If A is 2*2 skew symmetric matrix with $$A^2=A$$, Then A=0 $$$$ My attempt was by assuming the matrix A and substituting the squaring condition and the condition of skew symmetric and got system of equations but I couldn't solve it.I look forward to your comments.
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For skew symmetric matrices $I\pm A$ is invertible. – zwim Oct 27 '21 at 18:17
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2This is not true. Take the identity matrix over $\Bbb F_2$. Then $A^2=A$, and $A$ is skew-symmetric but nonzero. – Dietrich Burde Oct 27 '21 at 18:17
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As one of the Answers posted quickly shows, the steps you outlined should have led directly to solving the problem (for a real skew-symmetric matrix). Merely stating "but I couldn't solve it" is not the kind of context for the problem that is needed for good Questions at Math.SE. – hardmath Oct 27 '21 at 18:39
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Dietrich Burde....The identity matrix is not skew-symmetric – AK Math Oct 27 '21 at 18:40
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@AlKhuzai: In characteristic two the identity matrix is skew-symmetric. You failed to state that your matrix entries are drawn from the real numbers (or another domain that is not characteristic two). – hardmath Oct 27 '21 at 18:42
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$$A=A^2=(-A)^2=(A^T)^2=A^TA^T=(AA)^T=A^T=-A$$
This is valid in all cases where $\mathrm{char}(F)\ne2$. It can be easily prooved by defenition, and does not matter on size of $A$.
MH.Lee
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Let$$A=\begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix}$$
Then $A^2=A$ implies
$$\begin{bmatrix} -a^2 & 0 \\ 0 & -a^2 \end{bmatrix}=\begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix}$$
Hence $a=0$ and $A=O$
Lalit Tolani
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