Let a,b∈R
 and let u
, v
, w∈Rn
, where v≠0
.  Show that

projv(au+bw)=a(projvu)+b(projvw)

To show that

\[
\text{proj}_{\mathbf{v}}(a\mathbf{u} + b\mathbf{w}) = a \, \text{proj}_{\mathbf{v}}\mathbf{u} + b \, \text{proj}_{\mathbf{v}}\mathbf{w},
\]

where \( a, b \in \mathbb{R} \), \(\mathbf{u}, \mathbf{v}, \mathbf{w} \in \mathbb{R}^n \), and \(\mathbf{v} \neq 0\), we will use the definition of the projection of a vector onto another vector.

### Step 1: Recall the definition of the projection

The projection of a vector \(\mathbf{u}\) onto a non-zero vector \(\mathbf{v}\) is given by:

\[
\text{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v},
\]

where \(\mathbf{u} \cdot \mathbf{v}\) is the dot product between \(\mathbf{u}\) and \(\mathbf{v}\), and \(\mathbf{v} \cdot \mathbf{v}\) is the dot product of \(\mathbf{v}\) with itself.

### Step 2: Apply the projection to the left-hand side

We want to compute \(\text{proj}_{\mathbf{v}}(a\mathbf{u} + b\mathbf{w})\). Using the definition of the projection:

\[
\text{proj}_{\mathbf{v}}(a\mathbf{u} + b\mathbf{w}) = \frac{(a\mathbf{u} + b\mathbf{w}) \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.
\]

### Step 3: Expand the dot product

The dot product is linear, so we can distribute it over the sum:

\[
(a\mathbf{u} + b\mathbf{w}) \cdot \mathbf{v} = a (\mathbf{u} \cdot \mathbf{v}) + b (\mathbf{w} \cdot \mathbf{v}).
\]

Thus, we have:

\[
\text{proj}_{\mathbf{v}}(a\mathbf{u} + b\mathbf{w}) = \frac{a (\mathbf{u} \cdot \mathbf{v}) + b (\mathbf{w} \cdot \mathbf{v})}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.
\]

### Step 4: Separate the terms

We can now separate the terms involving \(a\) and \(b\):

\[
\text{proj}_{\mathbf{v}}(a\mathbf{u} + b\mathbf{w}) = a \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} + b \frac{\mathbf{w} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v}.
\]

### Step 5: Recognize the projections

From the definition of the projection, we recognize the terms on the right-hand side as projections:

\[
\text{proj}_{\mathbf{v}}(a\mathbf{u} + b\mathbf{w}) = a \, \text{proj}_{\mathbf{v}}\mathbf{u} + b \, \text{proj}_{\mathbf{v}}\mathbf{w}.
\]

Thus, we have shown that:

\[
\text{proj}_{\mathbf{v}}(a\mathbf{u} + b\mathbf{w}) = a \, \text{proj}_{\mathbf{v}}\mathbf{u} + b \, \text{proj}_{\mathbf{v}}\mathbf{w}.
\]

### Conclusion

We have successfully demonstrated that the projection operation is linear with respect to scalar multiplication and vector addition.

---

Would you like any further clarification on any steps or more details?

Here are 5 related questions:
1. How does the projection formula change if \(\mathbf{v}\) is a unit vector?
2. What properties must a vector \(\mathbf{v}\) satisfy for the projection operation to be well-defined?
3. Can we extend this result to complex vectors? How does the dot product change?
4. What is the geometric interpretation of the projection formula?
5. How does this projection relate to the concept of orthogonal decomposition?

**Tip:** The projection of a vector onto another vector minimizes the distance between the original vector and the projected vector along the direction of the second vector.

Let a, b ∈ R and let u, v, w ∈ Rn, where v ≠ 0. Show that projv(au+bw) = a(projvu) + b(projvw).

This solution demonstrates the proof that vector projections are linear: projv(au+bw) = a(projvu) + b(projvw). The problem explores key linear algebra concepts such as dot products, vector spaces, and the linearity of projections. Suitable for undergraduate students.

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