Linear Algebra question

[Steven G]

Let A be an nxn matrix such that A^17 = I.
Must A=1??

I know that A raised to a smaller power than 17 can't be I but I can't decide if A^17 can =I w/o A =I

 

[Otis]

Must A=1?

Hey Jomo. Did you intend to say A=I, or is A=1 meant to imply that n=1?

I know that A raised to a smaller power than 17 can't be I

I need help understanding what you have in mind because there are many square matrices A such that A^k=I_n with k<17. Here's one example, with n=2.

\(\displaystyle \begin{bmatrix} -9 & -10 \\ \;\;\;8 & \;\;\;\;9 \end{bmatrix}^2 = \text{I}_2\)

 

[Steven G]

Yes, I meant I and not 1

If A^17 = I and A is not I, Then A^k is not I for k=1,2,3,..., 17

 

[Steven G]

I need help understanding what you have in mind because there are many square matrices A such that A^k=I_n with k<17. Here's one example, with n=2.

\(\displaystyle \begin{bmatrix} -9 & -10 \\ \;\;\;8 & \;\;\;\;9 \end{bmatrix}^2 = \text{I}_2\)

The problem with your example is when you raise your matrix to the 17th power you do not get I

 

[Steven G]

In my opinion, I clearly defined A-- it is a nxn matrix such that A^17 = I.

 

[Otis]

If A^17 = I and A is not I, Then A^k is not I for k=1,2,3,..., 17

Hi. Thanks for repeating it, but I'm thinking that you meant to say k=1,2,3,...,16.

I know that A raised to a smaller power than 17 can't be I.

How did you learn that lesser powers of such matrices A cannot be the nth identity matrix? I suppose I'm wondering: why 17?

 

[Otis]

The problem with your example is when you raise your matrix to the 17th power you do not get I

Yes, I'd mistakenly switched to arbitrary square matrices for A, when reading the second paragraph. My goof.

 

[Cubist]

Let A be an nxn matrix such that A^17 = I.
Must A=1??

I know that A raised to a smaller power than 17 can't be I but I can't decide if A^17 can =I w/o A =I

Here's where thinking visually rather than algebraically can really help. Try thinking about rotations ?

 

[Steven G]

How did you learn that lesser powers of such matrices A cannot be the nth identity matrix? I suppose I'm wondering: why 17?

17 is a prime.

For example, if A^6 = I, then I = A^17 = A^6*A^6*A^5. So A^5 = I
I = A^17 = A^5*A^5*A^5*A^2 = A^2
I = A^17 = (A^2)^8*A = A.
So A=I

 

[Dr.Peterson]

17 is a prime.

For example, if A^6 = I, then I = A^17 = A^6*A^6*A^5. So A^5 = I
I = A^17 = A^5*A^5*A^5*A^2 = A^2
I = A^17 = (A^2)^8*A = A.
So A=I

Why are you assuming A^6 = I? And where does 17 being a prime come into this?

I think you must be omitting some contextual restriction, as it should be obvious that it is possible for A^17 to be I without being I itself. An example would be a rotation in 2- or 3-space by 1/17 of a turn.

Again, can you show the source of your claim here?

I know that A raised to a smaller power than 17 can't be I

Who says this, and under what conditions?

The best interpretation I can give to what you are saying is that if A^17 = I and also some lower power of A = I, then A = I. That may well be true. But that's not what you said in the OP.

 

[Otis]

if A^6 = I, then I = A^17

I clearly defined A-- it is a nxn matrix such that A^17 = I.

I know that A raised to a smaller power than 17 can't be I

There seems to be a contradiction between the first statement (A^6=I) and the last statement.

?