$\mathbf{U} = sp\{(a_1 a_2 a_3),(b_1 b_2 b_3),(c_1 c_2 c_3)\}$
$\mathbf{W} = sp\{(a_1 b_1 c_1),(a_2 b_2 c_2),(a_3 b_3 c_3)\}$
$\mathbf{U}$ and $\mathbf{W}$ are subspaces of $\mathbb{R}^3$

  • $\begingroup$ Hint The row-rank and the column-rank of any matrix are equal. $\endgroup$ – Yoni Rozenshein May 23 '13 at 9:49

$\def\rk{\mathop{\mathrm{rank}}}$We have, as $\rk A = \rk A^t$ holds for every matrix, that \begin{align*} \dim U &= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{pmatrix}\\ &= \rk \begin{pmatrix} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 & c_3 \end{pmatrix}^t\\ &= \rk \begin{pmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3 \end{pmatrix}\\ &= \dim W. \end{align*}

  • $\begingroup$ Thanks for the quick answer, but in the course im taking we havent defined rank as of yet, I read up on it and I understand that it does hold as proof but assuming you could not use rank, what other methods could you use to create this proof? $\endgroup$ – Bar Wachtel May 23 '13 at 11:23

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