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Linear Algebra - Topic: Basis & Subspaces

  1. May 25, 2008 #1
    I have a few questions here, my main problem is not understanding the notations used, hence not understanding the questions.

    1. The problem statement, all variables and given/known data

    1. Do the vectors [tex]j_{1}[/tex]= (1,0,-1,2) and [tex]j_{2}[/tex]= (0,1,1,2) form a basis for the space W = {(a,b,c,d) l a - b + c = 0, -2a - 2b + d =0} ?

    2. Find a basis for W = { (a,b,c,d) : a - b + 2d = 0 , 3a + c + 3d = 0 }

    3. Find the dimension of the following subspaces of [tex]\Re^{3}[/tex]
    a) span { (1,0,1),(0,1,1),(2,0,0) }
    b) span { (1,0,1),(2,2,4),(2,1,7),(-1,-1-2) }

    4. Suppose that a subspace W [tex]\subset[/tex] [tex]\Re^{3}[/tex] has a basis { (1,2,3),(1,0,1) }.
    a)Is { (-1,-2,-3), (3,2,5) } a basis for W? Why?

    2. Relevant equations
    How do you do this? Is there a working? or can you solved this by inspection.

    3. The attempt at a solution

    1. i substitute each vector in a,b,c,d equation and find the equality, so happen when i substitute in all returns 0, which implies that it does form a basis. Is there a proper working for this, cause i can do this by inspection.

    2. I was wondering forming a matrix and row reduced the matrix and get a linear dependence equation and plug in some value for c & d to find a & b.

    1 -1 0 2 0 ~ 1 0 1/3 1 0
    3 0 1 3 0 0 1 1/3 -1 0

    So: a = -1/3c - d & b= d - 1/3c

    Let c =1 and d =1

    a= -1/3 (1) - 1 = -4/3 & b = 1 - 1/3 (1) = 2/3

    so basis (-1/3,2/3,1,1)

    3 & 4. No idea what to do.
  2. jcsd
  3. May 25, 2008 #2


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    Homework Helper

    This is not a valid method. Imagine instead of giving you both j1 and j2, I gave you just j1 and ask if this vector alone spans the subspace W. Your method would indicate so, but in fact that is incorrect. To do this the 'proper way', note that if j1 and j2 span W, then all vectors in W can be expressed as linear combinations of j1 and j2. In other words,

    [tex]k_1 \left(\begin{array}{c}1\\0\\-1\\2\end{array}\right) + k_2 \left(\begin{array}{c}1\\0\\-1\\2\end{array}\right) = \left(\begin{array}{c}a\\b\\b-a\\2(a+b)\end{array}\right) [/tex]

    where [tex]k_1,k_2,a,b \in \Re[/tex]

    What you have to do is to express the above as a matrix, and then by row-reduction operations show that for any real a,b there will exist k1,k2 such that the above is satisfied.

    I have no idea what you are doing here.

    For qn 3. First, recall the definition of dimension. The dimension of a subspace is the number of linearly independent vectors which span the given subspaces, ie. the number of vectors in the basis of these subspaces. The vectors in any given basis are all linearly independent. Do you see what to do now?

    Qn 4. Suppose you are given a basis and another basis which may or may not span the same subspace as that of the first. Informally, if you can express every vector in the first group of vectors as a linear combination of the vectors in the second group, what does that tell you about the corresponding subspace spanned by the latter group with respect to the first?
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