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Interpretation of matrix multiplication?

  1. Aug 14, 2006 #1
    Matrix multiplication is clearly defined but is there a tangible or physical interpretation for it? Or is it just abstract formalism?

    I am thinking about each column of the matrix as vectors so matrix multiplication with two 2by2 matrices is about multiplying 4 vectors in a certain way. The result is a 2by2 matrix with all four vectors tangled up with each other. But does the two resultant vectors in this matrix have a meaning with respect to the original vectors if we gave the original vectors some meaning in the first place?

    I do see how an nbyn matrix multiplied with a nby1 matrix is. It is like dot product on a larger scale with one vector dotted with a set of vectors.

    So the question is about multidimentional matrix multiplication.
     
  2. jcsd
  3. Aug 14, 2006 #2
    A good interpretation is this: You can view your matrix as the matrix of a linear transformation (mapping) between two vector spaces (of finite dimension) in fixed basis for the two vector spaces. In the columns you express the image (under your linear map) of the basis vectors in terms of the new basis vectors.

    Let's look at an example: Let's look at [tex]\left[ \begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array} \right][/tex] and [tex]\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right][/tex]. Interpretation: Look at it as a two linear maps [tex]F,G: \mathbb{R}^2\rightarrow\mathbb{R}^2[/tex] and for simplicity fix the standard basis: [tex]e_1=\left( \begin{array}{c} 1\\0\end{array}\right), e_2=\left( \begin{array}{c} 0\\1\end{array}\right)[/tex]. It is important now to keep in mind that in both our domain and our target space (both [tex]\mathbb{R}^2[/tex]) we have chosen the same basis, which is of course a special case and must not always be like that.
    We then interpret the column vectors as the images of [tex]e_1[/tex] and [tex]e_2[/tex] under F or G. Under [tex]F:e_1\rightarrow \left( \begin{array}{c} 1\\3\end{array}\right), e_2\rightarrow \left(\begin{array}{c} 2\\4\end{array}\right)[/tex] again in the standard basis (which makes it a lot simpler). If you let [tex]A(F)=[a_{ij}][/tex] be the matrix of F, then it follows that [tex]F(e_1)=a_{11}\cdot e_1+a_{21}\cdot e_2, F(e_1)=a_{12}\cdot e_1+a_{22}\cdot e_2[/tex], which is exactly the multiplication of A(F) with e_1.
    Think about this a little: What happens if you take just some vector in [tex]\mathbb{R}^2[/tex], not one form the standard basis. How can you express it's image (maybe in terms of the standard basis?)? What happens, when you compose linear maps?
     
    Last edited: Aug 14, 2006
  4. Aug 14, 2006 #3

    mathwonk

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    the columns of AB are linear combinations of the columns of A whose coefficients are the entries in the columns of B.

    equivalently, as above, the columns of A are the images of the standard vectors ei under the map A, and the columns of AB are the images of the column vectors of B under the map A.

    so you can imagine a block spanned by the columns of B. then apply the map A to this block and you get the block spanned by the columns of AB.

    abstractly, matrices represent maps, and matrix multiplication corresponds to map composition.
     
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