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Subspace Orthogonality in Ax=b

  1. Oct 17, 2014 #1
    Let A be the matrix
    [2 0 1 0
    1 -1 4 3
    3 -1 5 3]

    Let b= [b1 b2 b3] transpose

    What equation must be satisfied by the components of b in order to guarantee that there will exists a vector x= [x1 x2 x3 x4] transpose
    satisfying the equation Ax=b. Justify your answer.

    I know C(A) is the orthogonal component of the left nullspace. Also I know that the rank of the matrix of is 2 and b must belong to the column space. Where do I go from here? This is not a homework question, I am simply trying to get a better understanding of the subspaces. I am a much better learner for things with physical applications and these ideas have been far to abstract for my liking.
     
  2. jcsd
  3. Oct 17, 2014 #2

    Fredrik

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    I used my limited superpowers to move your post to a homework forum because the rules tell us to treat all textbook-style problems as homework, even when the problem wasn't given as homework.
     
  4. Oct 17, 2014 #3

    Mark44

    Staff: Mentor

    You're making this much more complicated than it deserves.

    Set up an augmented matrix whose first four columns are the columns of A, and whose last column is your vector b.
    $$\begin{bmatrix} 2 & 0 & 1 & 0 & | & b_1 \\
    1 & -1 & 4 & 3 & | & b_2 \\
    3 & -1 & 5 & 3 & | & b_3 \end{bmatrix} $$

    Use row reduction to get the matrix on the left in reduced row-echelon form. As it turns out, one of the rows of the matrix turns into all zeroes. How does that affect the entry in the fifth column of this same row?
     
  5. Oct 17, 2014 #4

    BiGyElLoWhAt

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    I don't think you want the transpose on the b vector.
     
  6. Oct 17, 2014 #5

    Mark44

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    Not sure who your reply is directed to.

    The OP wrote "Let b= [b1 b2 b3] transpose" to represent this
    $$\vec{b} = \begin{bmatrix} b_1 \\ b_2 \\ b_3\end{bmatrix}$$
    A is 3 X 4, x is 4 X 1, so the result b will be 3 X 1, a column vector with three components. One way to write this that doesn't use LaTeX is <b1, b2, b3>T.
     
    Last edited: Oct 17, 2014
  7. Oct 17, 2014 #6

    Fredrik

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    A bit off topic, but I always think of (x,y,z) (written with commas between the components) as the transpose of (x y z). Now I just need to get the rest of the world to use my conventions. o0)
     
  8. Oct 17, 2014 #7

    BiGyElLoWhAt

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    I'm sorry all, I got carried away and mixed up columns/rows =/
     
  9. Oct 17, 2014 #8

    Mark44

    Staff: Mentor

    C
    O
    L
    U
    M
    N
    S
    are vertical. Think of the columns at the front of a building

    R O W S are horizontal.
     
  10. Oct 17, 2014 #9

    Mark44

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    I think that's going to be a hard sell...
     
  11. Oct 17, 2014 #10

    BiGyElLoWhAt

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    I meant in terms of nxm vs mxn, but whatever.
     
  12. Oct 17, 2014 #11

    Mark44

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    First number is number of rows, next is number of columns. Long ago I had trouble remembering this, but I came up with a mnemomic device - RC Cola (not sure they still make that soft drink). R comes first, then C.
     
  13. Oct 17, 2014 #12

    HallsofIvy

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    That's wonderful! (The southern United States would not be the same without RC Cola! (And moon-pies.))
     
  14. Oct 17, 2014 #13

    ImageUploadedByPhysics Forums1413585581.303393.jpg

    I've reduced it to this augmented matrix. Now what?
     
  15. Oct 17, 2014 #14

    Mark44

    Staff: Mentor

    Your work looks fine. Now, translate the augmented matrix into a system of equations. For example, the first row represents 1x1 + 0x2 + (1/2)x3 + 0x4 = b1/2.
    What do the other two rows represent?

    BTW, when you're working with augmented matrices, it's a good idea to draw a vertical dashed line to separate the coefficients of your matrix from the column that represents the vector b.
     
  16. Oct 18, 2014 #15

    Fredrik

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    Yes, it's impossible to get the world to settle on a notational convention, even when there's a good reason to prefer it over the alternative.

    3×1 matrices and 1×3 matrices are both elements of ##\mathbb R^3##. The only difference between them is the rule we use to multiply them with other matrices. ##(x,y,z)## (or if you prefer ##\langle x,y,z\rangle##) is a standard notation for an element of ##\mathbb R^3##. So ##(x,y,z)## is both a 3×1 matrix and a 1×3 matrix. This means that it's wrong to say that one of the equalities
    $$(x,y,z)=\begin{pmatrix}x & y & z\end{pmatrix},\qquad (x,y,z)=\begin{pmatrix}x & y & z\end{pmatrix}^T$$ is right and the other wrong. If one of these is to be considered an acceptable abuse of notation, it's only a matter of figuring out which one irritates us the least.

    The main reason to prefer the latter equality over the former is that if T is a linear operator on ##\mathbb R^3## and [T] is the matrix corresponding to T, then according to the former convention, the matrix equation that corresponds to ##T(x,y,z)=(a,b,c)## is
    $$[T]\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}a\\ b\\ c\end{pmatrix},$$ and according to the latter convention, it's
    $$\begin{pmatrix}x & y & z\end{pmatrix}[T]=\begin{pmatrix}a & b & c\end{pmatrix}.$$ (The [T] in the latter equation is actually the transpose of the [T] in the former. I didn't want to make the notation even uglier when we can just change the definition of [T]). So my abuse of notation preserves the order of the function and its argument, while yours reverses it.
     
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