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Powers of matrices equal to the identity matrix 
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#1
Nov1812, 02:36 PM

P: 783

I am curious about under what conditions the powers of a square matrix can equal the identity matrix.
Suppose that A is a square matrix so that [itex] A^{2} = I [/itex] At first I conjectured that A is also an identity matrix, but I found a counterexample to this. I noticed that the counterexample was an elementary matrix. So then I conjectured that A is an elementary matrix. Is this true? Can I prove this? What about for general powers of A? BiP 


#2
Nov1812, 03:54 PM

Engineering
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Thanks
P: 6,957

As a simple example think about 2x2 matrices.
If ##\displaystyle A = \begin{bmatrix}a & b \\ c & d \end{bmatrix}##, then ##\displaystyle A^2 = \begin{bmatrix}a^2 + bc & b(a+d) \\ c(a+d) & d^2 + bc \end{bmatrix} = I##. From the offdiagonal terms, ##b(a+d) = 0## and ##c(a+d) = 0##. Taking ##b = c = 0## isn't going to lead to anywhere interesting, so let's see what happens if ##d = a##. From the diagonal terms, ##a^2 + bc = 1##. You can satisfy that with matrices that are not elementary, for example ##\displaystyle A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}##. In fact the condition ##a^2 + bc = 1## here is the same as ##\det A = 1##, which isn't a complete coincidence  but things are not so simple for bigger matrices. 


#3
Nov1812, 09:06 PM

P: 783

I see. Thanks much [itex]\aleph_0[/itex]
BiP 


#4
Nov1912, 01:26 AM

P: 144

Powers of matrices equal to the identity matrix
A solution to A^{n}=I is obviously attained if A is a suitable diagonal or rotation matrix, and also for all similar matrices PAP^{1}, where P is invertible.



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