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To find a basis for the image (or range) of the linear transformation \( T \), we need to determine the column space of the matrix \( A \):

\[
A = \begin{pmatrix}
1 & 2 & 1 \\
0 & 1 & 3 \\
-1 & 3 & 4
\end{pmatrix}
\]

The image of \( T \) is the span of the column vectors of \( A \). To find a basis for this image, we need to determine which columns of \( A \) are linearly independent.

### Step 1: Set up the matrix with its columns

\[
\mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \quad \mathbf{v}_2 = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix}
\]

### Step 2: Determine linear independence

We need to check if the vectors \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \) are linearly independent. To do this, we form the matrix \( B \) with these vectors as columns and perform row reduction:

\[
B = \begin{pmatrix}
1 & 2 & 1 \\
0 & 1 & 3 \\
-1 & 3 & 4
\end{pmatrix}
\]

### Step 3: Perform row reduction on \( B \)

We'll row reduce \( B \) to its row echelon form (REF):

1. Add row 1 to row 3 to eliminate the first entry in the third row:

\[
\left[\begin{array}{ccc}
1 & 2 & 1 \\
0 & 1 & 3 \\
0 & 5 & 5 \\
\end{array}\right]
\]

2. Subtract 5 times row 2 from row 3 to eliminate the second entry in the third row:

\[
\left[\begin{array}{ccc}
1 & 2 & 1 \\
0 & 1 & 3 \\
0 & 0 & -10 \\
\end{array}\right]
\]

3. Finally, divide the third row by -10 to simplify:

\[
\left[\begin{array}{ccc}
1 & 2 & 1 \\
0 & 1 & 3 \\
0 & 0 & 1 \\
\end{array}\right]
\]

### Step 4: Interpret the results

Since we have three non-zero rows in the row-reduced matrix, all three vectors \( \mathbf{v}_1 \), \( \mathbf{v}_2 \), and \( \mathbf{v}_3 \) are linearly independent. Therefore, all three columns of \( A \) form a basis for the image of \( T \).

### Conclusion

The basis for the image of \( T \) is:

\[
\left\{ \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}, \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix} \right\}
\]

This means that the image of \( T \) spans a 3-dimensional space in \( \mathbb{R}^3 \), indicating that \( T \) is surjective (onto).

Would you like any further details or explanations?

Here are five follow-up questions that can further your understanding:
1. What is the geometric interpretation of the image of a linear transformation?
2. How does the rank of the matrix relate to the dimension of the image?
3. What are the implications if two columns were linearly dependent?
4. Can the image of \( T \) ever be the whole space \( \mathbb{R}^3 \)?
5. How can the rank-nullity theorem be applied to understand the kernel and image together?

**Tip:** Remember that the rank of a matrix is the dimension of its column space, which corresponds to the number of linearly independent columns.

Discover how to find a basis for the image of a linear transformation with a detailed solution. This problem explores concepts in linear algebra, including the column space and basis. Suitable for undergraduate students studying linear algebra.

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