The solution of "linear algebra" systems of linear equations

       Previous postThe 0 spaces of ax=0 and matrix A are described. here we discuss the solution of AX=B and the column space of Matrix A.

Ax=0 is definitely solvable, because the total presence of X is a full 0 vector. Make the equations set up. And ax=b is not necessarily a solution. We need to determine the Gaussian elimination element. We also use the previous article, which describes the solution of Ax=0, to illustrate:

We can get the augmented matrix of the above equations (the right side of the equation is not all 0 vectors, the extinction value changes, so we need to use an augmented matrix ) such as the following:

Then we do Gaussian elimination to get:

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

From the above matrix can be seen. The equation must be set up with :

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

If we have a B vector that satisfies the above conditions, for example: B=[1 5 1+5] and two free variables x2=0,x4=0, then we will write the matrix after the elimination to the form of the equation group, such as the following:

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

The resulting solution is:

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

XC is a special solution to this equation set. Because when x2,x4 takes a different value. Will get a different special solution.

So how do we get the same solution to the equation? How to represent all the special solutions in general form? the process of solving ax=b:
1, solving the special solution Xc2, solving ax=0 xnax=b solution is the special solution Xc+xn. The proofs are for example the following:

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

XC we've got it above, xn in the previous article . The general solution can be expressed as:

So far. We get the solution of the ax=b. Through the above analysis, we know that when B satisfies the formula. The equations have solutions:

In fact, the condition that the equation has solution is that vector B belongs to the column space of Matrix A. That is, vector B can be represented as a linear combination of the columns of matrix A.

Like the example above:

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

The solution of the equation is the coefficients in front of the columns in matrix A.

The following generalization to a more general case, we see the structure of the solution in the different cases of matrix A (if matrix A is a m*n matrix, the rank is R): 1, R=n<m, that is, column full rank (all columns have the main element)Since all columns have a principal, the number of free variables is 0. There are only 0 vectors in the 0 space of matrix A. The number of solutions for ax=b is 0 or 1.
To illustrate:

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

When b=[4 3 6 7], the only solution for Ax=b is x=[1 1].

2, R=m<n, will be full rank (all rows have the main element) because all rows have a main element, after the elimination of a total of 0 rows, the ax=b has an infinite number of solutions.

and the number of free variables is n-r, and the 0 space of matrix A is not only 0 vectors.

Like what:

watermark/2/text/ahr0cdovl2jsb2cuy3nkbi5uzxqvdgvuz3dlaxr3/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/ Dissolve/70/gravity/center ">

3, R=m=n. That is, columns, rows are full rank (matrix reversible)Because the columns and rows are full rank, there are some properties of the column full rank, row full rank: 0 space has only 0 vectors, the equation always has the solution and the solution is unique.

4, R<m,r<n, non-full rank matrix

Ax=b have infinitely many solutions or no solutions.
from the above discussion of four situations. We can summarize for example the following:Assuming that we want to see the solution of a linear equation group, we can get the simplest form of matrix A r,r by the Gaussian elimination method, for example, the following:
The corresponding solutions of these four cases are: 1, single solution or no solution 2, infinite solution 3, unique solution 4, no solution or infinite solution

Original:http://blog.csdn.net/tengweitw/article/details/40921003

Nineheadedbird

The solution of "linear algebra" systems of linear equations