If A is an nxn matrix such that rank(A) = 1 and the null space of A is the orthogonal complement of its column space, then how do I show that A is a symmetric matrix? I have proven that the null space of A and A-transpose are equal. How do I proceed now?
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Take an orthogonal basis $\{e_1, e_2, \dots, e_n\}$ of the linear space $V$ such that $\{e_1, e_2, \dots, e_{n-1}\}$ is a basis of $\ker A$. In this basis the matrix of $A$ is the zero matrix except the last column.
As the null space of $A$ is the orthogonal complement of its column space and the dot product is invariant under orthogonal transformations, the null space of $A$ is also the orthogonal complement of its column space in the basis $\{e_1, e_2, \dots, e_n\}$. This implies that the first $n-1$ coefficients of the last column vanish. Finally, $A$ is similar to the always vanishing matrix $B$ except the $(n,n)$ coefficient. We got the desired result as $B$ is symmetric.
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Can you elaborate on how the first n-1 coefficients of the last column of A vanish? I dont get that – Ritik Kumar Nov 10 '20 at 14:08
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The last column of the matrix can be written $v = \alpha_1 e_1 + \dots +\alpha_n e_n$. As the null space, i.e. $span{e_1, \dots, e_{n-1}}$ is the orthogonal complement of $v$, you have $e_i \cdot v=0$ for $1 \le i \le n-1$ and therefore $\alpha_i=0$ as ${e_1, \dots, e_i}$ is supposed to be an orthogonal basis. – mathcounterexamples.net Nov 10 '20 at 14:15