ND3G said:
I think you are missing the point of what mechanical thinking means in this context. By mechanical thinking the author assumes that the reader can understand the theorems as they stand with a level of clarity that they can then turn around and use them creatively on real problems.
That is not at all what I would consider "mechanical", but I won't argue about your choice of words.
What actually occurs is that the reader has to break down the theorem into smaller and smaller components until they understand each component on its own then piece them back together to discover what the theorem as a whole entails.
Now, maybe that sounds fantastic to you. Think of just how well that person is going to understand that theorem when they are done. Fine, but while that may have taken the person an hour or a couple of hours one or two practical examples probably could have offered the same insight in a fraction of the time.
No, not necessairily the same insight- working it out yourself, while, yes, longer and more difficult, produces much better understanding than seeing someone else work it out. That, I think, is the crux of our diagreement.
One approach is the mechanical approach which would have been to either just offer the theorems or just offer a bunch of examples. Both are extremely flawed. It would be like giving you all the parts to an automobile, an owners manual and asking you what it looks like assembled or giving you a picture of a car and asking you how it operates.
Systems thinking would have been a far better solution. Show me the car, show me the parts and show me how the parts come together to make it operate. At least then I have a general idea of how a car works instead of a complex mess of parts which I may or may not figure out over time or a device which operates for reasons unknown.
There is a reason why some people find math difficult while others find it pretty easy and it has nothing to do with the intelligence of each individual. It has everything to do with how they interpret information.
There are a few authors of math texts which that understand the distinction but sadly they are very much in the minority and rarely do their texts make it to the classroom where it could do the most good.
Back to the question...
X1 + 2X2 + 7X3 = 0
X1 + X2 + 2X3 = 0
-2X1 + 6X3 = 0
X2 - 10X3 = 0
4X1 -5X2 -7X3 = 0
First I reduced it to:
X1 + 2X3 -7X5 = 0
X2 -4X3 + 11X5 = 0
X4 + -2X5 = 0
X1 = -2s + 7t
X2 = 4s - 11t
X3 = s
X4 = 2t
X5 = t
X = [-2s 4s s 0 0]T + [7t -11t 0 2 1]T
The Basis being [-2 4 1 0 0]T and [7 -11 0 2 1]T.
The basis has two vectors, and a dimension of 2.
It just occurs to me that I have been misunderstanding what you said. Your matrix operator has, as I put in my first response, 5
columns and 3
rows. That means the null space is a subspace of R
5, not R
3 as I had thought. However, that is not entirely my fault! I was mislead by your writing the equations incorrectly. Applying the operator to a vector in R
5 and setting equal to 0 gives:
X1+ X2- 2X3+ 4X5= 0, 2X1+ X2+ X4- 5X5= 0, and 7X1+ 2X2+ 6X3- 10X4- 7X5= 0.
If we subtract the first equation from the second, we eliminate X2 and get X1+ 2X3+ X4- 9X5= 0. If we Subtract twice the first equation from the third, we get 5X1+10X3- 10X4- 15X5= 0. Subtract 5 times the first of those two equations from the second to eliminate X1 (which also happens to eliminate X3): - 20X4+ 30X5= 0 so X4= (3/2)X5. That means we can choose X5 and X3 to be anything we like- yes, the null space has dimension 2. Taking X5= 2 (for simplicity), X4= 3, then X1+ 2X3+ 3- 18= 0 so X1= 15- 2X3. If we also take X3= 0, X1= 15 and the the first equation becomes 15+ X2+ 8= 0, X2= -23. One basis vector for the nullspace is [15, -23, 0, 3, 2]T. If we take X3= 1 rather than 0, we have still that X5= 2 and X4 but now X1+ 1+ 3- 18= 0 so X1= 14. Then 14+ X2- 2+ 8= 0 so X2= -20. A second basis vector for the nullspace is [14. -20, 1, 3, 2]T. Those are not the same basis vectors as you give but a subspace may have many different bases.
