1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Determinant, subspace, linear transformation

  1. Apr 11, 2007 #1
    I am having some trouble with the following linear algebra problems, can someone please help me?

    1) Explain what can be said about det A (determinant of A) if:
    A^2 + I = 0, A is n x n


    My attempt:
    A^2 = -I
    (det A)^2 = (-1)^n
    If n is be even, then det A = 1 or -1
    But what happens when n is ODD, what is the detmerinant of A, then?


    2) If V is a subspace of R^n and W is a subspace of V, then W^| is a subspace of V^|. True or False? Justify. [Note: | means orthogonal complement]
    [By dimension theorem, dim dim V + dim V^| = n, but is it true that dim W + dim W^| = n also?
    If V is a subspace of R^n and W is a subspace of V, does this IMPLY that W is also a subspace of R^n? WHY or why not?
    I understand that if A is a subset of B and B is a subset of C, then A is a subset of C, but subspace and subset are not the same thing, so I am not too sure...]


    3) Let T: R^3->R^3 be a linear transformation.
    ker (T) = span {[1 1 1]^T, [1 -2 1]^T}, and T([3 2 1]^T)=[10 10 10]^T. Find the matrix of T.

    I have no idea how to do this one...


    Thanks for your help!:smile:
     
  2. jcsd
  3. Apr 11, 2007 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You have your formula: if n is odd then (det A)2= -1. What is det(A)? Now, are you working over the real numbers or the complex numbers? If over the real numbers, so that every determinant is a real number, what does that tell you about the number of 3 by 3 (or 5 by 5, etc.) matrices A such that A2+ I= 0?


    You have to be careful to distinguish between the orthogonal complement of W as a subspace of Rn and its orthogonal complement as a subspace of V. I interpret "W is a subspace of V" as meaning that W| is defined as all the vectors in V orthogonal to all vectors in W. Since you are now looking at V as your "base" vector space, dim W+ dim W|= dim V.

    Yes, it does. The distinction between "subset" and "subspace" is that the subspace must be closed under vector addition and scalar multiplication. If a set is a subpace of V it must be a subset of V and closed under those operations. Because it is a subset of V it is a subset of Rn and you already know it is closed under the operations. However, whether W is a subspace of Rn is irrelevant. You are asked to whether W| is a subspace of V|. Is it a subset of V|? Is it closed under the operations?


    The standard way to find the matrix of a linear transformation (in a given basis; here I assume that is the usual basis, [1 0 0], [0 1 0], [0 0 1]) is to apply the linear transformation to each basis vector in turn, writing the result in the basis. The coefficients form the columns of the matrix. What do you get if you apply this linear transformation to [1 0 0], [0 1 0], and [0 0 1]?
    (Hint: write each of those as a linear combination of [1 1 1], [1 -2 1], and [3 2 1] and use the information you are given. For example [1 0 0]= a[1 1 1]+ b[1 -2 1]+ c[3 2 1]. Find a, b, c. Actually all you need is c because the other two vectors are mapped to 0. The first column of your matrix will be [10c 10c 10c]) (All vectors above should have a ^T, of course.)


     
  4. Apr 11, 2007 #3
    Thanks!

    2) dim V + dim V^| = n (since V is a subspace of R^n)
    dim W + dim W^| = n (since W is a subspace of R^n) <--is this correct?

    W is a subspace of V => dim W<dim V
    Take the possibility that dim W<dim V
    => dim W|>dimV|
    => W| is not a subspace of V|

    Is this a correct way to do it?
     
  5. Apr 11, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    ?? In my previous post, I specifically said "no- that is not correct".

    No, it is not. dimension of both W and W| are less than or equal to dim V.
     
  6. Apr 11, 2007 #5
    But you said "Yes, it does. The distinction between "subset" and "subspace"..."
    Sorry if I am understanding incorrectly...
     
  7. Apr 13, 2007 #6
    2) If A is a subspace of B and B is a subspace of C, then is A is a subspace of C? Does anyone know the definite answer to this one?
     
  8. Apr 14, 2007 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, of course. I told you that in my first post.

    I said "No" when you were referring to W being a subspace of V and V a subspace of Rn- and I had already made the point that exactly what W| is depends upon whether you are thinking of W as a supspace of V or Rn.


    Take R3 and let V be the xy-plane: {<x, y, 0>}, let W be the x-axis: {<x, 0, 0>}. V has dimension 2 and W has dimension 1. W is, of course, a subspace of V and therefore a subspace of R3.

    If we are thinking of W as a subspace of V then W| is the y-axis: {<0, y, 0>} and has dimension 1: dim(W)+ dim(W|)= 1+ 1= 2= dim(V). Here W| is a subspace of V and so a subspace of R3

    But if we are thinking of W as a subspace of R3 then W| is the yz-plane: {<0, y, z>} and has dimension 2: dim(W)+ dim(W|)= 1+ 2= 3= dim(R3). Here W| is a subspace of R3 but NOT a subspace of V.
     
  9. Apr 18, 2007 #8
    Hi,

    1) I am working with real numbers throughout my entire lienar algebra course. For this question, if n is odd, I get (det A)2= -1, but what would det A be if I haven't learnt complex numbers yet? How should I answer this?


    2) The dimension theorem says "if U is a subspace of R^n, then dim U + dim U^| = n"
    And in the situtation of Q2, W is a subspace of V and V is a subspace of R^n, so W is a subsapce of R^n, so by the dimension theorem, dim W + dim W^| = n, what went wrong? I can't see any mistake in this logic...
     
  10. Apr 18, 2007 #9
    For question 3, I actually have the model solution to it, but I don't get what they are doing, I am so lost, can someone please help me?


    Let A=
    [a11 a12 a13
    a21 a22 a23
    a31 a32 a33]

    We know (matrix multiplication) A[1 1 1]^T = [0 0 0]^T, A[1 -2 1]^T = [0 0 0]^T, A[3 2 1]^T = [10 10 10]^T
    So the rows of A are identical <----I don't understand this step...WHY are the rows identical?
    We solve
    [1 1 1 | 0
    1 -2 1 | 0
    3 2 1 | 10] for the 1st row <---WHY should I choose to solve tihs matrix? WHERE did this matrix come from?
    Row reducing this matrix, we get a12=0, a11=5, a13=-5
    So A=
    [5 0 -5
    5 0 -5
    5 0 -5]
     
  11. Apr 18, 2007 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    .
    Since you are working with a vector space over the real numbers, you don't need to use complex numbers. In fact, you can't! If A2= -I means that (det A)2= -1, and that is IMPOSSIBLE, what does that tell about the existance of such an A? That is what I said in my first post: "what does that tell you about the number of 3 by 3 (or 5 by 5, etc.) matrices A such that A2+ I= 0?"


    All I can do is repeat, for the third time now: What is W|? Do you not see that what W| is depends on whether you are thinking of W as a subspace of V or of R3? Go back and look at the example I gave in my third response.

    A takes two independent vectors into [0 0 0]^T so its kernel has dimension 2 and its image has dimension 1. A must have rank 1. While that alone does not mean that all the rows are equal it means that they cannot be independent- they are all multiples of each other. It is then the fact that A[3 2 1]^T = [10 10 10]^2 where the '10's are identical that tells us the "multiple" is always 1.

    It said "for the first row" multiplying out the first row of A[1 1 1]^T gives a11+a12+ a13= 0. Multiplying out the first row of A[1 -2 1]^T gives a11- 2a12+ a13= 0. Multiplying out the first row of A[3 2 1]^T gives 3a11+ 2a12- a13= 10. The matrix here is the 'augmented' matrix for that system of three equations.
    Suppose you did the same thing for the second and third rows? Do you see that that you would get exactly the same three equations? That's another reason why the three rows must be the same.

     
  12. May 2, 2007 #11
    Hi,

    2) I re-read everything above and it seems to make more sense to me now. But how can you prove rigorously that the given statement is false? I am still having a terrible hard time doing this question...

    Thanks for your help!
     
  13. May 2, 2007 #12
    3) But is it always possible to express (1,0,0), (0,1,0), (0,0,1) as a linear combination of vectors like [1 1 1], [1 -2 1], and [3 2 1]? What can I do when this is not possible?
     
  14. May 2, 2007 #13
    should be the other way around.
     
  15. May 2, 2007 #14

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    W^| has vectors in V, and V^| doesn't. I don't think you even need dimensions

    Of course, if W=V, it actually turns out to be true
     
  16. May 2, 2007 #15

    radou

    User Avatar
    Homework Helper

    Why do you think so?

    As long as your three vectors in R^3 are linearly independent, you can express any vector from R^3 as a unique linear combination of these vectors.
     
  17. May 2, 2007 #16
    should be: it is always possible to express [1,1,1], [1,-2,1] and [3,2,1] as a linear combination of the standard basis vectors.
     
  18. May 2, 2007 #17

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Except that doesn't help us. We don't know what A[1 0 0] is, so we can't write out a vector v in terms of the standard basis and find Av

    Writing out v in terms of the three vectors that you DO know, however, gives Av trivially. And, coincidentally enough, the three vectors we have form a basis
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Determinant, subspace, linear transformation
Loading...