Mathematics - Linear Algebra System Examples

    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!

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