Linear Algebra Lecture8 Note

Compute Solution of Ax=b (X=XP+XN)

Rank r

R=M Solutions Exist

R=n Solutions Unique

Example

If we want to solve the equation, what conditions does b1,b2,b3 need to meet? The observation matrix shows that the third line is the first two rows and so the B1+B2=B3

Solvability Condition on B:

Ax=b is solvable when B was in C (A)

If a combination of Rows of a gives zero row, then the same combination of entries of B must give 0

Assuming that the above matrix becomes:

To find complete solution to ax=b:

1.Xp (particular): Set all-variables to zero, solve ax=b for pivot variable

In this example, the x2=0,x4=0

2.Xn (Nullspace) The previous section has been solved

3.X (complete) =xp+xn

The above actions can be interpreted as:

m by n matrix A of rank R (r<=m,r<=n)

Full column of rank (r=n):

All columns have a principal element;  no free variables; N (A) =zero vector; Solution to Ax=b are x=xp which means if solution exists then the solution is unique (0 or 1 solution)

The reality is that, in addition to the zero combination, the linear combination between columns cannot produce 0 columns

Full row of rank (r=m):

All rows have a principal element; No zero rows; Can solve ax=b for every B; Left with N-r (n-m) Free variables

Full Rank (r=m=n):

N (A) =zero vector; R (Row-simplest) =i (unit matrix)

Summary

The rank of the matrix determines the number of solutions of the equation Group

Linear Algebra Lecture8 Note