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Find all matrices $A$ such that $A = 2A^T$ and $A-A^T = I$.

I have been stuck on this question for quite some time:

For the first part, all I can think of are matrices with all entries being $0.$

For the second part, I don't understand how that is possible because regardless of the matrix, the diagonal entries of the matrix should be the same as its transpose matrix, meaning each entry of the difference of the two matrices should be 0. However, the diagonal entries of the identity matrix all have a value of 1. Am I misunderstanding the question or made a mistake midway?

amWhy
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Intaceraptor
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    I think, the first equation is only satisfied by the zero-matrix. Looking first at some diagonal element, and then at some non-diagonal element, it should be straightford to prove it. – Peter Feb 13 '20 at 14:52
  • The diagonal elements of $A-A^T$ must be $0$, hence the second equation cannot be satisfied. – Peter Feb 13 '20 at 14:54
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    The first depends on the characteristic of the field, to be precise whether the characteristic is $3$ (special case, all skew symmetric matrices are solutions) or not (only the zero matrix). – ancient mathematician Feb 13 '20 at 14:55

3 Answers3

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If $A = 2 A^T$ then taking transposes we have $A = {1 \over2 } A^T$ and so ${3 \over 2} A^T = 0$ from which we get $A^T = 0$

We must have $\operatorname{tr} (A-A^T) = 0$, so assuming that you are dealing with the reals, there are no solutions.

copper.hat
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Easier: If $A-2A^T=0$ and $A-A^T=I$, then $A^T=A-A^T=I$, and thus $A=I$. But clearly $I-I^T=0\neq I$, whatever the base ring is (unless the base ring is trivial), so there is no solution.

GreginGre
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Let $A=(a_{ij})$.

For the first part, we have $a_{ij}=2a_{ji}$ for all $i,j$.

When $i=j$, $a_{ii}=2a_{ii}$ and therefore $a_{ii}=0$.

When $i\ne j$, $a_{ij}=2a_{ji}=4a_{ij}$ and therefore $a_{ij}=0$ unless the field has characteristic $3$.

For the second part, we have $a_{ii}-a_{ii}=1$ for all $i$ which is impossible.