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Exercise with intersection and sum

  1. Mar 15, 2016 #1
    My professor did this exercise that I didn't quite get how she went through all of it.
    We have a ##U = {(x, y, z, t) : x+y+z+t = 0}## and ##B_{Im(f)} = \left[ \begin{pmatrix}
    7 \\
    -3 \\
    0 \\
    0
    \end{pmatrix},
    \begin{pmatrix}
    3 \\
    -3 \\
    0 \\
    0
    \end{pmatrix},
    \begin{pmatrix}
    5 \\
    0 \\
    1 \\
    -5
    \end{pmatrix}\right]## with ##dim(Im(f)) = 3##.
    The exercise asks the base and dimension of ##Im(f) \cap U## and of ##Im(f) + U##.

    Now she starts the exercise doing this:
    ##x = - y - z - t##
    and
    ##\left\{
    \begin{array}{l}
    y = a\\
    z = b\\
    t = c\\
    x = - a - b - c
    \end{array}
    \right.##
    After this she writes
    ##U = \left\{a \begin{pmatrix}
    -1 \\
    1 \\
    0 \\
    0
    \end{pmatrix} + b \begin{pmatrix}
    -1 \\
    0 \\
    1 \\
    0
    \end{pmatrix} + c \begin{pmatrix}
    -1 \\
    0 \\
    0 \\
    1
    \end{pmatrix} : a, b, c \in \mathbb{R}\right\}##
    then she writes ##dim(U) = 3## and ##B_U = \left[
    \begin{pmatrix}
    -1 \\
    1 \\
    0 \\
    0
    \end{pmatrix}, \begin{pmatrix}
    -1 \\
    0 \\
    1 \\
    0
    \end{pmatrix}, \begin{pmatrix}
    -1 \\
    0 \\
    0 \\
    1
    \end{pmatrix}\right]##
    So here they come my first questions:
    1) Why did she put ##x = - y - z - t##?
    2) Why did she put everything in a system and put ##y = a, z = b, t = c##?

    Then she starts with the intersection saying
    ##\vec v \in Im(f) \Rightarrow \vec 0##
    and then she writes
    ##\vec v = \alpha \begin{pmatrix}
    7 \\
    -3 \\
    0 \\
    0
    \end{pmatrix} + \beta \begin{pmatrix}
    3 \\
    -3 \\
    0 \\
    0
    \end{pmatrix} + \gamma \begin{pmatrix}
    5 \\
    0 \\
    1 \\
    -5
    \end{pmatrix} = \begin{pmatrix}
    7\alpha + 3\beta + 5\gamma \\
    -3\alpha - 3\beta \\
    \gamma \\
    -5\gamma
    \end{pmatrix}##
    then she put all these in a single equation
    ##7\alpha + 3\beta + 5\gamma -3\alpha - 3\beta + \gamma -5\gamma = 0##
    that becomes ##\gamma = -4\alpha##.
    After finding that, she substitutes to all the ##\gamma## in the last matrix and does the operations and she writes
    ##Im(f) \cap U = \left\{\begin{pmatrix}
    - 13\alpha + 3\beta \\
    -3\alpha - 3\beta \\
    -4\alpha \\
    20\alpha
    \end{pmatrix} = \alpha \begin{pmatrix}
    -13 \\
    -3 \\
    -4 \\
    20
    \end{pmatrix} + \beta \begin{pmatrix}
    3 \\
    -3 \\
    0 \\
    0
    \end{pmatrix} : \alpha, \beta \in \mathbb{R}\right\}##
    then she writes ##dim(Im(f) \cap U) = 2## and using a formula(that I know it's the Graussmann formula) writes ##dim(Im(f) + U) = dim(Im(f)) + dim(U) - dim(Im(f) \cap U) = 3 + 3 - 2 = 4## and then she writes ##Im(f) + U = \mathbb{R}^4##
    My questions are:
    3) Why did she put the vector ##\vec v## only belonging to ##Im(f)##? Shouldn't it belong to ##U## too in order to know the intersection between the two subspaces?
    4) Why didn't she write, in the ##\vec v## equation, the independent vectors of ##U## too?
    5) Why at one point she sum all the frist, second, third and fourth rows of the vectors in one matrix? And why after that she put everything in a normal equation?
    6) And in the end, what does ##Im(f) + U = \mathbb{R}^4## mean?

    Sorry for all these questions but really I can't understand how she did all this. I know how to solve it differently but I want to know why she did like this.
     
  2. jcsd
  3. Mar 15, 2016 #2

    haruspex

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    1) that's the condition for (x,y,z,t) to be a point in U.
    2) the idea is to express this point in terms of three parameters
    3) how do you think she got that 'single equation'? She had a column vector representing v, then somehow produced a scalar equation from it by adding up the elements of the vector and saying that sum is zero. That is the condition for v being an element of U, is it not?
    I'll leave it at that, since the rest may become clear to you now.
     
  4. Mar 15, 2016 #3
    Why? Is this the only way? What would have happen if it had been something like ##U = {(x, y, z, t) : x +2y +z = 0}##? I would have simply done ##x = -2y - z## and then put ##y = a, z = b, t = c##? Does it work like this?

    This in order to have three linearly independent vectors of ##U##, right? And does it have to be three parameters only?

    I know that the subspace of a general intersection ##U \cap V## is made of vectors ##\vec v## that belong to both subspaces and that you can write them as ##\vec v = \alpha_1 \vec u_1 + \alpha_2 \vec u_2 = \beta_1 \vec v_1 + \beta_2 \vec v_2##. This means that
    ##\alpha_1 \vec u_1 + \alpha_2 \vec u_2 - \beta_1 \vec v_1 - \beta_2 \vec v_2 = \vec 0##
    but in the exercise the professor didn't put the vectors of the other subspace, she only put those of ##Im(f)##.
    I know that we have to check if they are linearly independent. Like, for example, ##\vec u_1, \vec u_2## are independent and we have to check if ##\vec v_1## is dependent to ##\vec u_1, \vec u_2##, checking this with ##\vec v_1 = \alpha_1 \vec u_1 + \alpha_2 \vec u_2##. Some thing for ##\vec v_2##. This is what I know about the intersection. Am I wrong?
     
  5. Mar 15, 2016 #4

    Mark44

    Staff: Mentor

    Because U was defined like this: ##U = {(x, y, z, t) : x+y+z+t = 0}##
    Yes, pretty much. Alternatively you could do it like this:
    x = -2y - z
    y = y
    z = ..... z
    t = ......... t
    If you stare at this long enough, you start to see that a vector v has this form:
    ##\vec{v} = \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}##
    Here, the parameters are y, z, and t.
    The equation x + 2y + z = 0 has three variables, so two of the variables are going to be free (or parameters), plus your equation doesn't involve t at all, so it is free as well. I chose y and z (and t) to be the free variables, but you could choose any other two. With three free variables, geometrically the solution is going to be a 3-D space in ##\mathbb{R}^43##. This space takes three vectors to describe.
     
    Last edited: Mar 15, 2016
  6. Mar 15, 2016 #5

    haruspex

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    ##\vec{v} = \begin{bmatrix} x \\ y \\ z \\ 0 \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ 1 \\0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\1\end{bmatrix}##
     
  7. Mar 15, 2016 #6

    Mark44

    Staff: Mentor

    It took me a while to get the glitch out of my LaTeX, but I figured it out.
    I don't see any reason why the t coordinate of ##\vec{v}## should be 0. ##\vec{v} \in \mathbb{R}^4##, so it has four coordinates, but no restriction on t.
     
  8. Mar 15, 2016 #7

    haruspex

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    She wrote v as a linear sum of the basis vectors of Im(f). She then applied the condition for membership of U to get a restriction on the coefficients. Therefore v has been defined as an arbitrary member of the intersection.
     
  9. Mar 15, 2016 #8

    Mark44

    Staff: Mentor

    Gotcha. I got a little confused between the original problem and the other example the OP posted and asked about.
     
  10. Mar 16, 2016 #9
    So, in order to get ##\vec v## we can simply use one of the two subspaces? I mean, if I wanted to, could I have done something like this
    $$\vec v = \alpha \begin{pmatrix}
    -1 \\
    1 \\
    0 \\
    0
    \end{pmatrix} + \beta \begin{pmatrix}
    -1 \\
    0 \\
    1 \\
    0
    \end{pmatrix} + \gamma \begin{pmatrix}
    -1 \\
    0 \\
    0 \\
    1
    \end{pmatrix}$$
    and solve it like the professor did with the vectors of ##Im(f)##?
     
  11. Mar 16, 2016 #10

    haruspex

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    Not sure what you meant by the last phrase, but if you mean solve it by applying the conditions for being in Im(f), then yes.
     
  12. Mar 16, 2016 #11
    What do you mean "solve it by applying the conditions for being in ##Im(f)##"? Are you saying that ##U \subseteq Im(f)##?
     
  13. Mar 16, 2016 #12

    haruspex

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    No. You want v to be in both subspaces. Your prof started by taking it to be in Im(f), and expressed that by writing it as a sum of basis vectors. She then needed to make it a member of U as well. She did that by applying the condition for membership of U to the coefficients in the sum.
    You can equally do it the other way around: write it as a sum of basis vectors of U then apply the condition for membership of Im(f).
     
  14. Mar 16, 2016 #13
    I don't understand what is the condition of membership. Where did she use it? Is the equation? Because all I see is calculations with vectors of ##Im(f)##.
     
  15. Mar 16, 2016 #14

    haruspex

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    After "and then she writes", there is an equation showing v as a linear sum of the basis vectors of Im(f). That makes v a member of Im(f).
    Next, look at "and then she put all these in a single equation". To get from the equation for v to that equation she has simply added up the elements of v and set the result to zero. That is the condition for membership of U: x+y+z+t=0. So now v is in the intersection.
     
  16. Mar 16, 2016 #15
    Oooh! Now I got it!
    And what if I had, instead of ##U## written like the exercise I showed, something like this
    $$B_U = \left[\begin{pmatrix}
    2 \\
    -5 \\
    3 \\
    0 \\
    \end{pmatrix}, \begin{pmatrix}
    -7 \\
    0 \\
    9 \\
    3 \\
    \end{pmatrix}\right]$$
    This means that I don't know that, for example, ##x + y + z + t = 0## or anything else like that, so I can't do the final equation.
     
  17. Mar 16, 2016 #16

    haruspex

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    I don't understand... that second vector is not in U, and to form a basis for U you need three vectors.
     
  18. Mar 16, 2016 #17
    No, no! I mean like a new exercise!
    Let's say that I have ##U## and ##V##. ##dim(U) = 2## and ##B_U## with two vectors, then ##dim(V) = 3## and ##B_V## with three vectors. Let's say that I want to know ##U \cap V##. In this case none of the two has a description like ##x + y + z + t = 0##, right? So, how should you proceed in this case? Like I said here?
     
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