Hello it's a me again! Today we continue with **Linear Algebra** getting into the promised **Examples **for some stuff we covered last time. You may also need to check out more from our little series, but I think that people that read this post have already seen the other ones. So, without further do let's get started!

## Example 1:

Find the **dimension of the solution space** that the following linear system has:

*x + 3y - z + 2u + v = 0*

*x + 2y + 4z + 2u = 0*

*2x + 8y + z + 3u - v = 0*

When the number of rows/linear equations is less than the number of columns/variables we mostly know that the system has infinite solutions. But, how big is this solution space? Well, as we already know from last time the dimension of a linear space or subspace is equal to the number of basis vectors, and so we will only have to find how these solutions are build up.

The A matrix of this system looks like this:

*1 3 -1 2 1*

*1 2 4 2 0*

*2 8 1 3 -1*

After doing elementary operations we end up with the following echelon form:

*1 0 0 40/13 44/13*

*0 1 0 -5/13 -12/13*

*0 0 1 -1/13 -5/13 *

So, we have infinite solutions and let's say that u and v are arbitrary or parameters (as I like to call them)

That way:

x = -40/13 u - 44/13 v

y = 5/13 u + 12/13 v

z = 1/13 u + 5/13 v

u, v arbitrary

We can also write the solution in this form:

*(x, y, z, u, v) = (-40/13 u - 44/13 v, 5/13 u + 12/13 v, 1/13 u + 5/13 v, u, v)*

That breaks up to the following to vectors:

*s1 = (-40/13, 5/13, 1/13, 1, 0) u*

*s2 = (-44/13, 12/13, 5/13, 0, 1) v*

*X = N(A) = {s1, s2}* (this set is also a basis for the nullspace)

And so the dimension of the nullspace or solution space is 2.

## Example 2:

Find a basis for the nullspace and the rank of the following homegeneous systems in matrix form

i)

*2 1*

*1 -1*

After doing elementary operations we end up with the identity matrix I:

1 0

0 1

That way the linear system has only one solution the zero-solution (0, 0) and so the nullspace N(A) has no basis and is equal to the zero-set {0}. Because we have two non-zero rows the rank of this matrix is 2.

ii)

*1 -1 1*

*2 -2 2*

*-1 1 -1*

All the rows/linear equations are actually the same equation, cause they are multiplications of either one of the others and so the system actually it this one:

1 -1 1

0 0 0

0 0 0

So, this system has infinite solutions and supposing x1, x2, x3 are the variables then x2 and x3 are arbitrary and so the solution looks like this:

x1 - x2 + x3 = 0 => x1 = x2 - x3

So, the solution in vector form is:

(x1, x2, x3) = (x2 - x3, x2, x3) = (x2, x2, x3) + (-x3, x2, x3) = (1, 1, 0) x2 + (-1, 0, 1) x3

*s1 = (1, 1, 0) x2*

*s2 = (-1, 0, 1) x3*

*N(A) = {s1, s2}*

And so the set {s1, s2} is a basis of the nullspace N(A).

Because only one of the rows is non-zero this matrix has a rank of 1.

This also proofs the equation I told you last time: rank(A) + dimN(A) = dimR^n => 1 + 2 = 3.

(I guess that after this example you know understood the vector form better!)

## Example 3:

Suppose we have a matrix A that represents a linear system:

*3 -1 0*

*5 3 -1*

*-2 -4 1*

Let's also say that u = (1, 3, 14) is a basis of the nullspace N(A) of this matrix.

i) find the rank(A)

ii) find a basis of R(A)

iii) does the system AX = (-1, 2, -3) have atleast one solution?

Knowing that u is a basis we also know:

- the dimension of the nullspace that is dimN(A) = 1
- a solution of this system (1, 3, 14)

So, by knowing the dimension we can find the rank(A) and the dimension of R(A).

i) dimN(A) + rank(A) = dimR^n => 1 + rank(A) = 3 => **rank(A) = 2**

ii) R(A) has a dimension equal to the rank(A) and so dimR(A) = 2

So, because the Range contains the columns it will be a span of the set of all columns:

R(A) = span{(3, 5, -2), (-1, 3, -4), (0, -1, 1)}

That way the dimension is 3 tho and not 2 as we already know that it needs to be and so not all of them are linear independent. We can clearly see on our own that the last one is a linear combination of the others and so (3, 5, -2) and (-1, 3, -4) are 2 basis vectors of the range of A.

So, **R(A) = span{ (3, 5, -2), (-1, 3, -4)}**

iii)

Knowing that the rank(A) = 2 that is less than the number of variables (3) we know that the system has **infinite solutions** for sure, cause the number of equations is less then the number of variables.

Let's proof it.

Using the A/b augmented matrix and doing elementary operations we end up with:

14 0 -1 | -1

0 14 -3 |11

0 0 0 | 0

This means that the system has infinite solutions and supposing x, y, z are the variables and z is arbitrary the solutions have the following form:

(x, y, z) = N(A) =[(-1 + z)/14, (11+3z)/14, z] where z is arbitrary

I guess that these examples covered a lot about what you need to know and this is actually it!

Next time in Linear Algebra we will get into Linear Functions.

Bye!