Understanding Matrix Transpose and Examples | Learn about Matrix Transpose

[FrankJ777]

Hi

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

Thanks

 

[HallsofIvy]

If A is a linear transformation from vector space U to vector space V, then A^T is defined as the linear transformation from V to U such that < Au, v>_U= < u, A^Tv><sub>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 A^T 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 A^T 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, A^T(v- Au)>U= 0 and since w can be any vector in U we must have A^T(v- Au)U= A^Tv- A^TAu= 0 or A^TAu= A^Tv so u= (A^TA)^-1A^Tv. It is not necessarily true that A^TA has an inverse (for example A= 0) but it may have even when A does not and if A does have an inverse, then (A^TA)^-1A= A^-1. (A^TA)^-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:
\left[\begin{array}{cc} x_1 &amp; 1 \\ x2 &amp; 2\\ ... &amp; ... \\ x_n &amp; 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]

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

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

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