Matrices M such that M^2 = 0 ?

[geoffrey159]

Homework Statement

[/B]
What are the $n\times n$ matrices over a field $K$ such that $M^2 = 0$ ?

Homework Equations

The Attempt at a Solution

Please can you tell me if this is correct, it looks ok to me but I have some doubts. I have reused the ideas that I found in a proof about equivalent matrices.

• One possibility is $M = 0$

• Assume that $M$ represents the linear map $f: E \rightarrow F$ in basis ${\cal B}$ and ${\cal C}$ both with dimension $n$.
  
  Since $f^2 = 0$, then $\text{Im}(f) \subset \text{Ker}(f)$.
  Let $(e_1,...,e_p)$ be a basis of $\text{Ker}(f)$. This basis can be completed into a basis ${\cal B'} = (e_1,...,e_p,e_{p+1},...,e_n)$ of $E$.
  The family $(f(e_{p+1}),...,f(e_n))$ belongs to $\text{Im}(f) \subset \text{Ker}(f)$ and is linearly independent in $F$.
  • Linear indenpendence :
    $\sum_{k = p+1}^n \lambda_k f(e_k) = 0 \Rightarrow \sum_{k = p+1}^n \lambda_k e_k \in \text{Ker}(f) = \text{span}(e_{1},...,e_p)$, but ${\cal B'}$ being a basis of $E$, all the lambda's are zero.
    
  • Free families in a vector space have less vectors than a basis of that vector space, so $n-p \le p \Rightarrow p \ge n/2$. By the rank theorem, $f$ has rank less than $n/2$

The free family $(f(e_{p+1}),...,f(e_n))$ can be completed into a basis ${\cal C'} = (f(e_{p+1}),...,f(e_n), f_1,...,f_p)$ of $F$.​

So it follows from all this that in the basis $({\cal B',C'})$, the matrix of $f$ is zero everywhere but in the upper right corner where there is an identity matrix packed somewhere starting at line 1 and ending column $n$, the somewhere depending upon the rank of $f$.
​

-> My answer is 0 and matrices that are similar to a matrix zero everywhere but in the upper right corner, where there is an identity matrix packed somewhere starting at line 1 and ending column $n$.

 

[DEvens]

How does this fit into your answer?

1 -1
1 -1

 

[geoffrey159]

Take $P = \begin{pmatrix} 0 & 1 \\ 1 & - 1 \end{pmatrix}$ so that $P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$:

$\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} = P^{-1} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} P$

 

[Mark44]

Take $P = \begin{pmatrix} 0 & 1 \\ 1 & - 1 \end{pmatrix}$ so that $P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$:

$\begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} = P^{-1} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} P$

What does this work have to do with your problem? The question asks about square matrices M such that M² = 0. Your matrix P doesn't satisfy this requirement.

In post 1, your first point is that M = 0. Since M² = 0, n x n zero matrices clearly are included.
Your second point in that post doesn't include the matrix that DEvens gave. Also, it is much more general than it needs to me. In particular, in this part:

Assume that $M$ represents the linear map $f: E \rightarrow F$ in basis ${\cal B}$ and ${\cal C}$ both with dimension n.

The matrices in question are square, so the linear map f takes Rⁿ to a subspace of Rⁿ; namely, to the 0 vector in Rⁿ.

 

[geoffrey159]

What does this work have to do with your problem?

DrEvens asked me to illustrate my point on a simple example. I prove that the given matrix $M = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$, which satisfies $M^2 = 0$, is similar to $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, which is the only $2\times 2$ matrix 0 everywhere but in the upper right where there is an identity matrix.

Also, it is much more general than it needs to me.

The question just says that the coefficients are in the field $K$. You can't assume that you have a linear map from $\mathbb{R}^n \rightarrow \mathbb{R}^n$, or there is something I don't understand

 

[Mark44]

What does this work have to do with your problem?

DrEvens asked me to illustrate my point on a simple example. I prove that the given matrix $M = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix}$, which satisfies $M^2 = 0$, is similar to $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, which is the only $2\times 2$ matrix 0 everywhere but in the upper right where there is an identity matrix.

That's DEvens (no r). I didn't understand the point of your example. I though that you were giving P as an example of a matrix for which P² = 0.

Also, it is much more general than it needs to me.

The question just says that the coefficients are in the field $K$. You can't assume that you have a linear map from $\mathbb{R}^n \rightarrow \mathbb{R}^n$, or there is something I don't understand

You are correct. I should have said that the map is from Kⁿ to a subspace of Kⁿ.

You also said this in post #1:

So it follows from all this that in the basis $({\cal B',C'})$, the matrix of $f$ is zero everywhere but in the upper right corner where there is an identity matrix packed somewhere starting at line 1 and ending column $n$, the somewhere depending upon the rank of $f$.

What about this matrix?
$$\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$$
There is no identity matrix in the upper right corner...

 

[geoffrey159]

That's DEvens (no r)

I meant no offense

I should have said that the map is from Kⁿ to a subspace of Kⁿ.

Why not a more general vector space over the field $K$ ? I have no example in mind, maybe I'm having a lack of understanding here. If you explain ...

 

[geoffrey159]

What about this matrix?
$$\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$
There is no identity matrix in the upper right corner...

Yes there is : $P = P^{-1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, then $M = P^{-1} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} P$

 

[Mark44]

There is no identity matrix in the upper right corner in the example I gave, but that matrix is similar to one that has an identity matrix there.

Do your examples extend to 3 x 3 matrices or larger?

In the 2 x 2 case do you have a geometric feel for what this matrix does to an arbitrary input vector?
$$\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$

 

[geoffrey159]

No it does not have an identity matrix in the upper right corner, but it is not my point.
My point is :

$M^2 = 0 \iff M = 0 \text{ or } \exists P\in \text{GL}_n(K) :\ M = P^{-1} \begin{pmatrix} 0 & I_{\text{rk}(M)} \\ 0 & 0 \end{pmatrix} P$

I believe it extends to $n\times n$ matrices or larger (if you say so), it is what I've tried to prove, and I have no geometric feel whatsoever, for now

 

[WWGD]

Have you tried using orthogonality relations, i.e., a row, considered as a vector, must be orthogonal to each of the columns, considered as a vector? So each column must be in the ortho (complement) space of all of the row vectors. Of course, for general fields, this would be an abstract orthogonality relation.

 

[geoffrey159]

I haven't tried, but what I have already looks like a description, doesn't it ? It seems that It doesn't convince many people ...

 

[Mark44]

I meant no offense

No problem.

Why not a more general vector space over the field $K$ ? I have no example in mind, maybe I'm having a lack of understanding here. If you explain ...

The matrix is n X n. If the field is K, then the matrix represents a transformation from Kⁿ to itself.

 

[Dick]

I haven't tried, but what I have already looks like a description, doesn't it ? It seems that It doesn't convince many people ...

Maybe if you look at Jordan Normal Form you might be able to form a more convincing way of describing what you are trying to say.

 

[geoffrey159]

The matrix is n X n. If the field is K, then the matrix represents a transformation from Kⁿ to itself.

Along this conversation, I've had trouble understanding why you say that. For example, if $K = \mathbb{C}$, and your vector space over $K$ is the set of polynomials of degree less than $n$. Any endomorphism of that vector space comes with a $(n+1) \times (n+1)$ matrix with coefficients in $K$.
Why should the vector space be reduced to $K^n$ ?

Maybe if you look at Jordan Normal Form you might be able to form a more convincing way of describing what you are trying to say.

I don't know what Jordan Normal Form is. Do you have the solution to the problem ? How much does it cost ? However, it seems that it works on $2\times 2$ matrices.

 

[Dick]

I don't know what Jordan Normal Form is. Do you have the solution to the problem ? How much does it cost ? However, it seems that it works on $2\times 2$ matrices.

I meant you should look it up. It's free on here http://en.wikipedia.org/wiki/Jordan_normal_form It puts what are trying to say in clearer terms than 'an identity in the upper right corner'.

 

[Mark44]

Along this conversation, I've had trouble understanding why you say that. For example, if $K = \mathbb{C}$, and your vector space over $K$ is the set of polynomials of degree less than $n$. Any endomorphism of that vector space comes with a $(n+1) \times (n+1)$ matrix with coefficients in $K$.

No. A polynomial of degree less than n has at most n terms (c₀z⁰ + c₁z¹+ ... + c_{n - 1} z^{n - 1}), with exponents ranging from 0 through n - 1, inclusive. So the matrix would be n X n in size.

Why should the vector space be reduced to $K^n$ ?
I don't know what Jordan Normal Form is. Do you have the solution to the problem ? How much does it cost ? However, it seems that it works on $2\times 2$ matrices.

 

[geoffrey159]

It puts what are trying to say in clearer terms than 'an identity in the upper right corner'.

I have put it quite clearly in post #10, in symbolic language. I sense your annoyance, would you like to have the final word about this problem ?

 

[pasmith]

Why should the vector space be reduced to $K^n$ ?

All vector spaces of dimension $n$ over $K$ are isomorphic to $K^n$, whatever the actual objects: p-by-q matrices with entries in K where pq = n; polynomials of degree at most n - 1 with coefficients in K; functions $X \to K$ where $X$ is any set of cardinality $n$; linear functions $V \to K$ where $V$ is any n-dimensional vector space over K; and so forth.

 

[vela]

The free family $(f(e_{p+1}),...,f(e_n))$ can be completed into a basis ${\cal C'} = (f(e_{p+1}),...,f(e_n), f_1,...,f_p)$ of $F$.

So it follows from all this that in the basis $({\cal B',C'})$, the matrix of $f$ is zero everywhere but in the upper right corner where there is an identity matrix packed somewhere starting at line 1 and ending column $n$, the somewhere depending upon the rank of $f$.

-> My answer is 0 and matrices that are similar to a matrix zero everywhere but in the upper right corner, where there is an identity matrix packed somewhere starting at line 1 and ending column $n$.

Why not use the basis ${\cal C'} = (f_1,\dots,f_p,f(e_{p+1}),\dots,f(e_n))$? Then the ones will be on the diagonal, and you can avoid the awkward phrasing "an identity matrix packed somewhere starting at line 1 and ending column $n$."

This makes me think there's something wrong with your proof since by choosing the right bases, you can write any linear transformation in this block-diagonal form. You've argued that the 0-block will be bigger than the identity matrix, but is that enough to guarantee that $M^2=0$? And how do you know there is a similarity transformation that allows you to turn M into this diagonal form?

The latter concern is your main problem, I think. I recommend you reconsider Dick's suggestion to look into Jordan normal form.

 

[geoffrey159]

All vector spaces of dimension $n$ over $K$ are isomorphic to $K^n$, whatever the actual objects: p-by-q matrices with entries in K where pq = n

Ok, thanks. I understand now.

Why not use the basis ${\cal C'} = (f_1,\dots,f_p,f(e_{p+1}),\dots,f(e_n))$? Then the ones will be on the diagonal, and you can avoid the awkward phrasing "an identity matrix packed somewhere starting at line 1 and ending column $n$."

You can, but when you will show the converse: that if a matrix with 1's on the diagonal that is similar to a matrix $M$, then $M^2 = 0$, I think you will need the similarity between your matrix and my matrix.

[EDIT] And the 1's will be on the diagonal if $f$ has exactly rank $n/2$. [EDIT]

And how do you know there is a similarity transformation that allows you to turn M into this diagonal form?

Because the awkward matrix is an expression of $f$ in another basis than the original one. This means that there exist a 'change of basis' matrix $P$ such that $M = P^{-1} A P$, $A$ for awkward

 

[vela]

Because the awkward matrix is an expression of $f$ in another basis than the original one. This means that there exist a 'change of basis' matrix $P$ such that $M = P^{-1} A P$, $A$ for awkward

But if you simply reorder the vectors in C', you'd still have a basis, right? So the diagonal matrix is "an expression of $f$ in another basis than the original one," and according to your reasoning, there's a similarity transformation which diagonalizes M. The entries on the diagonal should be the eigenvalues of M, but this leads to a contradiction because 1 is not an eigenvalue of M.

 

[geoffrey159]

If you reorder the basis ${\cal B'}$ and/or ${\cal C'}$ such that $M$ is similar to a diagonal matrix that contains only 1's, then
$M = P^{-1} D P = P^{-1} D^2 P = M^2 = 0$.

[EDIT] But this is absurd because $M$ and the diagonal matrix should have the same rank. Does this mean there is no reordering possible ? I doubt this. It is getting very confusing.

 

[vela]

You're running into contradictions, which, to me, suggests that your claim that there is a similarity transformation relating M and D (or A) isn't true.

 

[geoffrey159]

[EDIT] Post deleted

 

[geoffrey159]

I thought about what you said, and I can see an error in my reasoning: I did not use correctly the 'change of basis' formula which is
$\text{Mat}_{\cal B,C}(f) = \text{Mat}_{\cal B,C}(\text{id}_F \circ f \circ \text{id}_E) = \text{Mat}_{\cal C',C}(\text{id}_F) \text{Mat}_{\cal B',C'}(f) \text{Mat}_{\cal B,B'} (\text{id}_E)$.
My reasoning shows rather that $M$ and $A$ are equivalent, but not similar ( $M = Q^{-1} A P$, and not $M = P^{-1} A P$). It is a real problem for me because I can't show the converse, it is not enough to have a matrix equivalent to $A$ to be sure that it squares to zero Also, using the space $F$ and basis ${\cal C, C'}$ just complicates matters because $f$ is a linear map from $E \rightarrow E$ since $\text{Im} (f) \subset \text{Ker} (f) \subset E$. It is possible to build a basis ${\cal B'}$ such that
$\text{Mat}_{\cal B,B}(f) = \text{Mat}_{\cal B',B}(\text{id}_E) \text{Mat}_{\cal B',B'}(f) \text{Mat}_{\cal B,B'} (\text{id}_E)$, such that $M = P^{-1} A P$.
Take $(e_1,...,e_p)$ a basis of $\text{Im}(f)$ that you complete in a basis $(e_1,...,e_p,...,e_{n-p})$ of $\text{Ker}(f)$ thanks to the inclusion $\text{Im} (f) \subset \text{Ker} (f)$ and the rank theorem. There are $(a_1,..,a_p)$ in $E$ such that $e_1 = f(a_1),...e_p = f(a_p)$, because $e_1,..,e_p\in \text{Im}(f)$. Now the family $(e_1,...,e_p,...,e_{n-p},a_1,...,a_p)$ is free in $E$, which has dimension $n$, so ${\cal B'} =(e_1,...,e_p,...,e_{n-p},a_1,...,a_p)$ must be a basis of $E$.
In this basis, $M = P^{-1} A P$

 

[vela]

Looks good. Note if you ordered the basis B' as $(e_1,a_1,e_2,a_2,\cdots,e_p,a_p,e_{p+1},e_{p+2},\cdots,e_{n-p})$, the matrix A would be in Jordan normal form.

 

[geoffrey159]

Thank you for your help ! (I'm glad to be done with this problem)