Let's consider two vectors in R

^{2}:

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 R

^{2}, 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 R

^{3}):

is said to be in the

*nullspace*of

because Ax = 0.

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