If $\displaystyle A\in\mathbb{R}^{4X8}$ then any 6 columns of $\displaystyle A$ are linearly independent.
True or False?
thankyou
He is saying that if you have $\displaystyle A\in \mathbb{R}^{4 \times 8} $ , then you have an arbitrary matrix, with column vectors in $\displaystyle \mathbb{R}^{4} $ .
ANY set of vectors containing the zero vector is linearly dependent, because:
$\displaystyle S = {u_{1} ... u_{p}} $ - A set of vectors, and the zero vector being in S, then:
$\displaystyle \vec{0} = c_{1}\cdot \vec{u_{1}} + ... + c_{j} \cdot \vec{0} + ... + c_{p} \cdot \vec{p} $
With all $\displaystyle c_{i} = 0 \mbox { except } c_{j} $
Hence there is a nontrivial solution for $\displaystyle A\vec{x} = \vec{b} $ .
Now, is the zero vector in the zero matrix ?