Square Matrix Proof: Diagonal Entries & Properties

[nacho-man]

Hi, I have two questions.

firstly;
1) If M is the coefficient matrix of a linear system, Mx = 0, what are the solutions? Describe the set of solutions geometrically.

Now this is just a trivial solution since x = 0,
but to describe the solution geometrically, how would you do that? Do you just describe it as the origin, or 0-space perhaps?

And, in regards to the topic title:

2) Let A be a square matrix satisfying A = $-A^T$
- prove that the diagonal entries of A are all zero,
- Prove that if B is any square matrix, then $ A = (B - B^T)$ satisfied the property that $A = -A^T$

 

[chisigma]

1) If M is the coefficient matrix of a linear system, Mx = 0, what are the solutions? Describe the set of solutions geometrically.

Now this is just a trivial solution since x = 0,
but to describe the solution geometrically, how would you do that? Do you just describe it as the origin, or 0-space perhaps?...

Let's suppo to have a 2 x 2 linear sysytem, that you can write as...

$\displaystyle m_{11}\ x_{1} + m_{12}\ x_{2} = 0$

$\displaystyle m_{21}\ x_{1} + m_{22}\ x_{2} = 0\ (1)$

In the $x_{1}, x_{2}$ plane the (1) is represented by two straigh lines having the origin as common point. The same is for systems like (1) of higher dimension [$\mathbb R^{3}$, $\mathbb R^{4}$ and so on...]

Kind regards

$\chi$ $\sigma$

 

[Deveno]

There is no reason to suppose the ONLY solution of:

$M\mathbf{x} = \mathbf{0}$

is $\mathbf{x} = \mathbf{0}$

unless $M$ is an INVERTIBLE matrix, if $M$ is square, or more generally, if $M$ is $m \times n$:

a) of full rank, if $m < n$

b) never, if $m > n$.

The set of all vectors $\mathbf{x}$ such that $M\mathbf{x} = \mathbf{0}$ is a subspace of the domain of $M$ (regarding an $m \times n$ matrix as a linear function:

$\Bbb R^n \to \Bbb R^m$, with $\mathbf{x} \mapsto M\mathbf{x}$), and this subspace is called, variously:

the kernel of $M$
the null-space of $M$
the solution space of the system $M\mathbf{x} = \mathbf{0}$.

For example, if

$M = \begin{bmatrix}1&0&0\\0&1&0\\0&0&0 \end{bmatrix}$

it is not hard to see the null-space consists of all vectors of the form $(0,0,a)$ for any real number $a$. This is equivalent to the system:

$x_1 = 0$
$x_2 = 0$

Where we are looking for solutions $\mathbf{x} = (x_1,x_2,x_3)$.

This type of system (indeed ANY system of type (b) above) is called "under-determined", systems of type (a) above are called "over-determined" and may not have any solutions beyond the trivial solution (but they might, because the equations may have "hidden dependencies" that we discover upon row-reducing $M$).

If our original space that $\mathbf{x}$ lies in is say, $\Bbb R^3$, the possibilities are:

1) Any $\mathbf{x}$ is a solution (if $M$ is the 0-matrix)
2) The solution set is a PLANE in $\Bbb R^3$
3) The solution set is a LINE in $\Bbb R^3$ (that is scalar multiples of a single vector)
4) The solution set is trivial (only the origin works).

Seeing as how in this "limited" example, your thought that only #4 can happen is only batting .250, you might want to re-think your approach.

**********

Your first question is vague, we need to know a bit more about the matrix $M$ to really say anything useful.

**********

For your second question, suppose $A = (a_{ij})$. Then $A^T = (a_{ji})$.

If $A = -A^T$ this means that for every entry $a_{ij}$ we have:

$a_{ij} = -a_{ji}$

The diagonal entries are $a_{ii}$ (that is: $i = j$). From the above, we have:

$a_{ii} = -a_{ii}$ for every $i$. What can you say about a real number $r$ for which:

$r = -r$?

For the second part, let $A = B - B^T$.

Then $-A^T = -[(B - B^T)^T] = -[B^T - (B^T)^T] = -B^T -(-B) = ?$

 

[Longines]

For your second question, suppose $A = (a_{ij})$. Then $A^T = (a_{ji})$.

If $A = -A^T$ this means that for every entry $a_{ij}$ we have:

$a_{ij} = -a_{ji}$

The diagonal entries are $a_{ii}$ (that is: $i = j$). From the above, we have:

$a_{ii} = -a_{ii}$ for every $i$. What can you say about a real number $r$ for which:

$r = -r$?
$

In extension to what has been said, it can also be shown that $a_{ii} = 0$ with the following:

If you have $a_{ii} = -a_{ii}$ for every $i$, you can simply add $a_{ii}$ to the LHS of the equation, meaning you have:
$2a_{ii} = 0$. Solve for $a_{ii}$ and voila, you have shown that the diagonals are equal to 0.

 

[nacho-man]

Thank you for the response guys!

I just realized how silly i was for question 1,

the previous parts for it determined that the matrix we are dealing with is:

1 2 -1 3 8
-3 -1 8 6 1
-1 0 3 1 -2 which i reduced to

1 0 -3 0 4
0 1 1 0 -1
0 0 0 1 2

and that is called the coefficient matrix M.

We are then told M is the augment matrix of a linear system Ax = b, so M = (A|b) and find the solutions, then describe the solutions geometrically.

So I let that reduce row echelon form = to
b1
b2
b3

and my solution was

x = (x1, x2, x3, x4, x5) = (b2, b2, 0, b3, 0) + s(3,1,1,0,0) + t(-4,-1,0,-2,1)
And I had let the free variables x3 = s and x5 = t where s and t are parameters.
then we arrive to the question i posted, that M is a coefficient matrix of a linear system, Mx = 0, and what are the solutions, and describe these solutions geometrically..
Finally, in addition, I have another question:

Let w = (1,0,1,1,1)$^T$. Find d such that Mw=d. What is the general solution of Mx =d and describe this set geometrically.

so am i correct in saying
d =
1
0
1
1
1

and Mx = d is just
x = d?

 

[Longines]

Thank you for the response guys!

I just realized how silly i was for question 1,

the previous parts for it determined that the matrix we are dealing with is:

1 2 -1 3 8
-3 -1 8 6 1
-1 0 3 1 -2 which i reduced to

1 0 -3 0 4
0 2 -8 -6 -1
0 0 0 1 2

Yeah, that's right. That is the Reduced Row Echelon Form (RREF) for that matrix.

Now in regards to Mx=0. You have to realize that we're dealing with a homogenous equation. Geometrically, this means the lines, planes or hyperplanes represented by the equations in a homogeneous system all pass through the origin. Also note that the linear combinations are also orthogonal as the dot product (that is, M $\cdot$ x) are equal to 0.

 

[nacho-man]

Hi again,

I have updated my last post on this page with what I am having difficulty with:

Firstly;
We are told M is the augment matrix of a linear system Ax = b, so M = (A|b) and find the solutions, then describe the solutions geometrically.

So I let that reduce row echelon form = to
b1
b2
b3

and my solution was

x = (x1, x2, x3, x4, x5) = (b2, b2, 0, b3, 0) + s(3,1,1,0,0) + t(-4,-1,0,-2,1)
And I had let the free variables x3 = s and x5 = t where s and t are parameters.then we arrive to the question i posted, that M is a coefficient matrix of a linear system, Mx = 0, and what are the solutions, and describe these solutions geometrically..

First of all, how do I express them a set of solutions geometrically.
Is there a clear cut procedure to follow? Someone mentioned that if it is a line, plane or hyperplane, it will be determined by the number of free variables in the system,
where 1 free variable = line,
2 = plane, 3+ = hyperplane.

is this so?

Secondly, how do I do:

Let w = (1,0,1,1,1)$^T$. Find d such that Mw = d. What is the general solution of Mx=d? Describe the set of solitions geometrically.