3
$\begingroup$

Given a matrix M such that its columns are the vectors of a new basis with respect to another basis B.

To find the coordinates of v in the other basis, we can simply take $M[v]_M = [v]_B$.

Let me give an example of M

$$\begin{bmatrix}1&2\\ 4&3\end{bmatrix}$$

I believe they are linearly independent(i just pulled out some random number off my head and tested), but the numbers aren't that important.

What i am confused about is we know that the columns of M form a set of basis vectors but when doing $M[v]_m$ matrix multiplication, we iterate within each $row_i$ of M for each value in the corresponding row of the output vector instead.

Now, i learn that, in my school's materials convention, we represent linear functionals as row vectors instead, since column vectors are for things like coordinate vectors and this makes sense to me at least here, but above, i am using a basis matrix's rows like linear functionals?

So yeah, is it just "it is how it is because matrix multiplication rules", or is there some special property or something about rows in matrices.

$\endgroup$
2
$\begingroup$

$$ \begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}= \begin{bmatrix} ax + by\\ cx + dy \end{bmatrix}= \begin{bmatrix} ax\\ cx \end{bmatrix} + \begin{bmatrix} by\\ dy \end{bmatrix} = x\begin{bmatrix} a\\ c \end{bmatrix}+ y\begin{bmatrix} b\\ d \end{bmatrix} $$

When you do ordinary matrix multiplication, you usually iterate over the rows to do the calculation, which is the first equality above. But this can always be rearranged to show that the answer is a linear combination of the columns your matrix.

So if you have a linear transformation $T$ between finite dimensional vector spaces $V$ and $ W$, say $T:V\to W$, represented by a matrix $M$, so that for any $v\in V$ you have $T(v)=Mv$, then the $\textit{column space}$ of $M$ is the subspace of $W$ spanned by the linearly independent columns of $M$ and is the subspace in which all possible results $Mv$ reside.

On the other hand, the linearly independent rows of $M$ span a subspace of $V$ called the $\textit{row space}$ of $M$. The orthogonal complement of the row space of $M$ is the $\textit{null space}$ of $M$ and is the subspace of $V$ that contains all $v\in V$ such that $Mv=0$.

All of these spaces have more formal definitions that need to be understood, but hopefully this gives a bit of a roadmap to sort out these things.

$\endgroup$
4
  • 1
    $\begingroup$ oooh. Thank you for the explanation! The part(and illustration) about being how they end up being linear combinations of the leftmost matrix really makes everything much clearer. I guess, the columns are like vectors, while the rows are like dimensions. When doing matrix multiplication e.g $MA$, can I take it as the rows or "dimensions" of A are based on the vectors of M. Like $row_i$ of A gets its input from the ith vector(column) of M? $\endgroup$
    – ZhengTay
    Oct 24 '20 at 13:16
  • 1
    $\begingroup$ Well, maybe. The things I detailed apply strictly to the matrix representation of a linear transformation of vector spaces. When multiplying matrices, if each matrix represents a linear transformation of vector spaces, then the matrix product can be interpreted as the matrix representing the composition of the transformations. $\endgroup$ Oct 24 '20 at 14:10
  • $\begingroup$ Don’t forget to upvote and accept my answer if you like it. ;) $\endgroup$ Oct 24 '20 at 14:11
  • 1
    $\begingroup$ Thank you! I think i have some intuitive understanding now. I did upvote! It's just i don't have enough reputation for it to be displayed according to the rules. It says it will still be recorded though! Once again, thank you very much! I got ignored alot when i was asking this elsewhere, because people think i was wasting time asking this. $\endgroup$
    – ZhengTay
    Oct 24 '20 at 14:43
2
$\begingroup$

First a note: as you correctly wrote, we have $M[v]_M=[v]_B$, but to obtain $[v]_M$ we need to calculate $M^{-1}[v]_B$, where $M^{-1}$ corresponds to the reversed basis transformation, i.e. its columns are just $[b_i]_M$.

Row vectors indeed act as linear functionals, and for any basis $m_1,\dots,m_n$, taking the $i$th coordinate with respect to this basis is a linear functional, and this is exactly what we get if only considering multiplication (from left) by the $i$th row of $M$.

$\endgroup$
1
  • $\begingroup$ thank you! I waited abit(saw this about an hour or less ago) before replying as I want to see if what understanding has somewhat stick. My intuition i got now is that each ith coordinate row of the transformation matrix while acting as a functional taking inputs from the to-be-transformed vector, i think of it as the reverse, where the vector's jth row element takes input from the jth column at the ith row/coordinate to output a scalar at the ith row/coordinate. Alternatively, a transformation matrix, M, transforms a vector, v, by changing its "basis(not)" kinda to that of M. $\endgroup$
    – ZhengTay
    Oct 24 '20 at 18:03

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.