Find Basis Of Inverse Image: Linear Transformation Example

Hey everyone! Today, we're diving into a cool problem in linear algebra: finding the basis of the inverse image of a linear transformation. Specifically, we'll be looking at a transformation ${\psi : {\Bbb R}^4 \to {\Bbb R}^3}$ defined as:

$$\psi (x_1, x_2, x_3, x_4) = \begin{bmatrix} x_1 + x_3 + x_4 \\ -x_2 - x_4 \\ x_1 + x_2 + x_3 + 2x_4 \end{bmatrix}$$

Our goal is to find a basis for the inverse image of the zero vector, which is often called the kernel of the transformation. This means we want to find all vectors in ${{\Bbb R}^4}$ that, when transformed by ${\psi}$, result in the zero vector in ${{\Bbb R}^3}$. This is a fundamental concept in understanding linear transformations, so let's break it down step-by-step.

Understanding Linear Transformations and Inverse Images

Before we jump into the calculations, let's make sure we're all on the same page with the key concepts. A linear transformation is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. In simpler terms, it's a way to map vectors from one space to another while maintaining the underlying linear structure. Think of it as a distortion that keeps straight lines straight and the origin fixed.

The inverse image (or preimage) of a vector ${v}$ under a linear transformation ${\psi}$ is the set of all vectors in the domain that map to ${v}$ in the codomain. In our case, we're interested in the inverse image of the zero vector, which is also known as the kernel or null space of ${\psi}$. The kernel tells us about the "stuff" that gets squashed down to zero by the transformation. It’s a subspace of the domain, and its basis is what we’re after.

Setting Up the Problem: Finding the Kernel

To find the basis for the inverse image of the zero vector, we need to solve the equation ${\psi(x_1, x_2, x_3, x_4) = \mathbf{0}}$, where ${\mathbf{0}}$ is the zero vector in ${{\Bbb R}^3}$. This means we need to find all vectors ${(x_1, x_2, x_3, x_4)}$ in ${{\Bbb R}^4}$ that satisfy the following system of linear equations:

$$\begin{cases} x_1 + x_3 + x_4 = 0 \\ -x_2 - x_4 = 0 \\ x_1 + x_2 + x_3 + 2x_4 = 0 \end{cases}$$

This system of equations comes directly from the definition of our linear transformation ${\psi}$. We're essentially setting each component of the transformed vector equal to zero and then solving for the variables ${x_1, x_2, x_3,}$ and ${x_4}$. Solving this system will give us the vectors that form the kernel of ${\psi}$.

Solving the System of Linear Equations

Now comes the fun part: solving the system of equations! We can use a variety of techniques, such as Gaussian elimination or substitution. Let's use Gaussian elimination, which involves putting the system into an augmented matrix and then performing row operations to get it into row-echelon form.

The augmented matrix for our system is:

$$\begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & -1 & 0 & -1 & 0 \\ 1 & 1 & 1 & 2 & 0 \end{bmatrix}$$

Let's perform the following row operations:

1. Replace Row 3 with Row 3 - Row 1: ${R_3 \rightarrow R_3 - R_1}$

   $$\begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & -1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 1 & 0 \end{bmatrix}$$

2. Replace Row 3 with Row 3 + Row 2: ${R_3 \rightarrow R_3 + R_2}$

   $$\begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & -1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

Now our matrix is in row-echelon form. We can see that the last row is all zeros, which means we have a redundant equation. This tells us that we'll have free variables in our solution.

From the row-echelon form, we can write the following system of equations:

$$\begin{cases} x_1 + x_3 + x_4 = 0 \\ -x_2 - x_4 = 0 \end{cases}$$

Identifying Free Variables and Expressing Solutions

Notice that ${x_3}$ and ${x_4}$ do not correspond to leading ones in our row-echelon form. This means they are free variables. Let's express the other variables in terms of these free variables.

From the first equation, we have: ${x_1 = -x_3 - x_4}$ From the second equation, we have: ${x_2 = -x_4}$

Now we can write our solution vector in terms of the free variables ${x_3}$ and ${x_4}$:

$$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} -x_3 - x_4 \\ -x_4 \\ x_3 \\ x_4 \end{bmatrix} = x_3 \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -1 \\ -1 \\ 0 \\ 1 \end{bmatrix}$$

This gives us a general solution for any vector in the kernel of ${\psi}$. Any vector in the kernel can be written as a linear combination of the two vectors we've found.

Determining the Basis of the Inverse Image

From our general solution, we can directly identify a basis for the kernel of ${\psi}$. The two vectors that scale ${x_3}$ and ${x_4}$ form a basis because they are linearly independent and span the solution space (the kernel).

Therefore, a basis for the inverse image (kernel) of ${\psi}$ is:

$$\left\{ \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ -1 \\ 0 \\ 1 \end{bmatrix} \right\}$$

These two vectors are linearly independent, meaning neither can be written as a scalar multiple of the other. They also span the kernel, meaning any vector in the kernel can be written as a linear combination of these two vectors. This confirms that they form a basis for the inverse image.

Verifying the Solution

It's always a good idea to verify our solution. We can do this by plugging our basis vectors back into the original transformation ${\psi}$ and making sure they map to the zero vector.

Let's check the first basis vector {egin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}}:

$$\psi\left(\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix}\right) = \begin{bmatrix} -1 + 1 + 0 \\ -0 - 0 \\ -1 + 0 + 1 + 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Now let's check the second basis vector {egin{bmatrix} -1 \\ -1 \\ 0 \\ 1 \end{bmatrix}}:

$$\psi\left(\begin{bmatrix} -1 \\ -1 \\ 0 \\ 1 \end{bmatrix}\right) = \begin{bmatrix} -1 + 0 + 1 \\ -(-1) - 1 \\ -1 - 1 + 0 + 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$

Both basis vectors map to the zero vector, which confirms that they are indeed in the kernel of ${\psi}$. This verification step increases our confidence in the correctness of our solution.

Conclusion: Key Takeaways

So, there you have it! We successfully found a basis for the inverse image (kernel) of the given linear transformation ${\psi}$. Here’s a quick recap of the key steps:

1. Set up the problem: We set ${\psi(x_1, x_2, x_3, x_4) = \mathbf{0}}$ to find the vectors in the kernel.
2. Solve the system of equations: We used Gaussian elimination to solve the resulting system of linear equations.
3. Identify free variables: We identified the free variables and expressed the other variables in terms of them.
4. Write the general solution: We wrote the general solution in terms of the free variables, which allowed us to extract the basis vectors.
5. Determine the basis: We identified the basis vectors from the general solution.
6. Verify the solution: We verified our solution by plugging the basis vectors back into the original transformation.

Finding the basis of the inverse image is a crucial skill in linear algebra. It helps us understand the behavior of linear transformations and the structure of vector spaces. This process allows us to see which vectors are transformed to the zero vector, giving us insights into the nullity and rank of the transformation. By mastering this technique, you'll be well-equipped to tackle more advanced topics in linear algebra.

I hope this explanation was helpful! Feel free to ask any questions you have. Keep practicing, and you'll become a pro at finding bases for inverse images in no time! Understanding these concepts opens doors to more advanced topics in mathematics, engineering, and computer science. Keep exploring the fascinating world of linear algebra, guys!