Please explain 1.all positive integers powers of a symmetric matrix is a symmetric matrix 2.odd positive integers power of skew symmetric matrix is a skew symmetric matrix and even positive integers powers of skew symmetric matrix is a symmetric matrix.
Positive Integral powers of symmetric matrix is a symmetric matrix?
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linear-algebra
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0It would be better for you if you could show your approach. – 2017-02-14
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0Use the identity $(AB)^t=B^tA^t$. – 2017-02-14
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0@BAYMAX But in this case it is justified because for the proof we can use similar technique..what you have presented.. – 2017-02-14
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0@Widawensen ok, my apologies. – 2017-02-15
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0@BAYMAX Interesting why this question was "put on hold" ? (and heavily down-voted) Maybe it's easy but what else is wrong with it ? - just a question .. – 2017-02-15
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0@Widawensen Yes i too agree,but can we do anything about it,i am little unaware of the technicalities regarding MSE.Also this may demotivate the user from asking questions freely! – 2017-02-15
1 Answers
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Suppose $A$ is the square matrix of order say $m$,then Observe that $(A^{n})^{t} = (A^{t})^{n}$ ,
so for your first case for Symmetric matrices $A^{t} = A$ , so $A^{n} = A.A.A.....A$ , $n$ times $= A^{t}.A^{t}.A^{t}....A^{t}$ , $n$ times $=(A^{t})^{n} = (A^{n})^{t}$.so $A^{n}$ is symmetric for any positive integer $n$.
For the second case also since $A^{t} = -A$,so $(A^{t})^{n} = -A.-A.-A....-A$ $n$ times,$=(-1)^{n} . (A)^{n}$ so if $n$ is odd it will be skew-symmetric and for $n$ even it will be symmetric.Hope this helps!