Basis Question

**TheMute** · May 29th 2010, 01:25 PM

Let Z be the set of all vectors fo the form $\begin{bmatrix} 2a - 3b - e \\ 0 \\ a + c +2d \\ -3b + c + d + e \end{bmatrix}$ , where $a, b, c, d, e \ \epsilon \ R$ . Z is a subspace of $R^4$ .

a) Find a set that spans Z.

b) Is the set from part (a) a basis for Z? Please explain.

c) Does Z have any two dimensional subspaces? If so give an example of such a subspace. If not, please explain why not.

Attempted Answers:
For part (a) I got: $\{(2, 0, 1, 0), (-3, 0, 0, -3), (0, 0, 1, 1) \} $ .This is the column space of W. Correct?

Part (b) seems like Yes but I don't know the reason. Part (c) I have no clue.

**davismj** · May 29th 2010, 05:26 PM

Originally Posted by TheMute

Let Z be the set of all vectors fo the form $\begin{bmatrix} 2a - 3b - e \\ 0 \\ a + c +2d \\ -3b + c + d + e \end{bmatrix}$ , where $a, b, c, d, e \ \epsilon \ R$ . Z is a subspace of $R^4$ .

a) Find a set that spans Z.

b) Is the set from part (a) a basis for Z? Please explain.

c) Does Z have any two dimensional subspaces? If so give an example of such a subspace. If not, please explain why not.

Attempted Answers:
For part (a) I got: $\{(2, 0, 1, 0), (-3, 0, 0, -3), (0, 0, 1, 1) \} $ .This is the column space of W. Correct?

Part (b) seems like Yes but I don't know the reason. Part (c) I have no clue.

For part (a), I would recognize that this vector is any combination of the vectors (2,0,1,0),(-3,0,0,-3),(0,0,1,1),(0,0,2,1),(-1,0,0,1). This is the easiest way to determine a set that spans Z. To be sure that your (a) is correct, you must recognize that the vectors (0,0,2,1) and (-1,0,0,1) are linear combinations of the other three, which I think you do.

For part (b), using the above set, we would need to verify that the five vectors are linearly independent (and it is fairly easy to see upon inspection that they are not).

Notice that

$\frac{1}{2}(2,0,1,0) + \frac{1}{3}(-3,0,0,-3) + \frac{3}{2}(0,0,1,1) = (0,0,2,1)$

and

$\frac{-2}{3}(2,0,1,0) - \frac{1}{9}(-3,0,0,-3) + \frac{2}{3}(0,0,1,1) = (-1,0,0,1)$

Therefore, since these vectors are linear combinations of the other three, they are linearly dependent. Removing the two vectors (-1,0,0,1) and (0,0,2,1) yields a list of three linearly independent vectors, {(2,0,1,0),(0,0,1,1),(-3,0,0,-3)} which span Z. Thus, the vectors form a basis.

**davismj** · May 29th 2010, 05:45 PM

For part (c), recall that if a subset $V \subset Z$ satisfies

(i) $0 \in V$
(ii) $x,y \in V \implies x + y \in V$
(iii) $x \in V, \lambda \in R \implies \lambda x \in V$

then V is a subspace.

Using any two of the bases above, we have that if

$V = \{ x \in R^{4} : x = \lambda_1(2,0,1,0) + \lambda_2(-3,0,0,-3), \lambda_1,\lambda_2 \in R\}$

Then $V \subset Z$ and

(i) $0 = 0(2,0,1,0) + 0(-3,0,0,-3) \in V$
(ii) Suppose

$\alpha_1,\alpha_2,\beta_1,\beta_2 \in R$
$x = \alpha_1(2,0,1,0) + \alpha_2(-3,0,0,-3) \in V$
$y = \beta_1(2,0,1,0) + \beta_2(-3,0,0,-3) \in V$

Then $x + y = [\alpha_1 + \beta_1](2,0,1,0) + [\alpha_2 + \beta_2](-3,0,0,-3) \in V$

(iii) $x = \alpha_1(0,2,1,0) + \alpha_2(-3,0,0,-3), \lambda x = \lambda\alpha_1(2,0,1,0) + \lambda\alpha_2(-3,0,0,-3) \in V$

Thus, V is a subspace. Since V is spanned by two linearly independent vectors, dim V = 2.

**dwsmith** · May 29th 2010, 05:53 PM

Originally Posted by davismj

For part (a), I would recognize that this vector is any combination of the vectors (2,0,1,-3),(-3,0,0,0),(0,0,1,1),(0,0,2,1),(-1,0,0,1). [/tex]

$(-3,0,0,0)$ should be $(-3,0,0,-3)$

**davismj** · May 29th 2010, 06:01 PM

Originally Posted by dwsmith

$(-3,0,0,0)$ should be $(-3,0,0,-3)$

Ugghhhh. Thanks for the heads up!

**TheMute** · May 29th 2010, 09:05 PM

So for part (c), would $V = \{(2, 0, 1, 0), (-3, 0, 0, -3) \}$ be a valid asnwer or is the whole $ $ $V = \{ x \in R^{4} : x = \lambda_1(2,0,1,0) + \lambda_2(-3,0,0,-3), \lambda_1,\lambda_2 \in R\} $ the answer? or is that just a minor technicality...

And are any 2 of the 3 vectors in $\{(2, 0, 1, 0), (-3, 0, 0, -3), (0, 0, 1, 1) \}$ a valid answer as long as I prove the (i), (ii), and (iii) you mentioned?

**davismj** · May 29th 2010, 09:19 PM

Originally Posted by TheMute

So for part (c), would $V = \{(2, 0, 1, 0), (-3, 0, 0, -3) \}$ be a valid asnwer or is the whole $ $ $V = \{ x \in R^{4} : x = \lambda_1(2,0,1,0) + \lambda_2(-3,0,0,-3), \lambda_1,\lambda_2 \in R\} $ the answer? or is that just a minor technicality...

And are any 2 of the 3 vectors in $\{(2, 0, 1, 0), (-3, 0, 0, -3), (0, 0, 1, 1) \}$ a valid answer as long as I prove the (i), (ii), and (iii) you mentioned?

1) It's pretty important that you write it the way I did.

V={(2,0,1,0),(-3,0,0,-3)} is a set of only two vectors.

Typically you write a set as

V={elements : conditions}

Where elements describes the type of elements in the set and conditions describes the conditions under which those elements are in the set. Mine would read "the set of all vectors x in R4 such that x =lambda1(2,0,1,0) + lambda2(-3,0,0,-3), lambda 1 and lambda 2 real."

2) Any two linearly independent vectors that span a set contained entirely in Z will work, by the definition.

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