# Math Help - Basis Vectors

1. ## Basis Vectors

Do:

$\left[\begin{array}{ccc}
1\\
1\\
0
\end{array}\right]
$
, $\left[\begin{array}{ccc}
1\\
0\\
1
\end{array}\right]
$
, and $\left[\begin{array}{ccc}
0\\
1\\
1
\end{array}\right]
$
form a basis for $\mathbb{R}^3$?

I just need some help on how to work these problems out. Thanks in advance for the help.

2. Originally Posted by Aryth
Do:

$\left[\begin{array}{ccc}
1\\
1\\
0
\end{array}\right]
$
, $\left[\begin{array}{ccc}
1\\
0\\
1
\end{array}\right]
$
, and $\left[\begin{array}{ccc}
0\\
1\\
1
\end{array}\right]
$
form a basis for $\mathbb{R}^3$?

I just need some help on how to work these problems out. Thanks in advance for the help.
Calculate the determinant of the $3 \times 3$ matrix formed by the three vectors.

$\left(\begin{array}{ccc}1&1&0\\1&0&1\\0&1&1\end{ar ray}\right)$

If the determinant of the matrix = 0, the vectors are linearly dependent and do not form a basis for $R^3$. The determinant of the matrix is -2. So,.....

3. Originally Posted by Aryth
Do:

$\left[\begin{array}{ccc}
1\\
1\\
0
\end{array}\right]
$
, $\left[\begin{array}{ccc}
1\\
0\\
1
\end{array}\right]
$
, and $\left[\begin{array}{ccc}
0\\
1\\
1
\end{array}\right]
$
form a basis for $\mathbb{R}^3$?

I just need some help on how to work these problems out. Thanks in advance for the help.
Maybe the following will serve as useful additional input.

Rather than compute a third-order determinant (which of course is hardly overwhelming, but still a bit expensive timewise), in just two
simple reduction steps (which you should be able to carry-out merely by eye) you'll see there are non-zero pivots along the main diagonal.
Hopefully, this will tell you the matrix with those three vectors as columns is non-singular,
which in turn should tell you that those three vectors are lineraly independent.

Of Course, all this presupposes some degree of understanding of notions underlying elementary linear algebra.

Essentially, the problem boils down to that of examining the solution space for Ax = 0.
If you find that the only x is x = 0, then you will have a satisfied definition of linear independence sitting right in front of you.
If there happen to be other (so-called non-trivial) solutions as well, then what might you conclude?