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Dimension of matrices problems

  1. Jun 26, 2005 #1
    I'm having trouble with some matrices questions, can someone please help me?

    A = \left[ {\begin{array}{*{20}c}
    1 & 2 & 3 & 1 \\
    2 & 4 & 8 & { - 2} \\
    { - 5} & { - 10} & { - 19} & 3 \\
    \end{array}} \right]

    B = \left[ {\begin{array}{*{20}c}
    1 & 2 & 0 & 7 \\
    0 & 0 & 1 & { - 2} \\
    0 & 0 & 0 & 0 \\
    \end{array}} \right]

    B is the row reduced echelon form of the matrix A.

    a) What is the rank of A.

    b) What down a basis for the column space of A.

    c) Find the dimension of the row space of A.


    d) Are the rows of A linearly independent? Explain your answer.

    No, rank(row space) > dimension(row space).

    e) Do the vectors (1,2,-5), (3,8,-19) and (1,-2,3) span R^3?

    The answer says something along the lines of "The 3 vectors are in the column space of A which has dim=2. This implies that the 3 vectors span a subspace with dim <= 2. So they do not span R^3."

    I'm not completely about how to get to the answer. I know that dim(R^3) = 3 but I don't see why the 3 vectors span a subspace which has dimension less than or equal to 2. Further, I can't understand why it follows that they cannot span a subspace greater than 2.

    Because I don't understand this and also due to the fact people seem to prefer seeing an attempt from the OP I will try to explain what I have tried. I wasn't really sure about this question from the beginning so I went back to the definitions.

    A set of vectors T spans a subspace of R^m, S, if every vector in S can be written as a linear combination of T(I think:D). In this case the set S is R^3 and T is the set of the 3 given vectors. So yes, I think I could just let v = (x,y,z) to be an arbitrary vector in R^3, equate coefficients, write the corresponding augmented matrix and deduce the answer. However, the way that the question is set out and the way that the answer is phrased suggests that I do not need to do that.

    In any case, looking at the answer I see that there is a mention of the dimension of the column space of A, which is equal to the rank of A. Ok well at this point I'm just confusing myself and its getting late so I'm finding it hard to think. So could someone please provide me with some assistance.
  2. jcsd
  3. Jun 26, 2005 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    How does that prove the rows are linearly dependent?

    Anyways, that's not right -- the (row) rank is defined to be the dimension of the row space.

    Think about this once you've awakened and have a clear head!
  4. Jun 27, 2005 #3
    I made a mistake in typing up the answer to part d. I don't know what I was thinking.

    Anyway, the answer(to part d) I've got with me is: Number of rows in A = 3 > dim(row space) = 2. This implies rows of A are not linearly independent.

    I'm pretty sure I've figured out why the answer is the way that it is. There's a theorem in my book related to this question. The question as a whole seems quite nit picky.

    As for part 'e' I'm not sure about it but the answer seems right. The dimension of R^3 is 3. But why does it follow that the since the vectors {(1,2,-5),(3,8,-19),(1,-2,3)} are in the column space of A, with has dimension 2, then the 3 vectors span a subspace with dimension less than or equal to 2?

    For those 3 vectors to span a subspace(call it S) then every vector in S can be written as a linear combination of the 3 vectors. Oh wait, that's why they span a subspace of dimension less than or equal to 2.
  5. Jun 27, 2005 #4


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    Homework Helper

    One way to look at this is that the row reduction from A to B tells you that the rows of A are not linearly independent. You can think of the rows of A as being three vectors in 4-space, but clearly all three of those vectors can be written as linear combinations of the two non-zero row vectors in B. If you lablel the row vectors as A1, A2, A3, A4, B1, B2 then

    A1 = B1 + 3B2
    A2 = 2B1 + 8B2
    A3 = -5B1 - 19B2

    Since there are only two independent vectors in the 4-space, they span a 2 dimensional sub-space. Matrix A has 4 column vectors. The first two are clearly dependent by inspection, with the second being twice the first. That leaves the three vectors of part e) as potentially independent. You can show that they are not independent by forming a row matrix from those three vectors and performing a row echelon reduction (could also be done directly as columns of course, but the calculator likes rows)

    [tex] C = \left[ {\begin{array}{*{20}c} 1 & 2 & {-5} \\ 3 & 8 & {-19} \\ 1 & {-2} & 3 \\\end{array}} \right][/tex]

    which has the row reduced echelon form

    [tex] D = \left[ {\begin{array}{*{20}c} 1 & 0 & {-1} \\ 0 & 1 & {-2} \\ 0 & 0 & 0 \\\end{array}} \right][/tex]


    C1 = D1 + 2D2
    C2 = 3D1 + 8D2
    C2 = D1 - 2D2

    If you transpose matrix D, the first two columns are a pair of independent vectors in 3-space that span the 2-dimensional subspace that contains all 4 of the column vectors of the original matrix A.

    I'm sure the theorems are fresher in your mind than in mine, but one of them says something like the number of independent rows and the number of independent columns is always the same and is called the rank of the matrix.
  6. Jun 27, 2005 #5
    Thanks for your detailed response OlderDan, I'll need to spend some time later on to go over it. For part 'e' I can think of a way longer method to get to the answer but the question is structured so that I don't need to do it that way, so I thought, why not take advantage of it?Thanks again for you help.
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