Let's consider two vectors in R2:
We can see from a plot that these vectors a and b are obviously not pointing in the same direction.
Also we notice that a is not a multiple (or a linear combination) of b (or vice versa). There is no constant such that when the elements of a are multiplied by the constant we get b.
Now consider a point P in R2, say
We can construct a linear combination of a and b that reaches this P or any other point.
How? Simply place one of the vectors at the origin and move along it (perhaps in a negative direction), and place the second vector at P and move along it, and find where the two lines meet.
Call the coordinates of the point where the lines cross x and y. There are actually two possibilities depending on which vector we choose for each role. Suppose we move in the reverse direction from the origin along a and from P along b. From the point-slope equation we know that:
We can solve these two equations pretty easily. From the first
and from the second:
The other solution is symmetrical (we're dealing with a parallelogram). We would need to go +2 across and +8/3 up from P to reach x, y = (4, -4/3).
Suppose we want to know the actual multipliers for a and b, i.e. the fraction of the length of a and b that we travel along each vector.
From the figure, we can estimate that the values will be about -0.7 and -1.25.
Call these multipliers u and v. In matrix language
One way we can solve the system (call it the Algebra 2 way) is as follows:
Solve the second equation for v:
Plug into the first equation:
Finally, consider P not just as a point but as a vector c:
Since we could have chosen P to be any point, c could be any vector in the x,y-plane and it would be constructed as a linear combination of a and b:
We say that c is in the column space of the matrix whose columns are the vectors a and b:
because we can get to c by taking a linear combination of the columns of M.
And of course, having a + b = c, we can also add -c to get:
If we think of another matrix as the column vectors a, b and c lined up:
The combination u = -2/3, v = -4/3, w = -1 solves this equation and brings us back to zero:
That solution (a vector in R3):
is said to be in the nullspace of
because Ax = 0.