Matrix Multiplication and Function Composition

  • Thread starter Septimra
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I am doing linear algebra and want to fully understand it, not just pass the class. I was recently taught matrix multiplication and decided to look up how it works. The good part is that I understand the concept. Matrices are a way of representing linear transformations. So matrix multiplication is actually a composition of functions. That is why it is not communicative and it is associative.

But i recently came across this article and I could not follow the math near the middle of the page.
http://nolaymanleftbehind.wordpress.com/2011/07/10/linear-algebra-what-matrices-actually-are/

the matrices that are being multiplied are

[ 2 1 ] [ 1 2 ]
[ 4 3 ] [ 1 0 ]

the basis are w1 and w 2

and w1 = [ 1 0 ]
and w2 = [ 0 1 ]

The author states that all that is needed it to see how the linear transformation affects the basis vectors.

Then it states that f(g(w1)) = f(w1+w2)
How does that work? Where on earth do you plug in the w1?
Please help
 

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  • #2
Simon Bridge
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That would depend on what the author sees as f and g ... but the basic principle is that a linear transformation can be represented as a transformation of the coordinate system. A square in an oblique coordinate system looks the same as an oblique shape in a rectangular coordinate system.
 
  • #3
Fredrik
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But i recently came across this article and I could not follow the math near the middle of the page.
http://nolaymanleftbehind.wordpress.com/2011/07/10/linear-algebra-what-matrices-actually-are/

the matrices that are being multiplied are

[ 2 1 ] [ 1 2 ]
[ 4 3 ] [ 1 0 ]

the basis are w1 and w 2

and w1 = [ 1 0 ]
and w2 = [ 0 1 ]

The author states that all that is needed it to see how the linear transformation affects the basis vectors.

Then it states that f(g(w1)) = f(w1+w2)
How does that work? Where on earth do you plug in the w1?
Please help
If we write elements of ##\mathbb R^2## as 2×1 matrices, the definition of ##g:\mathbb R^2\to\mathbb R^2## can be written as ##g(x)=Bx## for all ##x\in\mathbb R^2##. So
$$g(w_1)=Bw_1 =\begin{pmatrix}1 & 2\\ 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 0\end{pmatrix}=\begin{pmatrix}1\\ 1\end{pmatrix}=\begin{pmatrix}1\\ 0\end{pmatrix}+\begin{pmatrix}0\\ 1\end{pmatrix}=w_1+w_2.$$
You may find https://www.physicsforums.com/showthread.php?p=4402648#post4402648 [Broken] useful.
 
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  • #4
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Thank you a lot, I appreciate it. I now see what the author was saying.
But I still have one minor question. I thought the author was trying to prove that g(x) = Bx. I now see that I was mistaken. But could one of you prove this? How does g(x) = Bx if x is a vector?
 
  • #5
Fredrik
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How does g(x) = Bx if x is a vector?
x is an element of ##\mathbb R^2##. If we use the convention to write elements of ##\mathbb R^2## as 2×1 matrices, then we can just define ##g(x)=Bx## for all ##x\in\mathbb R^2##. If we instead use the convention to write elements of ##\mathbb R^2## in the standard ##(x_1,x_2)## notation for ordered pairs, the notation ##Bx## doesn't work, but we could e.g. define
$$g(x)=\left(\left(B\begin{pmatrix}x_1\\ x_2\end{pmatrix}\right)_1,\left(B\begin{pmatrix}x_1\\ x_2\end{pmatrix}\right)_2\right)$$ for all ##x\in\mathbb R^2##. This looks really awkward of course. This is why I chose to use the matrix notation instead of the ordered pair notation.

We could also define g by saying that it's the function defined by ##g(s,t)=(s+2t,s)## for all ##s,t\in\mathbb R##. The matrix of this function with respect to the standard ordered basis ##(e_1,e_2)## where ##e_1=(1,0)## and ##e_2=(0,1)##, has ##g(e_j)_i## on row i, column j, as explained in the FAQ post. This is the ith component of the vector we get when g takes e_j as input. For example, row 2, column 1, of this matrix is
$$g(e_1)_2=(g(1,0))_2=(1+2\cdot 0,1)_2=1.$$ Note that this is equal to ##B_{21}##, as it's supposed to be.

If you want to understand how matrix multiplication is really composition of linear functions, then you should study the FAQ post and do this exercise: Let A and B be linear functions from ##\mathbb R^n## to ##\mathbb R^n##. Let [A] and denote their matrix representations with respect to the standard basis for ##\mathbb R^n##. Let [AB] denote the matrix representation of AB with respect to the standard basis for ##\mathbb R^n##. Prove that for all ##i,j\in\{1,\dots,n\}##, we have
$$[A\circ B]_{ij}=[AB]_{ij}.$$ This result tells us that the matrix representation of ##A\circ B## is equal to the matrix product of the matrix representations of A and B.

Hint: The definition of matrix multiplication is ##(XY)_{ij}=\sum_k X_{ik}Y_{kj}##. You will also have to use the fact that every vector is a linear combination of basis vectors.
 
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