### Kernel and Null Space

##### 3. Structure of solutions of general linear equations

Let T: VW be a linear transformation. Then the equation (where bW is fixed and xV is unknown)

T(x) = b

is a linear equation. The following tells us the structure of solutions of linear equations.

The solutions of a linear equation T(x) = b are of the form x = x0 + xh, where

• x0 is one particular solution of T(x) = b.
• xh is any vector in kernelT.

Proof Let x0 be one particular solution of T(x) = b. Then for any x we have

T(x - x0) = T(x) - T(x0) = T(x) - b,

where the first equality is because T is linear, and the second equality is because x0 is one solution. Consequently,

T(x) = bT(x - x0) = 0xh = x - x0kernelT.

Geometrically, the result means that the general solution of T(x) = b is obtained by shifting the subspace kernelT = {all xh} by the amount x0. More specific geometric pictures can be found in this exercise.

The result explains what we saw in the three examples at the beginning of this chapter.

Example The system

 x1 + x2 - 5x4 = 3 x3 + 2x4 = 4

is linear because the left side is a linear transformation

T(x1, x2, x3, x4) = (x1 + x2 - 5x4, x3 + 2x4): R4R2.

By taking x2 = x4 = 0, we find a special solution x0 = (3, 0, 4, 0) of the system. Moreover, the kernel of T, which is the general solution of the homogeneous system

 x1 + x2 - 5x4 = 0 x3 + 2x4 = 0

consists of vectors of the form

xh = (- x2 + 5x4, x2, - 2x4, x4) = x2(-1, 1, 0, 0) + x4(5, 0, -2, 1), x2 and x4 arbitrary.

Therefore the general solution is

x = x0 + xh = (3, 0, 4, 0) + x2(-1, 1, 0, 0) + x4(5, 0, -2, 1).

We emphasize that the general solution is the sum of the fixed part x0 = (3, 0, 4, 0) and the free part xh = x2(-1, 1, 0, 0) + x4(5, 0, -2, 1), with the freedom in the free part given by x2 and x4. The free part is the solution of the homogeneous system.

Example The differential equation

f'' - 4f' + 3f = 3t - 1

is linear because the left side is a linear transformation. The general solution is

f = (t + 1) + c1et + c2e3t.

The fixed part f0 = t + 1 is a particular solution of the equation. The free part fh = c1et + c2e3t is all the solutions of the homogeneous equation

f'' - 4f' + 3f = 0.

We also note that, being all the linear combinations of et and e3t, the free part form a subspace for the same reason as in this exercise.

Example The matrix equation

 [ 1 -2 3 ]X = I, for 2 by 3 matrix X -2 3 -4

is linear because the left side is a linear transformation. The general solution is

 X = [ 1 -2 ] + c1[ 1 0 ] + c2[ 0 1 ]. 0 0 2 0 0 2 0 0 1 0 0 1

The fixed part

 X0 = [ 1 -2 ] 0 0 0 0

is a special solution, and the free part

 Xh = c1[ 1 0 ] + c2[ 0 1 ] 2 0 0 2 1 0 0 1

is all the solutions of the homogeneous equation

 [ 1 -2 3 ]X = O. -2 3 -4