1

Let $V$ be a vector space. Determine all linear transformations $T:V\rightarrow V$ such that $T=T^2$.

Suppose $x\in V$. Then we can write $x=T(x)+(x-T(x))$. Then $T(x)\in$ Range $T$, and $x-T(x)\in$ kernel of $T$. From this how I to proceed further?

vqw7Ad
  • 2,055
  • 2
    This may be helpful: https://en.wikipedia.org/wiki/Projection_(linear_algebra) – user3281410 Jun 07 '19 at 12:58
  • This may also be helpful: https://math.stackexchange.com/questions/261704/show-that-the-direct-sum-of-a-kernel-of-a-projection-and-its-image-create-the-or – J_P Jun 07 '19 at 13:38

2 Answers2

1

Given such a $T$, notice that $T(V)$ is a subspace of $V$, and that $T$ acts on $T(V)$ as the identity. Therefore, such linear transformations are a subset of projections from $V$ to subsets of $V$. Is it all of them?

Richard Jensen
  • 1,440
  • I don't quite understand the phrasing. Isn't it the other way around - that your argument shows that projections (i.e. those $T$ for which $T=T^2$) are a subset of such linear transformations (i.e. ones which act as the identity on $T(V)$)? – J_P Jun 07 '19 at 13:20
  • I would think of a projection as a linear map from a space to a subspace acting as the identity on that subspace. Therefore, a $T: V \rightarrow V$ where $T^2 = T$ is a projection to the subspace $T(V)$. Therefore, any such $T$ is a projection, so the set of those transformations is a subset of all projections.

    The definition of a projection is that $T^2 = T$, but I'm guessing this question is to identify the intuitive definition with the formal one.

     – Richard Jensen Jun 07 '19 at 13:27
  • Okay, I guess it depends what direction you approach this from but it shouldn't really matter since you establish equivalence anyway. Though maybe to avoid confusion and make it self contained, it would be helpful to add in the answer precisely which definition you are using. – J_P Jun 07 '19 at 13:37
0

There are lots of such linear maps. Given a decomposition of the space $V=R\oplus K$, we have a unique $T:V\to V$ such that $T(V)=R$ and $\ker(T)=K$ (which is the projection determined by the decomposition).

So what you really want to classify are all decompositions of $V$ as a (ordered) direct sum of subspaces. Given $0\leq k\leq \dim V$, the Grassmannian $Gr_k(V)$ is formed by all subspaces of $V$ with dimension $k$.

On the other hand, there is a nice way of finding all complements of a fixed subspace $R\subseteq V$, using a fixed complement $R^\perp$ (coming from a inner product, for example): every complement of $R\subseteq V$ is the graphic of a linear map $A:R^\perp\to R$.

In conclusion, there is a bijection between maps $T\in End(V)$ with $T^2=T$ and the set $$\bigcup_{k=1}^{\dim V}\left(\bigcup_{R\in Gr_k(V)} Hom(R^\perp;R)\right).$$

Considering that each $Gr_k(V)$ is a manifold and each $\bigcup_{R\in Gr_k(V)} Hom(R^\perp;R)$ is a bundle over $Gr_k(V)$, I'd say there are lots of solutions.

Hudson Lima
  • 111