Find all 10x10 Matrices such that ColA=NulA

[cocobobos]

Homework Statement

Find all 10x10 Matrices such that the column space is equal to the null space.

Homework Equations

Choose Function: n!/k!(n-k)!
where n is the total number of elements and k is the number k-cominations of the set.

rankA+dimNulA=n for a matrix in R^n

The Attempt at a Solution

There is a short and simple proof that we can use to determine that the rank of such a matrix should be 5.

Proof: Let A be a 10x10 matrix and suppose that ColA=NulA
=> rankA=dimNulA
=> rankA+dimNulA=10
=> rankA+rankA=10
=> 2rankA=10
=> rankA=5

Ok so from this I can deduce that I will have 5 linearly independent vectors in R^10 that will make up my matrix.

Now what I was thinking about this is that this must mean that to find ALL matricies I simply have to take permuations of my linearly independent vectors. For example a possibility for a matrix would be <e1,e2,e3,e4,e5,0,0,0,0,0> where e1,...,e5 are the basis vectors for R^10 and those 0's are the 0 vectors in r^10. Another possibility would be
<0,e2,e4,e5,0,0,0,e1,e3,0> and so on.

This seems like it'd be A LOT of possibilites so what I thought about doing was utilizing the choose function and from it's utilization I determined that because I have 10 columns total, with 5 vectors as a possibility to make up those columns, then I should have 252 possible combinations which are simply permutations of a 10x10 matrix with the vectors e1,...e5 as a basis.

Is my line of thinking correct or am I way off?

Thanks for the help!

 

[Office_Shredder]

You have more restrictions than just having rank 5. For example, the matrix
<e1,e2,e3,e4,e5,0,0,0,0,0> has as its null space span(e6,e7,e8,e9,e10) but has as its column space span(e1,e2,e3,e4,e5). If I tell you that the columns have to be e1,e2,e3,e4 and e5 can you identify which columns you can put them in in order to force the condition that the null space and column space are the same?

 

[micromass]

You seem to be focussing on finding all the 10x10 matrices that have rank 5. However, not every matrix with rank 5 will satisfy Col(A)=Nul(A). For example

$$\left(\begin{array}{cccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ \end{array}\right)$$

Will not have Col(A)=Nul(A). So you have a lot of redundancy.

Maybe you should think about what $A^2$ is.

 

[cocobobos]

I'm really confused what do you mean by A^2? Are you saying that A is the identity matrix? I guess I'm just confused in general about WHEN the column is EVEN equal to the null space at all.. How can you tell?

 

[micromass]

I'm really confused what do you mean by A^2? Are you saying that A is the identity matrix?

No, not at all.

Take an arbitrary x. What can you say about $Ax$ (think about Col(A))? What about $AAx$. What can you conclude about $A^2$??

 

[cocobobos]

No, not at all.

Take an arbitrary x. What can you say about $Ax$ (think about Col(A))? What about $AAx$. What can you conclude about $A^2$??

So what we can say about Ax is that it will be a linear combination of the column vectors of A?

 

[micromass]

So what we can say about Ax is that it will be a linear combination of the column vectors of A?

Yes,so it will be an element of Col(A) rigtht?? But Col(A)=Nul(A), so...

 

[cocobobos]

Yes,so it will be an element of Col(A) rigtht?? But Col(A)=Nul(A), so...

Ok so the column space of A consists of all possible products Ax for any x an element of R^n. So if the Col(A) is equal to the Nul(A) which is the set of all vectors x for which Ax = 0 then this tells us that Ax must be equal to zero? aka the linear combination must be equal to 0?

 

[micromass]

Ok so the column space of A consists of all possible products Ax for any x an element of R^n. So if the Col(A) is equal to the Nul(A) which is the set of all vectors x for which Ax = 0 then this tells us that Ax must be equal to zero? aka the linear combination must be equal to 0?

No.

The only thing you know now is that Ax is in Col(A). So Ax is in Nul(A). So...

 

[cocobobos]

No.

The only thing you know now is that Ax is in Col(A). So Ax is in Nul(A). So...

Well if Ax is in Nul(A) then isn't the null space of A the same as the solution set to the homogeneous system Ax=0?

 

[cocobobos]

So for a 10x10 matrix the null space of that matrix would consist of all vectors x in R^10 such that Ax=0 ?

 

[micromass]

Yes, y is in the nullspace if Ay=0.

Now you know that Ax is in the nullspace. So what must hold?

 

[cocobobos]

So Ax must equal 0?

 

[micromass]

So Ax must equal 0?

No. Ax is in the nullspace. That does not mean that it equals 0.

 

[cocobobos]

Ohh ok do you mean that if Ax is in the null space then:
Nul(A) must contain the zero vector
If x ∈ Nul(A) and y ∈ Nul(A), then x + y ∈ Nul(A) (aka closed under addition)
If x ∈ Nul(A) and c is a scalar, then cx ∈ Nul(A) (aka closed under scalar multiplication)

 

[micromass]

y is in Nul(A). What does that mean?? What is the definition??

Now apply this on y=Ax.

 

[cocobobos]

y is in Nul(A). What does that mean?? What is the definition??

Now apply this on y=Ax.

So if a vector y in is Nul(A) then this means that the equation y=c1v1+...+cnvn with cn scalars must equal 0?