1. ## Linear independence problem

If $A\in\mathbb{R}^{4X8}$ then any 6 columns of $A$ are linearly independent.

True or False?

thankyou

2. Originally Posted by Bucephalus
If $A\in\mathbb{R}^{4X8}$ then any 6 columns of $A$ are linearly independent.

True or False?

thankyou
Pick the zero matrix.

3. Also, in general.

$A\in\mathbb{R}^{mXn}$

The column vectors of A are in $\mathbb{R}^{m}$.
If there are more vectors than there are entries in each vector, the vectors cannot be linearly independent.

4. Pick the zero matrix?
What does that mean?
Are you saying that if the matrix is all zeros, then each column is a linear combination of another?

5. He is saying that if you have $A\in \mathbb{R}^{4 \times 8}$ , then you have an arbitrary matrix, with column vectors in $\mathbb{R}^{4}$ .

ANY set of vectors containing the zero vector is linearly dependent, because:
$S = {u_{1} ... u_{p}}$ - A set of vectors, and the zero vector being in S, then:

$\vec{0} = c_{1}\cdot \vec{u_{1}} + ... + c_{j} \cdot \vec{0} + ... + c_{p} \cdot \vec{p}$
With all $c_{i} = 0 \mbox { except } c_{j}$
Hence there is a nontrivial solution for $A\vec{x} = \vec{b}$ .

Now, is the zero vector in the zero matrix ?

6. Thanks for the explanation.
Yes it is.