# Math Help - Basis

1. ## Basis

Consider the set:
S=
in the vector space V=

Why isn't the set S a basis for the vector space V? (check all that apply)

A. The vectors are linearly dependent
B. The span of the vectors is not V
C. None of the above, it is a basis for V

Any help is appreciated. I'm still trying to understand these concepts.

2. Originally Posted by cdlegendary
Consider the set:
S=
in the vector space V=

Why isn't the set S a basis for the vector space V? (check all that apply)

A. The vectors are linearly dependent
B. The span of the vectors is not V
C. None of the above, it is a basis for V

Any help is appreciated. I'm still trying to understand these concepts.
If they are linearly dependent, then the equation
$a\begin{bmatrix}1 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}+ b\begin{bmatrix}1 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}+ c\begin{bmatrix}0 & 0 & 0 \\ 1 & 0 & 1\end{bmatrix}+ d\begin{bmatrix}0 & 0 & 0 \\ 1 & 1 & 0\end{bmatrix}+ e\begin{bmatrix}0 & 0 & 0 \\ 1 & 1 & 1\end{bmatrix}$ $= \begin{bmatrix}a+ b & b & a \\ c+ d+ e & d+ e & c+ e\end{bmatrix}$ $= \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$
will have a solution other than a= b= c= d= e= 0
.
Obviously $M_{23}$ has dimension 6 and there are only 5 matrices here.

3. Originally Posted by cdlegendary
Consider the set:
S=
in the vector space V=

Why isn't the set S a basis for the vector space V? (check all that apply)

A. The vectors are linearly dependent
B. The span of the vectors is not V
C. None of the above, it is a basis for V

Any help is appreciated. I'm still trying to understand these concepts.
Only one of the above applies, and that is B.

The vectors are linearly independent, and this can be seen by realising that there are two sets of vectors which are clearly linearly independent, those with entries in the top row, and those with entries in the bottom row. Both of these sets are linearly independent (can you see why?)

Now, the vectors do not span the vector space as, for example, the vector $\left(\begin{array}{ccc}
1 & 1 & 1\\
0 & 0 & 0 \end{array}\right)$
is not in its span. Again, can you see why this is?

I would also point out that a vector space which looks like $\mathbb{Z}_n$ has dimension n. Clearly, this looks like $\mathbb{Z}_6$ and so will have dimension 6. There are only 5 vectors here, so they cannot span your space.