Determine a basis and the dimension of the following subspace of R3R3:


span{[2,1,2],[1,−3,2],[4,−5,6]}span⁡{[2,1,2],[1,−3,2],[4,−5,6]}


Enter the basis as a list {a⃗ ,b⃗ ,…}{a→,b→,…} of vectors of the same form as above.

basis:  

dimension:

To determine the basis and the dimension of the subspace of \(\mathbb{R}^3\) spanned by the set of vectors \(\text{span}\{[2,1,2], [1,-3,2], [4,-5,6]\}\), we need to check if these vectors are linearly independent. If they are, they will form a basis for the subspace, and the dimension will be equal to the number of linearly independent vectors.

### Step 1: Form the matrix
Write the vectors as rows of a matrix \(A\):

\[
A = \begin{bmatrix}
2 & 1 & 2 \\
1 & -3 & 2 \\
4 & -5 & 6
\end{bmatrix}
\]

### Step 2: Perform row reduction (Gaussian elimination)

We will row reduce the matrix to its row echelon form to determine the number of pivot columns, which will give us the rank of the matrix (i.e., the number of linearly independent vectors).

Start with the matrix \( A \):

\[
\begin{bmatrix}
2 & 1 & 2 \\
1 & -3 & 2 \\
4 & -5 & 6
\end{bmatrix}
\]

**Row operations:**

1. Subtract \(\frac{1}{2}\) times the first row from the second row:

\[
\begin{bmatrix}
2 & 1 & 2 \\
1 - \frac{1}{2}(2) & -3 - \frac{1}{2}(1) & 2 - \frac{1}{2}(2) \\
4 & -5 & 6
\end{bmatrix}
=
\begin{bmatrix}
2 & 1 & 2 \\
0 & -\frac{7}{2} & 1 \\
4 & -5 & 6
\end{bmatrix}
\]

2. Subtract 2 times the first row from the third row:

\[
\begin{bmatrix}
2 & 1 & 2 \\
0 & -\frac{7}{2} & 1 \\
4 - 2(2) & -5 - 2(1) & 6 - 2(2)
\end{bmatrix}
=
\begin{bmatrix}
2 & 1 & 2 \\
0 & -\frac{7}{2} & 1 \\
0 & -7 & 2
\end{bmatrix}
\]

3. Divide the second row by \(-\frac{7}{2}\):

\[
\begin{bmatrix}
2 & 1 & 2 \\
0 & 1 & -\frac{2}{7} \\
0 & -7 & 2
\end{bmatrix}
\]

4. Add 7 times the second row to the third row:

\[
\begin{bmatrix}
2 & 1 & 2 \\
0 & 1 & -\frac{2}{7} \\
0 & 0 & 0
\end{bmatrix}
\]

### Step 3: Interpret the row echelon form

The row echelon form of the matrix is:

\[
\begin{bmatrix}
2 & 1 & 2 \\
0 & 1 & -\frac{2}{7} \\
0 & 0 & 0
\end{bmatrix}
\]

Since there are two non-zero rows, the rank of the matrix is 2. This means there are 2 linearly independent vectors, so the dimension of the subspace is 2.

### Step 4: Identify the basis vectors

The pivot columns correspond to the first and second vectors in the original set. Therefore, a basis for the subspace is given by:

\[
\left\{
\begin{bmatrix}
2 \\
1 \\
2
\end{bmatrix}, 
\begin{bmatrix}
1 \\
-3 \\
2
\end{bmatrix}
\right\}.
\]

### Final Answer

- **Basis**: \(\left\{
\begin{bmatrix}
2 \\
1 \\
2
\end{bmatrix}, 
\begin{bmatrix}
1 \\
-3 \\
2
\end{bmatrix}
\right\}\)
- **Dimension**: 2

Would you like to explore more about how this result was obtained?

### Relative Questions:
1. How would the process differ if the dimension of the space was larger?
2. Can you show how to determine if additional vectors are linearly independent?
3. How can you use the Gram-Schmidt process to find an orthonormal basis?
4. What happens to the basis if the span includes a linearly dependent vector?
5. How would the calculation change if you were working in \(\mathbb{R}^4\) instead of \(\mathbb{R}^3\)?

### Tip:
When working with spans and subspaces, reducing the corresponding matrix to row echelon form is a reliable method to determine linear independence and identify the basis vectors.

Discover how to determine the basis and dimension of a subspace in R^3 spanned by vectors [2,1,2], [1,-3,2], and [4,-5,6]. Learn about linear independence, row reduction techniques, and the Rank-Nullity Theorem. Suitable for advanced undergraduate students studying linear algebra.

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