I have a question:
In the vector space $\displaystyle A\in\mathbb{R}^{3}$, the set $\displaystyle S = \{(2,3,2)^T,(1,1,-1)^T\}$ is not a basis for the space $\displaystyle W=span\{(1,2,3)^T,(5,8,7)^T,(3,4,1)^T\}$.
True or false?
hi
$\displaystyle \left[\begin{matrix} 2 & 1 & 1 & 5 & 3 \\ 3 & 1 & 2 & 8 & 4 \\ 2 & -1 & 3 & 7 & 1 \end{matrix}\right] \mbox { is row equivalent to } \left[\begin{matrix} 1 & 0 & 1 & 3 & 1 \\ 0 & 1 & -1 & -1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right] $
Which gives that every vector in W, is a linear combination of the vectors in S. This implies that every linear combination of the vectors in W is a linear combination of the vectors in S, since the spanning vectors in W are themselves linear combinations of vectors in S! =)
Hi Bucephalus,
to show that the statement is false, which seems to be the case, all you have to do is show that span in W are linearly dependent of span in S.
What that means is:
(soz about the formatting...dunno how to write in column or write T of transpose)
a (2, 3, 2) + b (1, 1, -1) = (1, 2, 3) or (5, 8, 7) or (3, 4, 1)
it is rather obvious,
when a = 1 and b = -1 we get (1, 2, 3)
when a = 3, b = -1 we get (5, 8, 7)
when a = 1, b = 1 we get (3, 4, 1)
we just showed span in W are linearly dependent of that in S. Hence the statement is false.
now, if it is not instantly obvious to you, then you can set up like Twig showed you above. It might take a little longer but it is an excellent 100% working method.
In Twig's matrices, (i'm talking about second matrices that is in echelon form) as you can see the third row basically has 0's. Which means that there is no solution or the fact that there is linear dependence.
You can also look at column 1, 2, and 3...and notice the fact that column 3 is basically 1 (column 1) - 1 (column 2)...this is pretty much the dependence relationship that we found above.
hope that helps.
P.S. On contrary, if span in W were linearly independent of that in S, the statement would have been true.
This set of five vectors could never been linearly independent, yeah?
Because there are five columns and only 3 rows...??
So does that mean I should have been able to just look at the question and see that those in W can' t be linearly independent of those in S?
What if I were to put the vectors from W as the first three columns in Twig's matrix, followed by the vectors in S? Would I be able to come to the same conclusion?
$\displaystyle \left[\begin{matrix} 1 & 5 & 3 & 2 & 1\\ 2 & 8 & 4 & 3 & 1\\
3 & 7 & 1 & 2 & -1
\end{matrix}\right]
$
I will reduce that matrix now and see what I get.
Well I got
$\displaystyle
\left[\begin{matrix} 1 & 0 & -2 & -1/2 & -3/2\\ 0 & 1 & 1 & 1/2 & 1/2\\
0 & 0 & 0 & 0 & 0
\end{matrix}\right]
$
So really, I can take any two of these five vectors as the basis of the other three as long as the two I select are linearly independent of each other....
So in this case because I want to test that S is the basis, I would put these two vectors at the front of the matrix to be reduced? Is that right?
hi
Yes, since the question was if the vectors in S can be a basis for W, you want to test if all the linear combinations of the vectors in W can be written as a linear combination of the vectors in S. And the vectors in S are linearly independent, so they can form a basis for W.
exactly.
you can take first two rows (basis from S) and show that their linear combination results in last three rows (basis from W). In other words, last three rows are linearly dependent (or not linearly independent) of first two rows.
btw, the echelon form in this case is friendly. Because your last row is all 0, you can directly conclude that linearly independent is false.
Although, havenīt we drifted from th original question here?
Wasnīt the question just if the vectors in S form a basis for W?
For this itīs sufficient to see that every vector in W is a linear combination of the vectors in S.
And since the vectors in S do span all of W, and the two vectors in S are lin. independent -> They form a basis for W.
Hence the statement is false.