Matrix Transpose

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Hi

Could somebody please tell me what the use is for the transpose of a matrix, and maybe give an example if possible.

Thanks
 

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HallsofIvy
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If A is a linear transformation from vector space U to vector space V, then AT is defined as the linear transformation from V to U such that < Au, v>U= < u, ATv>V where u is any vector in U, v is any vector in V and < , >U is an inner product in U and < , > is an inner product in V.

Given a basis for U and a basis for V, of course A would be represented by a matrix and AT would be represented by the "transpose" matrix: swapping rows and columns. Notice that if U is n dimensional and V is m dimensional, the matrix representing A would have n columns and m rows while AT would have m columns and n rows.

Here's an application: Suppose A is not "onto" V- that is the image is a proper subspace of V. Then we cannot, in general, solve the equation Ax= v. Such an x exists only if v happens to be in the image of A (strictly speaking the image A(U)). If v is not, what we can do is find the u in U such that Au is "closest" to v. What do we mean by "closest"? Well, geometrically, the point on A(U) (visualize it as a plane in 3 dimensions) closest to v is the one at the base of a perpendicular from v to A(U).

That is, suppose Au is the point in A(U) closest to v. The v- Au is the vector from Au to v and we want that perpendicular to A(U). That means that if w is any vector in U, Aw is in A(U) and so <Aw, v- Au>V= 0. Now, using the definition of inner product, <w, AT(v- Au)>U= 0 and since w can be any vector in U we must have AT(v- Au)U= ATv- ATAu= 0 or ATAu= ATv so u= (ATA)-1ATv. It is not necessarily true that ATA has an inverse (for example A= 0) but it may have even when A does not and if A does have an inverse, then (ATA)-1A= A-1. (ATA)-1A is referred to as a "generalized inverse" of A.

Here's a specific application of that: Suppose we want to find a line, y= ax+ b that passes through the points (x1,y1), (x2,y2[/sup]), ..., (xn,yn). Of course, a line is determined by two points so in general you can't find a single line through n points. We can, however represent this as a matrix equation:
[tex]\left[\begin{array}{cc} x_1 & 1 \\ x2 & 2\\ ... & ... \\ x_n & 1\end{array}\right]\left[\begin{array}{c} a \\ b \end{array}\right]= \left[\begin{array}{c} y-1 \\ y_2 \\ ... \\ y_n \end{array}\right][/tex]

Of course that "2 by n" matrix has no inverse but AT is
[tex]\left[\begin{array}{cccc}x_1 & x_2 & ... & x_n \\ 1 & 1 & ... & 1\end{array}\right][/tex]
AT A is the 2 by 2 matrix
[tex]\left[\begin{array}{cc}\sum_{i=1}^n x_i^2 & \sum_{i=1}^n x_i \\ \sum_{i= 1}^n x_i & n \end{array}\right][/tex]

The equation ATAu= Av would be
[tex]\left[\begin{array}{cc}\sum_{i=1}^n x_i^2 & \sum_{i=1}^n x_i \\ \sum_{i= 1}^n x_i & n \end{array}\right]\left[\begin{array}{cc}a & b\end{array}\right]= \left[\begin{array}{cc} x_1 & 1 \\ x2 & 2\\ ... & ... \\ x_n & 1\end{array}\right]\left[\begin{array}{c} y_1 \\ y_2 \\ ... \\ y_n\end{array}\right][/tex]

You might recognize that as giving the formula for the "least squares" line, the whose total distance to the points is a minimum.
 

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