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So I'm studying for an exam and solving this problem. I've been watching countless online tutorials and reading books but I'm still not 100% if I'm doing this correctly since there's many different methods and specifics with different matrices. Could someone please double check this and tell me if this is the correct way to go about these problems?

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2a)

Rref(a) has 3 pivot columns of 3. Therefore the columns are linearly independent

Rref(b) has 1 pivot column of 2. Therefore the columns are linearly dependent

Rref(c) has 3 pivot columns of 4. Therefore the columns are linearly dependent

2b) Rref(c) =

              1 0 -9/4 0   
              0 1 -4/3 0   
              0 0  0   1   

which is X3 * the vector

9/4 
4/3 
1 
0

Therefore it spans R4 and not R3

*I feel like ^ that's definitely not how you do it, but I can't seem to get a grasp on how to

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An alternate approach to this is suppose that $v_1, \ldots, v_n$ are the columns of your matrix. You can show linear independence by supposing $a_1 v_1 + \ldots + a_n v_n = 0$ for some scalars $a_1, \ldots, a_n$. If the ONLY choice of scalars that satisfy this is $a_1 = \ldots = a_n = 0$, then your vectors are linearly independent. Otherwise they are dependent. There are also some shortcuts you can use for part a:

  1. Your method of finding pivots in matrix $A$ works just fine.
  2. For matrix $B$ you can observe that column $2$ is just $-3$ times column $1$. Since they are a scalar multiple of each other they are dependent. No calculation needed.
  3. Recall that the rank of a matrix is the number of independent columns, which can be proven to be equal to the number of independent rows of the matrix. Also, the rank of a matrix is equal to the number of pivot columns after you've done Gaussian elimination. Since matrix $C$ is $3\times 4$ then it can have only at most rank 3. Therefore the columns must be dependent.

For part b, we observe that the columns of matrix $C$ (by part a) are linearly dependent thus they can't span $\mathbb{R}^4$. You did however show that $C$ does in fact have 3 pivot columns, hence it has rank 3 therefore 3 of the columns are independent and span $\mathbb{R}^3$.

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  • $\begingroup$ Hey, thanks for this! But for part b why can't they span R4 if they are linearly dependent? Is the number of pivot columns always the rank? And the span? So if a matrix has 5 pivot columns does it have rank 5 and therefore spans R5? $\endgroup$ – Mark Mar 30 '15 at 4:45
  • $\begingroup$ The columns of $C$ are certainly elements of $\mathbb{R}^4$, but you need 4 linearly independent vectors in order to span the space. If your space is $n$-dimensional then you need at least $n$ vectors to span the space though this is not a guarantee. The number of pivots is the rank, but remember that span is the collection of all linear combinations of a set of vectors. Also, just to be clear, matrices don't span spaces but vectors do. If a matrix has rank 5 then 5 of its columns (or rows) span $\mathbb{R}^5$ (the ones with pivots). $\endgroup$ – Mnifldz Mar 30 '15 at 4:51

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