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I'm having trouble understanding the question's answer from #19. How does the combination of a vector v and its negative fill a half space? Doesn't it only fill a line?

John
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  • It seems to me that it would fill a whole line, but yes. Where is this from? – Calle Aug 08 '15 at 15:23
  • The whole line, I think – Zhanxiong Aug 08 '15 at 15:24
  • I think, but am not sure, that the region in question is not the linear span of $v$ and $w$ but rather it is the region obtained by sweeping the half line $tv;;t≥0$ towards the half line $tw;;t≥0$. that gets the first quadrant in the example given, and it gets a half space if $w=-v$. – lulu Aug 08 '15 at 15:28
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    @lulu, that seems a bit far-fetched to me... And why would it be the upper half space and not the lower half space? – Calle Aug 08 '15 at 15:31
  • You're right, it's the whole line. But how that can be called a "half-space"? It's from the solutions to Introduction to Linear Algebra by Strang. – John Aug 08 '15 at 15:31
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    If $v$ and $w$ are linearly dependent (e.g, if $w = -v$), then the statements made in both 18 and 19 are false. You are quite right that the apparent answer to the question is wrong. – Rob Arthan Aug 08 '15 at 15:34
  • @Calle I agree it is a stretch...just trying to come up with something to justify the answer. You get the "upper" half space if you specify the direction of the rotation (clockwise, counterclockwise). If it really means the linear span then it is clear you just get the line. – lulu Aug 08 '15 at 15:37
  • @RobArthan The actual question in the book illustrates that v and w are linearly independent. My mistake for not including it. For 19 for w to equal -v they must be linearly dependent though so it still seems wrong. – John Aug 08 '15 at 15:38
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    You can see where the conclusion is tempting even if false as stated.. If $v=(1,0)$ and $w=(-1,\epsilon)$ for some small $\epsilon >0$ then you get the upper half space less a thin wedge in the second quadrant. Letting $\epsilon\rightarrow0$ sure looks like you get the upper half space. But, of course, if you look at the strict linear combinations of $v$ and $w$ you just get the $x$-axis when $\epsilon = 0$ – lulu Aug 08 '15 at 15:44
  • Perhaps you could edit your question to include some more relevant context. (And by the way if $v$ is non-zero and $w = -v$, then the points $cv+dw$ for non-negative $c$ and $d$ comprise the whole line spanned by $v$, not just a half-line.) – Rob Arthan Aug 08 '15 at 15:49
  • Interesting. It seems as though the author was imagining the wedge opening into the half space, like @lulu described, and missed this corner case. – Colm Bhandal Aug 08 '15 at 16:14

1 Answers1

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I asked Gilbert Strang and he wrote back:

Thank you !!

My cone is certainly wrong and I am happy that you sent the correction.

Best wishes

Gilbert Strang

John
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  • I just have to say, this is quite wholesome. I really started to love linear algebra after Professor Strang's lectures, and it's heartwarming that he still responds to emails even with all of his mathematical fame. – Julian L Mar 13 '20 at 06:40