Homogeneous Systems The Null Space We have seen that the solution to the homogeneous system of equations          Ax  =  0 is a subspace of Rn.  We will now begin a discussion of how to find a basis for this system.  The approach we will take is by an illustrative example.   Example Find a basis for the null space of the matrix         Solution We have seen before that the null spaces of row equivalent matrices are the same.  Hence this question is equivalent to that of finding the null space of          Now lets rewrite the system in equation form         x1       -5x3 + 4x4 + 3x5  =  0              x2 + 5x3 - 3x4 - 2x5  =  0 We can move the last three variable (the ones that are not corner variables) to the right hand side of the equations and add identity equations to get         x1  =  5x3 - 4x4 - 3x5         x2  =  -5x3 + 3x4 + 2x5         x3  =     x3         x4  =     x4         x5  =     x5 It is useful to introduce parameters here         s1  =  x3        s2  =  x4        s3  =  x5 so that         x1  =  5s1 - 4s2 - 3s3         x2  =  -5s1 + 3s2 + 2s3         x3  =     s1         x4  =              s2         x5  =                       s3 and we can write this in vector form         We can see that the null space is represented by triplets (s1, s2, s3).  This is equivalent (isomorphic) to the space R3.  We select the standard basis          (1,0,0), (0,1,0), (0,0,1) and come up with the basis for the null space         {(5,-5,1,0,0), (-4,3,0,1,0), (-3,2,0,0,1)} Example Let          Find the null space of A.     Solution As before, we find rref the matrix.         The corresponding equations are         x1  =  0         x2  =  0         x3  =  0 and we see that the null space is the subspace containing only 0.  Nonhomogeneous Systems Now that we know how to solve the homogeneous equation         Ax  =  0 we move on to nonhomogeneous systems         Ax  =  b We use the technique of rref as with homogeneous systems.  The next example illustrates.   Example Solve         Solution We solve the augmented matrix         and find the rref of the augmented matrix.  We get         This gives us the solution         Notice that this is not a vector space (it does not contain the zero vector) so it does not make sense to ask for a basis for a null space. The above answer shows that  The solution to          Ax  =  b  can be written in the form          x  =  xp + xh  Where          xp is a particular solution to the nonhomogeneous equation         xh represents the null space of A (the solution to the homogeneous equation)   Back to the Linear Algebra Home Page