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I am currently studying linear algebra. Yet, I found discussions about linear algebra usually explain things in a geometric fashion.

I am quite confused on how to link up these two topics.

Can anyone kindly recommend some books/readings for me as an introduction to the concept of combining these two topics?

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One may define Euclidean geometry as the study of transformations of $\mathbb{R}^2$ by expressions of the form $Ax+b$ where $A$ is invertible (and often with determinant $\pm 1$. –  Fredrik Meyer Aug 31 '11 at 20:19
    
Yes, but how about higher space? This is what I am asking. –  Benjamin Sep 1 '11 at 5:24

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The classic introductions to the connection between linear algebra and classical geometry are Emil Artin's Geometric Algebra and Irving Kaplansky's Linear Algebra And Geometry. The latter now has a cheap Dover edition.

But the best book on the subject is probably Thomas Banchoff and John Werner's Linear Algebra Through Geometry. The Banchoff/Werner book gives an entire course in linear algebra primarily in three-space where all concepts are given geometric interpretations, such as projection, inner product constructions and linear maps.It's extremely visual with lots of pictures.At the end of the book,abstract vector spaces are considered as generalizations of all that's gone before. I think you'll find that book to be exactly what you're looking for.

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Thanks!! I found a copy in my library and I am going to grab it. –  Benjamin Sep 1 '11 at 5:22

Most linear algebra books do an OK job of showing the connection. This leads me to believe that you may be treating this connection with more reverence than it merits. If you don't mind here's a quick explanation that may show the beginnings of the connection. Please forgive me if I've misunderstood you and this is too elementary for what you need:

If there's one common thread in linear algebra it's the solving of this equation:

$$A \mathbf{x} = \mathbf{b}$$

where $\mathbf{x} \in \mathbb{R}^n$ and $\mathbf{b} \in \mathbb{R}^m$ are vectors and $A$ is a linear function with domain in $\mathbb{R}^n$ and range in $\mathbb{R}^m$. Much of your time in linear algebra is spent exploring various ways of solving this equation. Many of these methods depend on simplifying $A$. This is very algeraic in nature. So where is the geometry?

Nestled somewhere in the linear algebra curriculum you always learn that the function $A$ can always be represented by a matrix and once you've done this you can write our equation above in much more detail. It's here that the geometry becomes apparent. Here's a quick and dirty example:

Suppose we've already figured out a matrix for $A$ and we have the following:

$$\left( \begin{array}{cc} 1 & 0 \\ 2 & 1 \\ \end{array} \right) \left( \begin{array}{c} x \\ y \\ \end{array} \right) = \left( \begin{array}{c} 4 \\ 0 \\ \end{array} \right)$$

Somethings going on with my TEX here but I hope you can make out what that's supposed to be. Once we do the matrix multiplication we get: $$ x + y = 4$$ $$ 2x + y = 0$$

These are both equations for lines. A solution of the original problem $\mathbf{x} = (x,y)$ is then a point $(x,y)$ satisfying both equations. We know this to be the intersection of the two lines. And there's your geometry plain and simple. For larger matrices and vectors the geometric picture is more complicated. For instance if we were to try a 3 by 3 matrix A the corresponding problem viewed geometrically would be the intersection of 3 planes.

Hope this helps.

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My feeling is exactly the opposite: solving systems of linear equations is very much algebra, but linear transformations of $\mathbb R^n$ are geometry! After all, $\mathbb R^2$ is the plane, $\mathbb R^3$ is 3D space, and so on. –  Rahul Aug 31 '11 at 18:59
    
Thank you for your explanation. But I am looking for geometric interpretation on higher space. –  Benjamin Sep 1 '11 at 5:25
    
I can probably help you with that, but can you be more specific? –  Schmitty Sep 1 '11 at 15:55

Here is one book at this link: which I actually enjoy reading a lot.

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That does seem like a nice book; unfortunately, it appears to be out-of-print –  ItsNotObvious Aug 31 '11 at 15:19
    
@3phere: It is in print. Please see here: flipkart.com/books/… –  user9413 Aug 31 '11 at 15:27
    
Thanks for your recommendation :D –  Benjamin Sep 1 '11 at 5:26

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