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The terminology that is used everywhere when applying a matrix to a "vector" is considered is this: the matrix represents a linear transformation and there is a row or column vector. But a matrix can also be a vector from a vector space because that's the definition for vectors. What happens when matrix multiplication is considered? Can someone explain this mess?

LearningMath
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  • Please explain what you mean by "But a matrix can also be a vector from a vector space because that's the definition for vectors". Could you state the definition that you're talking about? – Ben Grossmann May 14 '15 at 03:13

2 Answers2

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Matrices transform coordinate vectors. That is, if you want to consider linear transformations on the vector space of $n \times m$ matrices, you should

  1. choose a basis for your vector space
  2. represent the elements of your vector space (in this case $n\times m$ matrices) as coordinate vectors AKA column matrices
  3. represent your linear transformation as a $k\times (nm)$ matrix
  4. perform your matrix multiplication as usual

Let's look at an example:

Say your space is $2\times 2$ matrices with real entries. This is a $4$-dimensional vector space. Let's choose an easy basis for this, how about $$e_1 = \pmatrix{1 & 0 \\ 0 & 0} \\ e_2 = \pmatrix{0 & 1 \\ 0 & 0} \\ e_3 = \pmatrix{0 & 0 \\ 1 & 0} \\ e_4 = \pmatrix{0 & 0 \\ 0 & 1}$$

Now say you are considering the linear transformation $T$ defined by $T(ae_1 + be_2 + ce_3 + de_4) = \pmatrix{a+b & b-c-d \\ c & 0}$.

Then the matrix $M$ which represents this transformation should always obey the equation:

$$M\pmatrix{a \\ b \\ c \\ d} = \pmatrix{a+b \\ b-c-d \\ c \\ 0}$$

Then we can see by the linearity of matrix multiplication that this is just

$$aM_1 + bM_2 + cM_3 + dM_4 = a\pmatrix{1 \\ 0 \\ 0 \\ 0} + b\pmatrix{1 \\ 1 \\ 0 \\ 0} + c\pmatrix{0 \\ -1 \\ 1 \\ 0} + d\pmatrix{0 \\ -1 \\ 0 \\ 0}$$

where $M_i$ is the $i$th column of the matrix $M$. Thus

$$M = \pmatrix{1 & 1 & 0 & 0 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0}$$

And finally if you want to use matrix multiplication to find out how $M$ transforms the matrix $x = \pmatrix{1 & 1 \\ 1 & 1}$, then just multiply $M$ by the coordinate vector of the matrix $x$:

$$M[x] = \pmatrix{1 & 1 & 0 & 0 \\ 0 & 1 & -1 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0}\pmatrix{1 \\ 1 \\ 1 \\ 1} = \pmatrix{2 \\ -1 \\ 1 \\ 0}$$

Therefore $$\color{red}{T(x) = \pmatrix{2 & -1 \\ 1 & 0}}$$

I should also note, though, that matrix multiplication can have a different interpretation as well. Multiplying two matrices results in a matrix that represents the composition of linear transformations.

For instance, say $S: \Bbb R^3 \to \Bbb R^2$ and $T: \Bbb R^3 \to \Bbb R^3$ are two linear transformations. Further let $A$ be the matrix which represents $S$ in the basis we're concerned with and likewise $B$ be the matrix representing $T$.

Then what's the matrix which represents the composition of these two linear transformations, $S \circ T$? Well it's just $AB$.

This is easy to show. By definition $S(x) = Ax$ and $T(y) = By$, therefore $$(\color{red}{S \circ T})(y) = S(T(y)) = S(By) = A(By) = (\color{red}{AB})y$$

  • Excellent answer. I have some questions though: Why are you allowed to make mn x 1 matrix from mxn matrix? Is this because the spaces they are in are isomorphic? Why you have to rewrite the matrix as a mn x 1 matrix and then do what you do? Do matrices ONLY operate on column/row vectors? And one more question: What is the difference between a tuple and mx1 and nx1 matrix? Isn't the tuple a different and special entity and should be considered as different from matrices with one column or row? Why and how you can mix them? Thanks a lot again. – LearningMath May 16 '15 at 12:26
  • That's a lot of questions. OK here goes: $(1)$ There is an isomorphism between any $n$-dimensional vector space over $\Bbb R$ and the "coordinate space" $\Bbb R^n$. That means that as long as we restrict ourself to just scalar multiplication, vector addition, and linear mappings, we can always decide to switch between those different vector spaces as we see fit. It just turns out that the space of column matrices is often the most useful such vector space so we often "switch to that one" to do our operations. –  May 16 '15 at 16:24
  • $(2)$ Yes. $(3)$ Because matrices only act as linear operations on row or column matrices. By convention we usually use column matrices. $(4)$ Matrices can of course act as operations on other objects -- in particular other non-column matrices -- but then they are not necessarily linear mappings. –  May 16 '15 at 16:28
  • $(5)$ The difference is that the set of all k-tuples is the set ${(a_1, \dots, a_k) \mid a_1, \dots, a_k \in \Bbb R}$, the set of all column matrices is the set ${\pmatrix{a_1 \ \vdots \ a_k} \mid a_1, \dots, a_k \in \Bbb R}$, and the set of all row matrices is ${\pmatrix{a_1 & \cdots & a_k} \mid a_1, \dots, a_k \in \Bbb R}$. As you can see these are all different spaces. But they're very similar. Again, we use the fact that they're isomorphic as vector spaces to just switch between them whenever we want -- but in this case they're so similar already that we ... –  May 16 '15 at 16:30
  • ... often don't even distinguish between them. We just call them all $\Bbb R^n$ and switch freely between them without even really thinking about it. In fact we do the same thing with points in Euclidean space (endowed with a coordinate system), and translation (arrow) vectors in Euclidean space. We just conflate all $5$ of these technically different spaces because the math works exactly the same for each of them. –  May 16 '15 at 16:36
  • Thanks a lot. How can you show that there is isomorphism between the set of k-tuples and column/row vectors(mx1 and 1xn matrices)? – LearningMath May 16 '15 at 17:24
  • There's probably already a question here on that, but if not, it would be worth asking that as a new question. –  May 16 '15 at 17:39
  • Yes, there is and it's posted by me. And I didn't get a good explanation and I still don't understand this, so if you want to help on this it's here http://math.stackexchange.com/questions/1283038/how-is-the-vector-space-of-abstract-tuples-isomorphic-to-vector-space-of-n-t. Thanks a lot again. – LearningMath May 16 '15 at 23:38
  • @notorious If I answered your question, you can accept it by clicking on the green arrow to the left of my answer. If not, what exactly are you still having trouble with? –  May 20 '15 at 23:42
  • Yes, you answered my question perfectly. I don't know why it's not already accepted (I did it). Thanks again, it was a beautiful answer. – LearningMath May 23 '15 at 21:22
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If I understand correctly, you want to know where multiplication of matrices, by matrices, stands when considering a certain space $\mathcal M$ of matrices as a vector space. In this case you simply have a vector space over $\Bbb R$, whose elements happen to form a ring. In fact you have an $\Bbb R$-algebra.

Another example of this is the space of continous functions $\mathcal C(\Bbb R,\Bbb R)$. It is an $\Bbb R$ vector space, but function multiplication is also defined by $(fg)(x)= f(x)g(x)$. This operation also makes it an $\Bbb R$-algebra.

GPerez
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