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Finding the Basis?

  1. Dec 15, 2007 #1
    Find a basis for and calculate the dimension of nullA:

    A = [1 2 7]T, [1 1 2]T, [-2 0 6]T, [0 1 -10]T, [4 -5 -7]T

    Like most algebra texts mine has pages and pages of proofs with hardly a single example tying it together.

    Here is what I think I know:

    If the determinant does not equal 0 than the vectors span R^n and are linearly independent. Therefore, they can makeup a basis for A.

    In this case since it is not a square matrix I cannot use determinant to find the solution so I opted to find whether it not it spans R^n such as:

    X1 + 2X2 + 7X3 = 0
    X1 + X2 + 2X3 = 0
    -2X1 + 6X3 = 0
    X2 - 10X3 = 0
    4X1 -5X2 -7X3 = 0

    The above has a non-trivial solution so the vectors are linearly dependent and do not form a basis.

    Now I could add a remove vectors until I find a set of vectors with only the trivial solution but it is a pretty time intensive process, especially for a large matrix.

    Is there a faster way?
  2. jcsd
  3. Dec 15, 2007 #2


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    so the matrix is
    [tex]\left[\begin{array}{ccccc}1 & 1 & -2 & 0 & 4 \\ 2 & 1 & 0 & 1 & 5 \\7 & 2 & 6 & -10\end{array}\right][/tex]?

    Yeah, it as if they expected us to think!

    Are you sure of that? Subtracting the second equation from the first, we get X2+ 5X3= 0. Subtracting the fourth equation from that, 15X3= 0 so X3 must be 0. Then the third equation gives X1= 0 and the fourth X2= 0. There is NO non-trivial solution to those equations so the null space consists only of the 0 vector.

  4. Dec 15, 2007 #3
    "Yeah, it as if they expected us to think!"

    By "think" I can only assume you are referring to mechanical thinking. Believe me, I know exactly why the book is written the way it is and why it is such a flawed approach. The author, like most mechanical thinkers, over assumes and is incapable of realizing that most people simply do not interpret information as they do.

    Most people learn by example, not by reading owners manuals.

    Btw, I misunderstood the question at first. I need to first find the null, then determine the basis of the null which is [-2 4 1 0 0]T [7 -11 0 2 1]T with a dimension of 2.
  5. Dec 15, 2007 #4


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    Are you serious? You complain about there not being any examples to follow and then talk about "mechanical thinking". Obviously if I relied on "mechanical thinking" I would rely heavily on examples that I could follow mechanically. If, instead, I think creatively, applying the concepts in the book in a variety of different ways, I would not need examples.

    Now please explain to me why you think the system of equation you gave had a non-trivial solution. Also please explain why you think the null space of A is two dimensional.
  6. Dec 16, 2007 #5
    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.

    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.

    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.
    Last edited: Dec 16, 2007
  7. Dec 16, 2007 #6


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    That is not at all what I would consider "mechanical", but I won't argue about your choice of words.

    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.

    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 R5, not R3 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 R5 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.
    Last edited by a moderator: Dec 17, 2007
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