A 2x2 matrix $M$ exists. Suppose $M^3=0$ show that (I want proof) $M^2=0$

Question

I'm sure I've done this before in abstract algebra. Regardless it's escaped me now.

I have proved that for $T:U\rightarrow V$, with $dim(U)=m$ and $dim(V)=n$ that $rank(T)\le m$ which is obvious, but I have proved it none the less.

I want to.... this is where I get stuck. I'm not quite sure how to say it.

Suppose we have a $T$ for example that takes $U\subset\mathbb{R}^3\rightarrow V\subset\mathbb{R}^3$, we could still have a $T$ that maps a plane to something (a plane, line or point) in this case rather than $rank(T)\le 3$ I can state $rank(T)\le 2$.

(Correct my notation here, I don't like writing subset, I mean number of dimensions of the space!)

To show this I can say $F:P\subset\mathbb{R}^2\rightarrow U\subset\mathbb{R}^3$.

Then let $G:P\rightarrow V$ and that $G=TF$, now $rank(G)\le dim(P) = 2$

Now, if $U,P,Q,R$ are spaces of dimension $\le 2$ and P,Q,R are subspaces of U

$A:U\rightarrow P$

$B=A:P\rightarrow Q$

$C=A:Q\rightarrow R$

We can now say $A^2=BA$ then $A^3=CBA$

We know that for $A$ rank(A)$\le 2$, this thus provides an upper-bound for rank($A^3$)

I want to show now that if I keep applying a linear transformation that the rank is a monotonically decreasing ($\le$) sequence. I am unsure on a proof that the rank of a transformation cannot be greater than the domain's dimensions. Although this is trivial.

(NOTE:

I have proven that if a set of n vectors, R, span a vector space V, and you take a set W of m linearly independent vectors in V that m$\le$n)

If rank(A)=2 then rank($A^2$)=2 and so rank($A^3$)=2

if rank(A)=1 then rank($A^2$)=0 and rank($A^3$)=0

if rank(A)=0 then rank($A^2$)=0 and rank($A^3$)=0

But again, I can't prove this, or at least I am unsure of how to write it.

something to clarify:

For a map $T:U\rightarrow V$ how do we distinguish (using notation) whether or not we actually use all dim(U)?

For example we might have a T that maps a plane in R^3 to a line in R^3, this can be expressed as a composition of maps, one that takes a 2 dimensional vector space (coordinates of the plane in 3 space) to a 3 dimensional point in T's domain, which T then maps to a line.

We have rank - the dimensions of the image - to make this distinction on the map's target (the rank of T is 1 if it maps something to a line, but the dimensions of the target is 3)

How do we write this? What do I say to describe this?

How do I distinguish between the number of vectors in a basis and the number of dimensions the space that basis is in has?

Sorry for the length of this! I tried to talk though my problem in the hope of seeing it myself, no such luck, thanks.

Answer 1

Here's a naive approach to the question in the title:

If $M$ has rank 2, then it's invertible and $M^3$ cannot be $0$. If $M$ has rank 0, then $M^2=0$ trivially. So the only interesting case is when $M$ has rank $1$.

In this case there's a vector $v$ such that $Mv$ is nonzero. Since $M^3v=0$, it must be that either $Mv$ or $M^2v$ is a nonzero vector that $M$ maps to zero. So the kernel and the image of $M$ has a nontrivial intersection -- but both of these subspaces have dimension $1$, so they must coincide.

In particular, because the image is contained in the kernel, $M^2v$ must be $0$ for all $v$, so $M^2=0$.

Answer 2

I had a hard time following all the details of the arguments in the OP's question, but based upon the Hamilton-Cayley theorem there is a pretty simple solution to the query expressed in the title.

Hamilton-Cayley tells us that we have

$M^2 + bM + c = 0 \tag{1}$

for some scalars $b, c$. If $b= c = 0$, (1) reduces to

$M^2 = 0, \tag{2}$

and we are done. If $c \ne 0$, multiply (1) by $M^2$ yielding

$M^4 + bM^3 + cM^2 = 0, \tag{3}$

and using $M^4 = M^3 = 0$, we have

$cM^2 = 0,\tag{4}$

and we are done. If $c = 0$, $b \ne 0$, multiply (1) by $M$ to obtain

$M^3 + bM^2 = 0, \tag{5}$

or

$bM^2 = 0, \tag{6}$

and once again we are done. We have thus covered all cases, and we are done, and I mean done! QED.

Hope this helps. Cheerio,

and as always

Fiat Lux!!!

Note added in edit: a similar technique was used in my answer to this question: nilpotent and linear transformation

Answer 3

Here is another proof that I find more direct:

We know that $M$ cannot be of full rank or of rank $0$. Hence $M$ is of rank one and there exist vectors $u, v$ such that $M = uv^T$. Then

$$ M^3 = uv^Tuv^Tuv^T = (v^Tu)^2uv^T = (v^Tu)^2M. $$

Therefore $v^Tu = 0$ and so

$$ M^2 = uv^Tuv^T = (v^Tu)uv^T = (v^Tu)M = 0. $$

Answer 4

Consider the subspaces $V_i=\ker(M^i)$ for $i=0,1,2,3$; by convention $M^0=I$ so $V_0=\{0\}$, and $M^3=0$ so $V_3=\Bbb R^2$; you need to show that $V_2=\Bbb R^2$ too. Clearly $V_0\subseteq V_1\subseteq V_2\subseteq V_3$, since a vector killed by $M^i$ will also be killed by $M^{i+1}$. The key observation is that if $V_i\neq V_{i+1}$, which necessarily means $\dim(V_i)<\dim(V_{i+1})$, then once also has $\dim(V_j)<\dim(V_{j+1})$ for all $0\leq j<i$. This is because $\dim(V_i)<\dim(V_{i+1})$ means there are vectors $v$ with $M^i\cdot v\neq 0$ but $M^{i+1}\cdot v= 0$, but for such a vector, $w=M^{i-j}\cdot v$ is a vector with $M^j\cdot w\neq 0$ but $M^{j+1}\cdot w= 0$.

Thus $V_2\neq V_3$ would imply $0=\dim(V_0)<\dim(V_1)<\dim(V_2)<\dim(V_3)=2$, and there are not enough intermediate integer values to have this. So in fact $V_2=V_3$ as you wanted to prove. In the same way you can see that for an $n\times n$ matrix $A$ the equation $A^k=0$ for any $k\in\Bbb N$ implies $A^n=0$. This is something that would be hard to obtain from explicit case-by-case considerations.

Answer 5

There is an elementary proof. Write $$M=\left[\begin{array}{cc} a & b\\ c& d \end{array}\right].$$ Since $M^{3}=0$, taking the determinant of both sides and using the determinant of a product is the product of the determinants, gives $\text{det}\,M=0$. Hence, $$ad-bc=0.$$ By matrix multiplication, the following is found. $$M^{2}=\left[\begin{array}{cc} a^2+bc & b(a+d)\\ c(a+d) & bc+d^2 \end{array}\right],$$ and $$ M^{3}=\left[\begin{array}{cc} a^3+2abc+bdc & a^2b+b^2c+bd^2+abd\\ a^2c+adc+bc^2+d^2c & abc+2bdc+d^3 \end{array}\right].$$ But by factoring and using $ad-bc = 0$, it is seen that $M^3$ can be rewritten in the following form. $$ M^{3}=\left[\begin{array}{cc} (a^2+bc)(a+d)-a(ad-bc) & b(a+d)^2-b(ad-bc)\\ c(a+d)^2-c(ad-bc) & (bc+d^2)(a+d)-d(ad-bc) \end{array}\right] = \left[\begin{array}{cc} (a^2+bc)(a+d) & b(a+d)^2\\ c(a+d)^2 & (bc+d^2)(a+d) \end{array}\right] = (a+d)M^2.$$ Since $M^3 = 0$, we now have that either $M^2=0$ or $a+d=0$. If $a+d=0$, then $0=ad-bc=-a^2-bc=-d^2-bc$. Looking again at each component of $M^2$, we see that each component is 0.