A problem on matrices : Powers of a matrix

Question

If $ A= \begin{bmatrix} i & 0 \\ 0 & i \\ \end{bmatrix} , n \in \mathbb N$, then $A^{4n}$ equals?

I guessed the answer as $ A^{4n}= \begin{bmatrix} i^{4n} & 0 \\ 0 & i^{4n} \\ \end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$ which actually was the answer.

Can you provide me with the correct method of getting the answer?
Thank you.

What you did is correct, just make the observation that $i$ raised to an even power is -. I am not sure if they wanted you to prove this, use induction. — Amzoti, May 24 '13 at 12:26
There is nothing wrong with your method, as long as you realize it works only for diagonal matrices; if $A=\pmatrix{x&0\cr0&y\cr}$, then $A^m=\pmatrix{x^m&0\cr0&y^m\cr}$. — Gerry Myerson, May 24 '13 at 12:26

score 5 · Accepted Answer · edited May 24 '13 at 12:32

5

you can see $A^2=\begin {pmatrix}i^2&0\\0&i^2 \end{pmatrix}$

you assume that it satisfy fo n-1 so :

$A^{n-1}=\begin {pmatrix}i^{n-1}&0\\0&i^{n-1} \end{pmatrix}$

then :

$A^n=A*A^{n-1}=\begin {pmatrix}i^n&0\\0&i^n \end{pmatrix}$

edited May 24 '13 at 12:32

Amzoti

56,093

answered May 24 '13 at 12:31

Somaye

2,730

So should i take n=4n? – chndn May 24 '13 at 12:39
this satisfy for any n and so for 4n too – Somaye May 24 '13 at 12:40
okay i understood – chndn May 24 '13 at 12:41

score 3 · Answer 2 · answered May 24 '13 at 12:32

Well you're in luck because what you did actually work with diagonal matrices. Let's look at an example :

Let $A = \begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix}$ be a diagonal matrix then $ A^{2}= \begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix} = \begin{bmatrix} a^2& 0 \\ 0 & b^2 \\ \end{bmatrix}$ and more generally :

$$A^n = \begin{bmatrix} a^n & 0 \\ 0 & b^n \\ \end{bmatrix}$$

So you guess wasn't wrong at all, it's just that when you have a diagonal matrix, raising it to a power $n$ equals to raising each of it's coefficients to the power $n$.

But keep in mind that this only works for diagonal matrices. If you have a matrix $B$ that is not diagonal, the best way to find $B^n$ is to diagonalize it (if possible), raise it's coefficients to the power $n$ and go back to your first basis.

your answer is equally correct thank you +1 – chndn May 24 '13 at 12:42 — chndn, May 24 '13 at 12:42

score 2 · Answer 3 · answered May 24 '13 at 13:35

2

Since $A=iI$ and scalars commute with matrices you get

$$A^{4n}=(i)^{4n}I^{4n}=I \,.$$

answered May 24 '13 at 13:35

N. S.

132,525

score 0 · Answer 4 · answered May 24 '13 at 13:32

0

Isn't it enough to show $A^4=I$ ?

answered May 24 '13 at 13:32

Sugata Adhya

3,979

A problem on matrices : Powers of a matrix

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