1
$\begingroup$

Given $$sp\left \{ \begin{pmatrix} a_1\\ a_2\\ a_3 \end{pmatrix} ,\begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix} ,\begin{pmatrix} c_1\\ c_2\\ c_3 \end{pmatrix} \right \} = \mathbb{R}^3$$

What is $$sp\left \{ \begin{pmatrix} a_1\\ b_1\\ c_1 \end{pmatrix} ,\begin{pmatrix} a_2\\ b_2\\ c_2 \end{pmatrix} ,\begin{pmatrix} a_3\\ b_3\\ c_3 \end{pmatrix} \right \}\subseteq \mathbb{R}^3$$

I'm thinking of row spaces and column spaces but with not much luck, a leading hint would be appreciated more than the final answer.

$\endgroup$
1
$\begingroup$

Hint: The row rank and the column rank of a matrix is the same.


Another solution

Hint: The set of $n$ vectors spans $\mathbb{R}^n$ if and only if the determinant is non-zero.

Hint: The determinant of matrix $A$ is equal to the determinant of matrix $A^t$.

$\endgroup$
  • $\begingroup$ Could it be solved without speaking of ranks as I believe we haven't learned that yet. $\endgroup$ – Georgey May 25 '13 at 9:15
  • $\begingroup$ @Georgey That is the most direct way of doing it. You should also easily see that this is equivalent to your other question about the null space of $A$ and $A^t$ having the same dimension. $\endgroup$ – Calvin Lin May 25 '13 at 9:18
  • $\begingroup$ Yeah and both answered by using ranks which we haven't learned yet so maybe that's why I'm not confident with the answers given... $\endgroup$ – Georgey May 25 '13 at 9:19
  • $\begingroup$ @Georgey At the very least, do you see why showing that the dimension of the null space are equal is equivalent to showing that the dimension of the range space (span) is equal? $\endgroup$ – Calvin Lin May 25 '13 at 9:23
  • 1
    $\begingroup$ @Georgey Alright, try the other hint. You want to show that the transpose also spans the entire space. $\endgroup$ – Calvin Lin May 25 '13 at 9:29
1
$\begingroup$

You cannot say much about the relation between the two spans, except that they have the same dimension (for instance if one is the whole space, the other must be so as well). One way to see that you cannot expect a strict relation is to change the vectors in a way that certainly leaves their span unchanged, for instance permuting them or adding a multiple of one to another. This modifies the "transposed" vectors in a way that is not guaranteed to leave their span unchanged. To simplify your example, start with two vectors equal to zero.

Now that the question says that the original vectors $\vec a,\vec b,\vec c$ do indeed span the space, let me hint at an argument that the transposed ones do so as well. The vectors $\vec a,\vec b,\vec c$ form a basis of $\Bbb R^3$. Thus the map $\phi:\Bbb R^3\to\Bbb R^3$ that sends $(x,y,z)$ to the linear combination $x\vec a+y\vec b+z\vec c$ is an isomorphism (invertible linear map). If you interpret the transposed vectors as $1\times 3$ matrices $(a_i~b_i~c_i)$ (that is, you undo the transposition) then they describe the linear maps $\Bbb R^3\to\Bbb R$ defined by applying $\phi$ and then taking coordinate$~i$ (for $i=1,2,3$, repectively). Given this, can you argue that any $1\times 3$ matrix must be a linear combination of these three matrices, or equivalently that any linear map $\Bbb R^3\to\Bbb R$must be a linear combination of these three linear maps? (This is certainly true for the three coordinate maps themselves.)

$\endgroup$
  • $\begingroup$ I got the subset and equal symbols messed up, I believe that now it says much more... $\endgroup$ – Georgey May 25 '13 at 9:25
  • $\begingroup$ Ok so the $det$ of the matrix of the first set is non-zero for sure, meaning that the $det$ of the matrix of the second set is non zero for sure. Thus (1) if the system is homogeneous then it has the trivial solution, (2) if the system is non-homogeneous than it has a unique solution. Thus the second set spans $\mathbb{R}^3$, $\blacksquare$. How's this proof? $\endgroup$ – Georgey May 25 '13 at 9:56
  • $\begingroup$ If you are willing to use $\det$ (which I tried to avoid) then your first sentence should suffice. I don't see the point of considering systems (I don't see any) and certainly not non-homogeneous ones. Just leave out the second sentence and you're fine. $\endgroup$ – Marc van Leeuwen May 25 '13 at 10:16
  • $\begingroup$ I tried to think of your way of solution but since I don't really know much about the coordinates system. All I know about it is the if $V$ is a space over $F$ and $B$ is a set of vectors which is an arranged basis of $V$ than there's a unique presentation of any $v \in V$ as a linear combination of $B$ elements. Is that enough to use your proof direction? $\endgroup$ – Georgey May 25 '13 at 10:23
  • 1
    $\begingroup$ I supposed that you knew the definition of dimension, which implies that if $d$ vectors span a space of dimension $d$ then they are linearly independent (and therefore a basis). The problem is to show that if $3$ original vectors span $\Bbb R^3$ then the three transposed vectors also span $\Bbb R^3$; this is true but not obvious. $\endgroup$ – Marc van Leeuwen May 25 '13 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.