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A = [1 0 1 0] = row 1 [1 2 0 1] = row 2 (2 * 4 matrix)

and [0 0] = row 1 [2 1] = row 2

(2 * 2 matrix)

I know that Column of matrix of m*n dimension spans if rank of matrix is equal to m. If columns of A span then m<= n but if columns are linearly independent m>=n. And matrix is invertible if columns span and are linearly independent. So if a matrix is invertible matrix is a square matrix.

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    $\begingroup$ I'm confused by your opening statements. What is A? How are you identifying rows? $\endgroup$
    – jameselmore
    Commented Jan 31, 2015 at 3:20

1 Answer 1

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Well assuming that

$$A = \left(\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 1 & 2 & 0 & 1 \end{array}\right) \quad \text{and} \quad B = \left(\begin{array}{cc} 0 & 0 \\ 2 & 1 \end{array}\right)$$

are the two matrices you're talking about, and assuming the space which the columns may or may not span is $\mathbf{R}^2$, then one can see that the columns of $A$ span $\mathbf{R}^2$ since among the set of vectors making up the columns we have the two standard basis vectors $(1,0)$ and $(0,1)$ (that is, columns $3$ and $4$), of $\mathbf{R}^2$. Whereas, to see that the columns of $B$ don't span $\mathbf{R}^2$ notice that the first column is a scalar multiple of the second, namely, $(0,2) = 2\cdot (0,1)$, thus the two vectors making up the columns of $B$ span the $y$-axis, not all of $\mathbf{R}^2$.

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  • $\begingroup$ can you do a rigorous proof that's what I don't know how to do $\endgroup$
    – seeds
    Commented Jan 31, 2015 at 12:31
  • $\begingroup$ The columns of the first span because given any vector $(a,b) \in \mathbf{R}^2$ we have that $a(1,0) + b(0,1) = (a,b)$. Whereas, the columns of the second don't because (for example) any vector of the form $(a, 0)$ is not in the span, that is, for no scalars $c$ and $d$ do we have that $c(0,2) + d(0,1) = (a,0)$ $\endgroup$
    – Andrew Ross
    Commented Jan 31, 2015 at 17:32
  • $\begingroup$ why do you want a vector of form (a,0) for the second matrix. why not of form (a,b) $\endgroup$
    – seeds
    Commented Jan 31, 2015 at 18:37
  • $\begingroup$ Well technically the vector $(a,0)$ is of the form $(a,b)$ just with $b = 0$. The reason for the difference is because in the first case you want to show that the columns span, thus you need to show that for any vector in $\mathbf{R}^2$ some linear combination of the columns will yield that vector. In the second case you want to show that the columns do not span, and one way to do this is to produce a vector in $\mathbf{R}^2$ that does not lie in the span (i.e. no linear combination of the columns equals that vector). My point was that anything that looks like $(a,0)$ is not in their span. $\endgroup$
    – Andrew Ross
    Commented Jan 31, 2015 at 19:17
  • $\begingroup$ Is (0,b) in the span for matrix two $\endgroup$
    – seeds
    Commented Jan 31, 2015 at 19:25

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