Introduction

A linear transformation is a mapping between two vector spaces that preserves the operations of vector addition and scalar multiplication. In this discussion, we will explore the concept of linear transformations by finding the pre-images of certain vectors under a given transformation.

C31: Problem Statement

Consider the linear transformation \(S: \mathbb{C}^3 \rightarrow \mathbb{C}^3\) defined by:

\[ S \left( \begin{bmatrix} a \\ b \\ c \end{bmatrix} \right) = \begin{bmatrix} a - 2b - c \\ 3a - b + 2c \\ a + b + 2c \end{bmatrix} \]

We aim to find the pre-images of the following vectors:

  1. \(\begin{bmatrix} -2 \\ 5 \\ 3 \end{bmatrix}\)
  2. \(\begin{bmatrix} -5 \\ 5 \\ 7 \end{bmatrix}\)

Solution for the First Vector

To find the pre-image of \(\begin{bmatrix} -2 \\ 5 \\ 3 \end{bmatrix}\), we set up the system of equations given by:

\[\begin{align*} a - 2b - c &= -2, \\ 3a - b + 2c &= 5, \\ a + b + 2c &= 3. \end{align*}\]

We represent this system as an augmented matrix and perform row reduction to find the reduced row echelon form (RREF):

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

Steps to RREF for the First Vector

  1. Start with the original matrix.
  2. Subtract 3 times the first row from the second row.
  3. Subtract the first row from the third row.
  4. Swap the second and third rows for convenience.
  5. Divide the new second row by its leading coefficient to get a leading 1.
  6. Add the appropriate multiple of the second row to the first row to clear the second column.
  7. Add the appropriate multiple of the second row to the third row to clear the second column.
  8. Observe the contradiction in the third row, indicating no solution.

The RREF of this matrix is:

\[ \begin{bmatrix} 1 & 0 & 1 & |0 \\ 0 & 1 & 1 & |0 \\ 0 & 0 & 0 & |1 \\ \end{bmatrix} \]

The third row corresponds to the equation \(0 = 1\), which is a contradiction. Thus, the system has no solution, and there is no pre-image for the given vector under the transformation \(S\).

Solution for the Second Vector

For the vector \(\begin{bmatrix} -5 \\ 5 \\ 7 \end{bmatrix}\), we set up a similar system of equations:

\[\begin{align*} a - 2b - c &= -5, \\ 3a - b + 2c &= 5, \\ a + b + 2c &= 7. \end{align*}\]

The augmented matrix and its RREF are:

\[ \begin{bmatrix} 1 & -2 & -1 & |-5 \\ 3 & -1 & 2 & |5 \\ 1 & 1 & 2 & |7 \\ \end{bmatrix} \]

Steps to RREF for the Second Vector

  1. Start with the original matrix.
  2. Subtract 3 times the first row from the second row.
  3. Subtract the first row from the third row.
  4. Swap the second and third rows.
  5. Scale the new second row to get a leading 1.
  6. Add the appropriate multiple of the second row to the first row to clear the second column.
  7. Add the appropriate multiple of the second row to the third row to clear the second column.
  8. The last row becomes all zeros, which is expected for a free variable in the solution.

The RREF of this matrix is:

\[ \begin{bmatrix} 1 & 0 & 1 & |3 \\ 0 & 1 & 1 & |4 \\ 0 & 0 & 0 & |0 \\ \end{bmatrix} \]

From the RREF, we can deduce that \(c\) is a free variable and \(a\) and \(b\) can be expressed in terms of \(c\):

\[\begin{align*} a + c &= 3, \\ b + c &= 4. \end{align*}\]

Thus, we have:

\[\begin{align*} a &= 3 - c, \\ b &= 4 - c. \end{align*}\]

Any value for \(c\) will give us a pre-image of \(\begin{bmatrix} -5 \\ 5 \\ 7 \end{bmatrix}\) under \(S\), forming a line of pre-images in \(\mathbb{C}^3\).

Conclusion

The process of finding pre-images under a linear transformation involves solving a system of linear equations. The solution set can vary from no solution to a unique solution, or even infinitely many solutions, depending on the transformation and the vector in question. This exercise has demonstrated the power of linear algebra in understanding such transformations and their potential solution spaces.