Suppose $A$ is an $n \times n$ matrix, and write $O(A)$ for $\{\vec x : A \vec x = \vec 0 \}$. Prove or disprove that if $O(A) = O(A^T)$ then $A = A^T$.
Any help would be appreciated
Matrices, transposes and ranks
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linear-algebra
matrices
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1I have changed the math formatting to use set notation, among other small changes. Please review to check that I did not unintentionally change your meaning. – 2017-01-19
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1The set $O(A)$ is often called the *nullspace* of $A$. – 2017-01-19
2 Answers
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Hint: For example: if $A$ is invertible, then $O(A) = O(A^T) = \{0\}$. Find an invertible matrix for which $A \neq A^T$.
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$$A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ Has only the zero solution to $Ax=0$, yet $A^T\neq A$.