What is the dimension of the subspace formed by two vectors in R3?

[flyingpig]

Homework Statement

http://img6.imageshack.us/img6/5017/69430037.th.png

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The Attempt at a Solution

a) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP10519f5ffahf530d2ei00005867aii1c038ibgc?MSPStoreType=image/gif&s=29&w=178&h=56

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP10819f5ffahf530d2ei0000437856bcgaif2657?MSPStoreType=image/gif&s=29&w=73&h=56

So the vectors that form the basis are <1,-2,7> and <4,1,4>

The dimension of the subspace is 2 right? Because the two vectors <1,-2,7> and <4,1,4> form a plane in R³? Is it a coincidence that this is the same as finding a basis for the column space?

b) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP125919f5fc52791ag4c6000014fi7d54aa6g5d94?MSPStoreType=image/gif&s=31&w=167&h=56

[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP299519f5f8h0d007afe50000540gfh59c2d627f5?MSPStoreType=image/gif&s=40&w=62&h=56

So there is a pivot in every row/column, then <1,1,1> is indeed in W

 

[Mark44]

Homework Statement

http://img6.imageshack.us/img6/5017/69430037.th.png

Uploaded with ImageShack.us

The Attempt at a Solution

a) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP10519f5ffahf530d2ei00005867aii1c038ibgc?MSPStoreType=image/gif&s=29&w=178&h=56

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP10819f5ffahf530d2ei0000437856bcgaif2657?MSPStoreType=image/gif&s=29&w=73&h=56

So the vectors that form the basis are <1,-2,7> and <4,1,4>

Sure, but I would pick the first column and the third column, which is what the solution to your reduced matrix is telling you - namely, that the 2nd column equals the 1st + the 3rd.

The dimension of the subspace is 2 right?

Of course. You have two vectors in your basis, so the dimension of that subspace is 2.

Because the two vectors <1,-2,7> and <4,1,4> form a plane in R³? Is it a coincidence that this is the same as finding a basis for the column space?

No.

b) [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP125919f5fc52791ag4c6000014fi7d54aa6g5d94?MSPStoreType=image/gif&s=31&w=167&h=56

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP10819f5ffahf530d2ei0000437856bcgaif2657?MSPStoreType=image/gif&s=29&w=73&h=56

So there is a pivot in every row/column, then <1,1,1> is indeed in W

That's not what I get. I think you have an error in your row reduction (or might have typed in an incorrect value, since you seem to have done it with some computer software).

What you're doing is finding a solution of the equation a<1, -2, 7> + b<4, 1, 4> = <1, 1, 1>.
To do that, form the augmented matrix below, and row reduce it.
\left[ \begin{array} {c c c c} 1&amp;4&amp;|&amp;1\\-2&amp;1&amp;|&amp;1\\7&amp;4&amp;|&amp;1\end{array}\right]

Click the matrix above to see what I did.

I end up with no solution, as it's telling me that b = 1/3 and b = 1/4, which can't happen. Thus <1, 1, 1> is not in the span of the other two vectors.

 

[flyingpig]

Sure, but I would pick the first column and the third column, which is what the solution to your reduced matrix is telling you - namely, that the 2nd column equals the 1st + the 3rd.

Oh...if I overlook the fact some columns can be a linear combination of the other ones, is there a way to recover? Like I've mistaken the column space = basis of subspace right now. So the pivots doesn't tell you a thing about the basis of the subspace (W is the subspace of R³, it got cutt off in the question)

 

[flyingpig]

Also that last matrix should have been

[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP299519f5f8h0d007afe50000540gfh59c2d627f5?MSPStoreType=image/gif&s=40&w=62&h=56

Copied the wrong matrix

 

[Mark44]

Also that last matrix should have been

[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP299519f5f8h0d007afe50000540gfh59c2d627f5?MSPStoreType=image/gif&s=40&w=62&h=56

Copied the wrong matrix

Is this what you get after row reduction?
\left[ \begin{array} {c c c c} 1&amp;0&amp;|&amp;0\\0&amp;1&amp;|&amp;0\\0&amp;0&amp;|&amp;1\end{array}\right]

If you understand that this is an augmented matrix, then there is clearly no solution, as the last line is saying that 0*a + 0*b = 1, which is impossible.

 

[flyingpig]

OH right yes I forgot lol, it is suppose to be augmented, but I forgot. So yes the vector <1,1,1> is not in W. Can you correct my first answer as well?

 

[Mark44]

OH right yes I forgot lol, it is suppose to be augmented, but I forgot. So yes the vector <1,1,1> is not in W. Can you correct my first answer as well?

I'm not sure what you're asking for. Any two of the three column vectors would serve as a basis for the span of these three vectors. You picked the first and second, and I would pick the first and third. Either answer is correct.

 

[flyingpig]

I'm not sure what you're asking for. Any two of the three column vectors would serve as a basis for the span of these three vectors. You picked the first and second, and I would pick the first and third. Either answer is correct.

No, now I believing that I am wrong because my second column tells me that the second column vector is a linear combination of the first and third, so it is redundant and doesn't make the set linearly independent, so I must throw it away

 

[Mark44]

That's basically what I said in post #2. Since the 2nd vector is the sum of the 1st and 3rd vectors, it can be omitted. You omitted the 3rd vector. Either set of two vectors is a basis for the set spanned by the three vectors.

My choice is slightly better than yours because it came directly from the solution of the row-reduced matrix. Other than that, your two vectors and my two vectors all lie in exactly the same plane in R³.

 

[flyingpig]

Yes, but the third vector can't be a linear combination of my second and my first, so my answer is incorrect. ALso mine also came from the rref(A)

 

[Mark44]

Yes, but the third vector can't be a linear combination of my second and my first, so my answer is incorrect.

Yes it can, and no it isn't. (Apologies for using "it" to mean two different things.)
-1<1, -2, 7> + 1<4, 1, 4> = <3, 3, -3>.
This shows that the 3rd vector is a linear combination of the first two.

ALso mine also came from the rref(A)

 

[flyingpig]

So the pivots don't tell really tell you the basis of the subspace? Or is this a special case where it doesn't?

 

[Mark44]

Speaking for myself, I don't think about them in terms of pivots. Instead, I focus on what system of equations the matrix represents, and whether I'm dealing with an augmented matrix (for a system Ax = b, with b != 0). When I get the matrix reduced, I think about what the reduced matrix means in terms of the system of equations it represents.

If you want some constructive criticism, you seem to be doing a lot of this stuff on auto pilot, with minimal understanding why you're doing these operations. The matrices are a means to an end, not an end in themselves. You seem to be missing that, at least that's how it seems to me.