About $P_{{L},{M}}$, projection transformation onto subspace $L$ along subspace $M$ .

I need help to study following theorem:

For every idempotent matrix $E\in\mathbb{C}^{n\times n}$, $R(E)$ and $N(E)$ are complementary subspaces with $E = P_{{R(E)},{N(E)}}$. Conversely, if $L$ and $M$ are complementary subspaces, there is a unique idempotent $P_{{L},{M}}$ such that $R(P_{{L},{M}}) = L$, $N(P_{{L},{M}}) = M$ where $R(E)$ and $N(E)$ denotes the range and null space of $E$, respectively while $P_{{L},{M}}$ denote the transformation that carries any $x\in\mathbb{C}^{n\times n}$ into its projection on $L$ along $M$.

I am not very much clear about the term $P_{{L},{M}}$. All i know is that there is a one to one correspondence between idempotent matrices of order $n$ and the projectors $P_{{L},{M}}$ where $L\oplus M =\mathbb{C}^{n}$ is the direct sum.

I want to know in what context we are terming $P_{{L},{M}}$ as a 'Projection'. What is the significance of projecting a vector onto subspace? How to calculate projector matrices? Can we interpret it geometrically?

Any kind of help would be very much helpful for me. Thanks

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First of all, let's prove the theorem, then talk about what it means. The first direction, let $E$ be idempotent, that is $E^2 = E$. This, at least to me, seems at like a kinda opaque condition, it's not very geometric. Well, let's make it geometric.

First off, idempotence implies that anything inside the range is fixed by $E$, that is $E|_{R(E)} = id$. To see this, suppose $y \in R(E)$, write $y = E(x)$, then we have $$E(y) = E(E(x)) = E(x) = y$$

Observe that $N(E) \cap R(E) = 0$, indeed, suppose $y$ lies in both, we have $y = E(y)$ by $y \in R(E)$, and $E(y) = 0$, by $y \in N(E)$!

So far, no linear algebra. Now it comes in.

Okay, so the kernel and range only touch at 0, and by rank-nullity the sum of their dimensions is $n$, so if we choose a basis $e_1, \ldots e_k$ for the kernel, and a basis $f_1, \ldots f_{n-k}$ for the range, then $$e_1 \ldots e_k, f_1, \ldots f_{n-k}$$form a basis for $\mathbb{C}^n$ (this really has nothing to do with $\mathbb{C}$, you could replace it with $\mathbb{R}$, or $\mathbb{Q}$, etc.), and our map $E$ now looks like $$\sum \lambda_i e_i + \sum \lambda_j f_j \mapsto \sum \lambda_j f_j$$i.e. we forget our coordinates in the kernel. As a matrix, this looks like 0's everywhere except for 1s along the diagonal for the coordinates $f_j$. For example, if $k = n-k = 1$, we have $$\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$or if say $k = 1, n-k = 2$, we have $$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$The reason why it's called projection is that we take a vector $v$ and send it to its shadow in $R(E)$, for example if $k = n-k = 1$ as above, we take a vector $(a,b)$ and send it to $(a,0)$. Pictorially, the vector forms the hypotenuse of the right triangle with vertices $\{(0,0), (a,0), (a,b)\}$, we're "squishing" the hypotenuse down to one leg with vertices $\{(0,0), (a,0)\}$.

In the other direction, about $L$ and $M$ complementary subspaces, we can form an idempotent by writing down a matrix just like those above, namely take a basis $e_1, \ldots e_k$ for $L$ and $f_1, \ldots f_{n-k}$ for $M$, and write down the same matrix.

So the takeaway is this: the condition $E^2 = E$ isn't that mysterious, it just amounts to choosing a subspace $M$, to "cast shadows onto", and a subspace $N$, to determine "where the sun sits in the sky", i.e. which complementary subspace casts no shadow.

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Have no words to express my thanks. Your last para tells everything i was looking for. –  srijan May 16 '12 at 16:45
dawww you're very welcome! –  uncookedfalcon May 17 '12 at 3:07