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### Problem:
Show that unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are linearly independent and that they span the space \( \mathbb{R}^3 \).

### Step 1: **Definition of Unit Vectors**
In three-dimensional space, the unit vectors are defined as:
\[
\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \quad
\mathbf{j} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \quad
\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\]
These are standard basis vectors in \( \mathbb{R}^3 \).

### Step 2: **Linear Independence**
A set of vectors is linearly independent if the only solution to the equation
\[
c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k} = \mathbf{0}
\]
is \( c_1 = c_2 = c_3 = 0 \), where \( \mathbf{0} \) is the zero vector in \( \mathbb{R}^3 \).

Substitute the unit vectors into the equation:
\[
c_1 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} +
c_2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} +
c_3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\]

This gives the system of equations:
\[
\begin{aligned}
c_1 &= 0 \\
c_2 &= 0 \\
c_3 &= 0
\end{aligned}
\]
The only solution is \( c_1 = c_2 = c_3 = 0 \), so the vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are linearly independent.

### Step 3: **Spanning \( \mathbb{R}^3 \)**
A set of vectors spans \( \mathbb{R}^3 \) if any vector \( \mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \in \mathbb{R}^3 \) can be written as a linear combination of \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). In other words, we need to show that there exist scalars \( c_1, c_2, c_3 \) such that:
\[
\mathbf{v} = c_1 \mathbf{i} + c_2 \mathbf{j} + c_3 \mathbf{k}
\]

Substituting the unit vectors, this becomes:
\[
\begin{pmatrix} x \\ y \\ z \end{pmatrix} = 
c_1 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} +
c_2 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} +
c_3 \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}
\]
This gives the system:
\[
\begin{aligned}
x &= c_1 \\
y &= c_2 \\
z &= c_3
\end{aligned}
\]
Thus, for any vector \( \mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \in \mathbb{R}^3 \), the scalars \( c_1 = x, c_2 = y, c_3 = z \) provide the required linear combination. Therefore, \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) span \( \mathbb{R}^3 \).

### Conclusion:
The unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are linearly independent, and they span the space \( \mathbb{R}^3 \).

---

Would you like further clarification on any step? Here are five related questions to expand this concept:

1. How do you prove linear independence for more than three vectors?
2. What happens if the vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are scaled by non-unit values? Do they still span \( \mathbb{R}^3 \)?
3. Can a set of two vectors span \( \mathbb{R}^3 \)? Why or why not?
4. What is the geometric interpretation of the span of vectors in \( \mathbb{R}^3 \)?
5. How would you determine if a different set of vectors spans a subspace of \( \mathbb{R}^3 \)?

### Tip:
For any set of vectors to span \( \mathbb{R}^n \), you need exactly \( n \) linearly independent vectors.

Show that unit vectors i, j, and k are linearly independent and they span the space R^3.

Understand how the unit vectors i, j, and k in R^3 are linearly independent and span the entire space. This detailed solution breaks down key concepts such as linear independence and the spanning of vector spaces. Ideal for students studying linear algebra.

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