The previous article describes the solution of Ax = 0 and the zero space of matrix,
Here we will discuss the solution of Ax = B and the column space of matrix.
Ax = 0 is certainly a solution, because the total existence of x is the whole zero vector, making the equations true. While Ax = B does not necessarily have solutions. We need Gaussian elimination elements to determine. The previous article uses matrix A, which describes the solution of Ax = 0 as an example:
We can obtain the augmented matrix (The right side of the equation is not a fully zero vector, and the value of the elimination element changes. Therefore, the augmented matrix is required.) As follows:
Then we perform Gaussian elimination to get:
As shown in the preceding matrix,The equation must have:
Assume a vector B that satisfies the preceding conditions, for example, B = [1 5 1 + 5], and make two free variables x2 = 0, x4 = 0, then, we will write the matrix after the elimination element into a equations in the following form:
The solution is as follows:
Xc is a special solution of this equations, because when X2 and X4 take different values, different special solutions will be obtained. So how can we get the same solution of the equation? That is, how to express all the special solutions in the general form?
The process of solving Ax = B:
1. The solution for solving Xc2 and XnAx = B for solving Ax = 0 is the exclusive Xc + Xn. The proof is as follows:
Xc we have obtained above, Xn obtained in the previous article, thenThe general solution can be expressed:
So far, we have obtained the Ax = B solution. Through the above analysis, we know that when B meets the following formula, the equations are solved:
In fact, the condition for the equation to be resolved is that vector B belongs to the column space of matrix A, that is, vector B can be expressed as A linear combination of columns of matrix.For example:
The solution of the equation is the coefficient before each column in matrix.
The following is A general example. We will look at the structure of the solution based on the different situations of matrix A (assuming that matrix A is m * n and the rank is r ):
1. r = n <m, that is, full column rank (all columns have principal element)Since all columns have principal components, the number of free variables is 0, and the zero space of matrix A contains only zero vectors. The number of solutions for Ax = B is 0 or 1.
Example:
When B = [4 3 6 7], the unique solution of Ax = B is x = [1 1].
2. r = m <n, that is, full rank of rows (all rows have principal components)Because all rows have the principal element, there will be no rows with the value 0 after elimination, and there will be infinite solutions for Ax = B. The number of free variables is n-r, and the zero space of matrix A is not only A zero vector.
For example:
3. r = m = n, that is, columns and rows are full-rank (matrix reversible)Because columns and rows are full-rank, they have full-rank columns and some properties of full-rank rows: the zero space only has zero vectors, and the equations are always solutions and solutions are unique.
4. r <m, r <n, non-full rank matrix
Ax = B has infinite multiple solutions or no solutions.
We can summarize the above four situations as follows:If you want to see the solution of A linear equations, we can use Gaussian elimination method to obtain the simplest form of R of matrix A. The possible situation of R is as follows:
The solutions corresponding to these four situations are: 1. Unique solution or no solution 2. Infinite solution 3. Unique Solution 4. No solution or infinite solution.
Original article: http://blog.csdn.net/tengweitw/article/details/40921003
Nineheadedbird
[Linear algebra] solutions to linear equations
