Question

(i) Let \(M\) be a subspace of a vector space \(V\). Show that there is a linear projection of \(V\) onto \(M\) (i.e., \(\left.P\right|_{M}=I_{M}\) and \(\left.P(V)=M\right)\). (ii) Show that if a linear map \(P: V \rightarrow V\) satisfies \(P^{2}=P\), then \(V=P(V) \oplus\) \(\operatorname{Ker}(P)\). Moreover, \(Q=I_{V}-P\) is a projection such that \(Q(V)=\operatorname{Ker}(P)\) and \(\operatorname{Ker}(Q)=P(V)\).

Short answer

For (i), define \(P\) such that \(P^2 = P\) and \(P(V) = M\). For (ii), \(P^2 = P\) implies \(V = P(V) \ \text{Ker}(P)\) with \(Q = I_V - P\) projecting onto \(\text{Ker}(P)\).

Step-by-step solution

Understand the Definitions and Given Information

We'll work with a vector space \(V\) and a subspace \(M\). We need to show the existence of a linear projection \(P\) such that \(P|_M = I_M\) and \(P(V) = M\).

Define a Linear Projection

A projection \(P: V \rightarrow V\) is a linear map that satisfies \(P^2 = P\). This means applying \(P\) twice is the same as applying it once.

Construct the Projection

For each vector \(v otin M\), consider its decomposition \(v = m + k\) where \(m \ M\) and \(k \ M\)'s complement in \(V\). Define \(P(v) = m\). This ensures that \(P(m) = m\) for all \(m \ M\).

Verify the Properties of P

We need to verify that \(P|_M = I_M\), so for any vector \(m \ M\), \(P(m) = m\). Also, \(P(V) = M\), because \(P(v) = m\) gives vectors only in \(M\).

Show that P^2 = P

For a vector \(v \ V\), write \(v = m + k\) again. Then, \(P(v) = m\) and applying \(P\) again to \(m\) gives \(P(m) = m\). So, \(P(P(v)) = P(m) = m = P(v)\) implying \(P^2 = P\).

Showing Direct Sum Decomposition for Part (ii)

Given \(P^2 = P\), let \(W = \text{Ker}(P)\). We have \(V = \text{Im}(P) \ W\). For any \(v \ V\), write \(v = P(v) + (v - P(v))\) where \(P(v) \ \text{Im}(P)\) and \(v - P(v) \ W\). Thus, \(V = P(V) \ W\).

Define Q and Show It is a Projection

Let \(Q = I_V - P\). Then, \(Q^2 = (I_V - P)(I_V - P) = I_V - P - P + P^2 = I_V - P = Q\). Since \(P + Q = I_V\), we have \(Q(V) = \text{Ker}(P)\) and \(\text{Ker}(Q) = \text{Im}(P)\).

Key concepts

Vector Space Subspaces

A vector space is a collection of vectors where vector addition and scalar multiplication are defined. A **subspace** is a subset of this vector space that is also a vector space on its own, under the same operations.

For example, consider a vector space **V**. A subset **M** is a subspace if it meets these conditions:

• The zero vector is in **M**.
• **M** is closed under vector addition. This means if **u** and **v** are in **M**, then **u + v** is also in **M**.
• **M** is closed under scalar multiplication. This means if **u** is in **M** and **c** is a scalar, then **c * u** is also in **M**.

In simpler terms, any linear combination of vectors in **M** must still be within **M**.

Linear Maps

A **linear map** (or linear transformation) between two vector spaces is a function that preserves the operations of vector addition and scalar multiplication. If **V** and **W** are vector spaces, a function **T** from **V** to **W** is linear if for all vectors **u, v ∈ V** and all scalars **c**, the following hold:

• **T(u + v) = T(u) + T(v)**. This shows that **T** is additive.
• **T(c * u) = c * T(u)**. This shows that **T** is homogeneous.

Projection is a special linear map. A **projection** **P** on **V** is a linear map that satisfies **P^2 = P**. This means, applying **P** twice is the same as applying it once. **Projections** are useful in breaking down vector spaces into simpler parts.

Direct Sum Decomposition

The idea of direct sum decomposition involves splitting a vector space into two smaller, easier-to-manage pieces. If **V** can be written as a sum of two subspaces **A** and **B**, we say **V = A ⊕ B** (direct sum notation) if every vector **v** in **V** can be uniquely written as **v = a + b** where **a ∈ A** and **b ∈ B**.

In our exercise, for a projection **P** on **V**, we use the image (range) and kernel (null space) of **P**:

• **Im(P)**, the image of **P**, is the subset of **V** consisting of all vectors that **P** maps onto. Basically, every **P(v)** is in **Im(P)**. If **P** projects onto **M**, then **Im(P) = M**.
• **Ker(P)**, the kernel of **P**, is the subset of **V** consisting of all vectors that **P** maps to zero. That is, all vectors **v** such that **P(v) = 0** are in **Ker(P)**.

We can decompose **V** as a direct sum of **Im(P)** and **Ker(P)**, denoted **V = Im(P) ⊕ Ker(P)**, when **V** can be explicitly broken down into these components. Importantly, any vector **v** in **V** can be uniquely written as **v = P(v) + (v - P(v))** where **P(v)** is in the image and **(v - P(v))** is in the kernel.