5. Prove that if the set of $$ n $$-vectors $$ \left\{x_{i}\right\} $$ is a basis for a vector space $$ V $$, then the set $$ \left\{c_{i} x_{i}\right\} $$ is also a basis for $$ V $$, where the $$ c_{i} $$ are arbitrary nonzero scalars. Interpret this geometrically.

Question

UpStudy AI · Accepted Answer

To prove that scaling each vector in a basis by a nonzero scalar yields another basis for the vector space $$ V $$, let’s proceed step by step. We will also provide a geometric interpretation to solidify the understanding.

## Proof

**Given:**
- A set of $$ n $$-vectors $$ \left\{x_{1}, x_{2}, \dots, x_{n}\right\} $$ forms a basis for an $$ n $$-dimensional vector space $$ V $$.
- Scalars $$ c_{1}, c_{2}, \dots, c_{n} $$ are nonzero.

**To Show:**
- The set $$ \left\{c_{1} x_{1}, c_{2} x_{2}, \dots, c_{n} x_{n}\right\} $$ also forms a basis for $$ V $$.

**Proof Steps:**

1. **Basis Criteria:**
   For a set of vectors to be a basis for $$ V $$, two conditions must be satisfied:
   - **Linear Independence:** No vector in the set can be written as a linear combination of the others.
   - **Spanning:** Any vector in $$ V $$ can be expressed as a linear combination of the set.

2. **Original Basis Properties:**
   Since $$ \left\{x_{1}, x_{2}, \dots, x_{n}\right\} $$ is a basis:
   - They are **linearly independent**.
   - They **span** $$ V $$.

3. **Scaled Vectors:**
   Consider the set $$ \left\{c_{1} x_{1}, c_{2} x_{2}, \dots, c_{n} x_{n}\right\} $$.

4. **Linear Independence of Scaled Vectors:**
   Suppose a linear combination equals the zero vector:
   $$
   a_{1} (c_{1} x_{1}) + a_{2} (c_{2} x_{2}) + \dots + a_{n} (c_{n} x_{n}) ) = 0
   $$
   Factor out the scalars:
   $$
   c_{1} a_{1} x_{1} + c_{2} a_{2} x_{2} + \dots + c_{n} a_{n} x_{n} = 0
   $$
   Since $$ \left\{x_{1}, x_{2}, \dots, x_{n}\right\} $$ is linearly independent, the only solution is:
   $$
   c_{1} a_{1} = c_{2} a_{2} = \dots = c_{n} a_{n} = 0
   $$
   Given that each $$ c_{i} \neq 0 $$, it follows that:
   $$
   a_{1} = a_{2} = \dots = a_{n} = 0
   $$
   Thus, the scaled vectors are **linearly independent**.

5. **Spanning $$ V $$:**
   Any vector $$ v \in V $$ can be expressed as:
   $$
   v = d_{1} x_{1} + d_{2} x_{2} + \dots + d_{n} x_{n}
   $$
   We can rewrite this in terms of the scaled vectors:
   $$
   v = \frac{d_{1}}{c_{1}} (c_{1} x_{1}) + \frac{d_{2}}{c_{2}} (c_{2} x_{2}) + \dots + \frac{d_{n}}{c_{n}} (c_{n} x_{n})
   $$
   Let $$ a_{i} = \frac{d_{i}}{c_{i}} $$. Since $$ c_{i} \neq 0 $$, each $$ a_{i} $$ is well-defined. Therefore:
   $$
   v = a_{1} (c_{1} x_{1}) + a_{2} (c_{2} x_{2}) + \dots + a_{n} (c_{n} x_{n})
   $$
   This shows that the scaled vectors **span** $$ V $$.

6. **Conclusion:**
   Since the set $$ \left\{c_{1} x_{1}, c_{2} x_{2}, \dots, c_{n} x_{n}\right\} $$ is both linearly independent and spans $$ V $$, it is a **basis** for $$ V $$.

## Geometric Interpretation

Geometrically, scaling each basis vector by a nonzero scalar $$ c_{i} $$ stretches or shrinks the vector but does not change its direction (unless $$ c_{i} $$ is negative, in which case it also reverses the direction). Here's what this means in different dimensions:

1. **In Two Dimensions ($$ \mathbb{R}^2 $$):**
   - Consider the standard basis vectors $$ e_{1} = (1, 0) $$ and $$ e_{2} = (0, 1) $$.
   - Scaling them by nonzero scalars $$ c_{1} $$ and $$ c_{2} $$ yields $$ c_{1} e_{1} $$ and $$ c_{2} e_{2} $$.
   - The new vectors still span the plane and remain independent; essentially, the grid defined by the basis is stretched or compressed along the axes but retains its grid-like structure without collapsing into a line or a point.

2. **In Three Dimensions ($$ \mathbb{R}^3 $$):**
   - Similar to the 2D case, scaling each of the three standard basis vectors along their respective axes stretches or compresses the space along those axes.
   - The orthogonality (if present in the original basis) is maintained if scalars are positive; otherwise, some vectors may reverse direction.

3. **General $$ n $$-Dimensions:**
   - Each basis vector is scaled independently, affecting the "unit lengths" along each basis direction.
   - The overall structure of the space remains intact because scaling does not introduce linear dependencies or remove the ability to span the entire space.

**Key Point:** Scaling basis vectors by nonzero scalars changes the "scale" along each basis direction but preserves the fundamental structure of the vector space. The new vectors remain independent and can still reach every point in the space through their linear combinations.
If you scale each vector in a basis by a nonzero scalar, the new set of vectors is also a basis for the same space. This means the vectors remain independent and can still span the entire space. Geometrically, scaling stretches or compresses the space along each basis direction without changing the overall structure.

Prove that if the set of -vectors is a basis for a vector space , then the set
is also a basis for , where the are arbitrary nonzero scalars. Interpret
this geometrically.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Other Questions

Prove that if the set of -vectors is a basis for a vector space , then the set is also a basis for , where the are arbitrary nonzero scalars. Interpret this geometrically.