If $T: \mathbb{R}^3\rightarrow \mathbb{R}^3$ is a linear transformation of rank $2$ , then rank $T^2 \le 1$ .

Give proof if true, else give a counter example.

Question

If  $T: \mathbb{R}^3\rightarrow \mathbb{R}^3$  is a linear transformation of rank  $2$ , then  rank $T^2 \le 1$ .

Give proof if true, else give a counter example.

zzr0ck3r · Answer

Looking over old masters exams and I am not sure on this...

ganeshie8 · Answer

I think we may choose T such that T^2 = T ?

kainui · Answer

Yeah I was gonna say the same thing, specifically just pick some vector and use it as the projection operator since that's easy to visualize as going from 3D space to a 2D surface and then doing the same mapping is essentially the identity operation now.

kainui · Answer

Oh I should clarify since using a single vector as a projection operation in 3D space is rank 1 and you want rank 2. Luckily every vector defines a unique 2D surface in 3D space so:
Let's say the projecion operator is $P = \frac{vv^\top}{v^\top v}$ quick to show $P^2=P$. And easy to pick some random vector v to make it work out. Then you just need to produce a counter example so no rigorous proof is necessary once you have the thing, just gotta show that $T^2$ is not rank 1 or less after showing T is rank 2.

So as I was talking about with the vector defining a plane: 
$$T = I-P$$
$$T^2 = (I-P)^2 = I^2 -2P + P^2 = I- 2P+P = I-P=T$$

So $T^2$ and $T$ have the same rank.

ganeshie8 · Answer

Nice nice. so all projection matrices which project a vector onto a 2D plane are counterexamples.

bobo-i-bo · Answer

An explicit example is a linear map with this corresponding matrix:
$$ \left( \begin{array}{ccc}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0 \end{array} \right)$$

zzr0ck3r · Answer

ha, sorry. After I posted this I could not get back online.