\(\displaystyle A^1=A\),

\(\displaystyle A^n=A^{n-1}A\).

To prove that the formula you've been given is correct, you could use induction : Prove

* that the formula works when n=1 (or n=0 or whatever)
* that assuming it works for n-1, then it also works for n.

Another approach : if you have learned about eigenvalues and eigenvectors (or when you eventually do), you can diagonalise the matrix - that is, find P and D such that \(\displaystyle A=PDP^{-1}\) and D is a diagnoal matrix. Then \(\displaystyle A^n=PD^nP^{-1}\). You'll have found P and D, it's easy to raise a diagonal matrix to any power, so you multiply this out and you have your formula for A[sup:259ryefv]n[/sup:259ryefv]

 

[erndoglai]

THANKS DR MIKE that A^k = PD^KP-1 was really helpful i totally forgot that... i knew i wrote it down on my notes i just forgot about it but thanks so much =]