Can You Find a Matrix with Ranking Requirements?

[Petrus]

Hello MHB,
I would like to have a tips for this problem.
Find a matrice $$A$$ of the order $$4 x 4$$ satisfying that
rank $$A=3$$, rank $$A^2=2$$, rank $$A^3=1$$ and
rank $$A^4=0$$

I have no idé how I should think and to try guess the matrice don't fel correct..
$$|\pi\rangle$$

 

[Evgeny.Makarov]

Re: Find a matrice with rank

Hint: Wikipedia claims that "any triangular matrix with 0s along the main diagonal is nilpotent".

 

[Fernando Revilla]

I have no idé how I should think and to try guess the matrice don't fel correct..
$$|\pi\rangle$$

Choose $A=\begin{bmatrix}{0}&{1}&{0}&0\\{0}&{0}&{1}&0\\{0}&{0}&{0}&1\\{0}&{0}&{0}&0\end{bmatrix}$

 

[Petrus]

Thanks both of you for taking your time and helping! Just learned for 10 minutes ago a Really important thing! If determinant =0 Then the rank is lower Then n for $$nxn$$ matrice where n is a integer and if determinant $$\neq 0$$ the rank is n which is a big key for this problem! Thanks once again!

Regards,
$$|\pi\rangle$$

 

[Deveno]

One way to proceed might be to start with a matrix you know will be of rank 3:

$\begin{bmatrix}\ast&\ast&\ast&\ast\\ \ast&\ast&\ast&\ast\\ \ast&\ast&\ast&\ast\\0&0&0&0 \end{bmatrix}$

Of course, if we row-reduce this matrix, we can bring it into upper-triangular form without changing its rank, so we may as well start with:

$\begin{bmatrix}\ast&\ast&\ast&\ast\\ 0&\ast&\ast&\ast\\ 0&0&\ast&\ast\\0&0&0&0 \end{bmatrix}$

Now we know that the image of any vector will be of the form (x,y,z,0), so we would like to fill in the third row so that we wind up with a vector of the form (u,v,0,0).

That is, we want to choose a and b such that:

$\begin{bmatrix}\ast&\ast&\ast&\ast\\ 0&\ast&\ast&\ast\\ 0&0&a&b\\0&0&0&0 \end{bmatrix} \begin{bmatrix}x\\y\\z\\0 \end{bmatrix} = \begin{bmatrix}\ast\\ \ast\\0\\0 \end{bmatrix}$

Clearly, if we choose a = 0, any value will work for b, and if we pick b non-zero, we can perform the row-operation of multiplying the third row by 1/b without affecting the rank, so we may as well choose b = 1.

Next, we want to pick a,b,c such that:

$\begin{bmatrix}\ast&\ast&\ast&\ast\\ 0&a&b&c\\ 0&0&0&1\\0&0&0&0 \end{bmatrix} \begin{bmatrix}u\\v\\0\\0 \end{bmatrix} = \begin{bmatrix}\ast\\ 0\\0\\0 \end{bmatrix}$

Again, setting a = 0 will do the trick, and if we pick any non-zero value for b, we can multiply the 2nd row by 1/b, giving us:

$\begin{bmatrix}\ast&\ast&\ast&\ast\\ 0&0&1&d\\ 0&0&0&1\\0&0&0&0 \end{bmatrix}$

Clearly subtracting d times the third row from the second row does not change the rank. I hope you can see now where this is going...Of course, we wind up with the self-same matrix Fernando Revilla displayed, which shouldn't be surprising...the point is, we didn't "luckily guess" it, we DEDUCED it.

 

[Fernando Revilla]

Of course, we wind up with the self-same matrix Fernando Revilla displayed, which shouldn't be surprising...the point is, we didn't "luckily guess" it, we DEDUCED it.

Well, there a sinuous boundary between what we must consider primary or not primary fonts of information. :)

 

[Deveno]

I agree. You or I might look at a problem like this, and use our prior experience to exhibit such a matrix which, once displayed, can easily be verified to be a satisfactory example.

The question is: how does someone without such prior knowledge go about discovering such matrices on their own?

Evgeny.Makarov's post is actually more along these lines: he suggests a possible theorem (quoted from Wikipedia) which hopefully the OP has within his power to prove. Having done so, the OP should be able to create many such matrices with the property he sought.

Don't get me wrong: I have nothing against telling a poster the answer to his (or her) question on any moral or philosophical grounds- knowledge is (or should be, in my opinion) free, and freely given. But mathematics also offers a person learning it the chance to EXPLORE...to leverage what they have previously learned and discover something they didn't know they already knew.

So in my posts, I try to encourage that exploration process. One should (in my opinion) play around with stuff, and see for yourself how things work. Sometimes, of course, this process comes down to remembering what has been shown to you.

 

[Fernando Revilla]

So in my posts, I try to encourage that exploration process.

All right. :)

 

[Petrus]

Re: Find a matrice with rank

Hint: Wikipedia claims that "any triangular matrix with 0s along the main diagonal is nilpotent".

I did recently checked clearly the nilopotent deffination and it make sense what we Was aiming for:) $$A^k=0$$, $$k \geq 1$$ which we Was aiming for! That make more sense now! The matrice Will get more zero to it Will be complety zero!

Regards,
$$|\pi\rangle$$