Question

If \(A\) is \(3 \times 3\) and invertible, show that the image of the plane through the origin with normal \(\mathbf{n}\) is the plane through the origin with normal \(\mathbf{n}_{1}=B \mathbf{n}\) where \(B=\left(A^{-1}\right)^{T}\). [Hint: Use the fact that \(\mathbf{v} \cdot \mathbf{w}=\mathbf{v}^{T} \mathbf{w}\) to show that \(\mathbf{n}_{1} \cdot(A \mathbf{p})=\mathbf{n} \cdot \mathbf{p}\) for each \(\mathbf{p}\) in \(\mathbb{R}^{3}\).]

Short answer

The image is the same type of plane, with normal \( \mathbf{n}_1 = B\mathbf{n} \), maintaining origin passage.

Step-by-step solution

Understand the Plane Description

The plane through the origin can be described with the points \( \mathbf{p} \) such that \( \mathbf{n} \cdot \mathbf{p} = 0 \). Here, \( \mathbf{n} \) is normal to the plane.

Image of the Plane

To find the image of this plane under the transformation by a matrix \( A \), we need to look at the points that map to \( A \mathbf{p} \) and satisfy \( \mathbf{n} \cdot \mathbf{p} = 0 \).

Transformed Normal Vector

The problem states that the normal vector to the transformed plane is \( \mathbf{n}_1 = B \mathbf{n} \), where \( B = \left(A^{-1}\right)^T \). We need to verify that this transformation keeps the plane through the origin.

Use the Dot Product Identity

Recall that the dot product \( \mathbf{v} \cdot \mathbf{w} = \mathbf{v}^T \mathbf{w} \). For the plane through the transformed origin, we want \( \mathbf{n}_1 \cdot (A \mathbf{p}) = 0 \).

Show the Transformation Property

Substitute \( \mathbf{n}_1 = B \mathbf{n} \) into the expression. We need to show that \( \mathbf{n}_1 \cdot (A \mathbf{p}) = \mathbf{n} \cdot \mathbf{p} \) for any \( \mathbf{p} \). Start by calculating:\[ \mathbf{n}_1 \cdot (A \mathbf{p}) = (B \mathbf{n}) \cdot (A \mathbf{p}) = (B \mathbf{n})^T (A \mathbf{p}) = \mathbf{n}^T B^T A \mathbf{p}. \]

Simplify the Expression

Since \( B = (A^{-1})^T \), thus \( (B^T) = A^{-1} \). This gives us:\[ \mathbf{n}^T A^{-1} A \mathbf{p} = \mathbf{n}^T \mathbf{p}. \]The expression simplifies to demonstrate that \( \mathbf{n} \cdot \mathbf{p} \), confirming that this is still the same plane description.

Conclusion

From the calculation, it shows that the image of the plane under the transformation by \( A \) is another plane through the origin with normal vector \( \mathbf{n}_1 = B \mathbf{n} \), where \( B = (A^{-1})^T \), maintaining the plane's definition.

Key concepts

Invertible Matrices

Invertible matrices play a crucial role in linear algebra. An invertible or non-singular matrix is one that has an inverse. If you have a square matrix, like our 3x3 matrix \(A\), and it is invertible, it means there exists another matrix \(A^{-1}\) such that multiplying \(A\) and \(A^{-1}\) yields the identity matrix. Here’s why this is important:

• The process of finding the inverse allows us to reverse the effect of a linear transformation represented by the matrix.
• For any invertible matrix \(A\), any transformation it performs is reversible, making the image and pre-image aspects particularly interesting in geometric transformations.

When you apply such transformations to a plane, the properties of the plane can be preserved, which is exactly what happens in our earlier problem: the direction of the normal vector is transformed while the plane's equations still hold.

Normal Vector

A normal vector to a plane is a vector that is perpendicular to every vector in that plane. It provides crucial information as it completely determines the orientation of the plane. The normal vector \(\mathbf{n}\) in our initial plane serves a key role:

• It helps define the plane as all points \(\mathbf{p}\) must satisfy the condition \(\mathbf{n} \cdot \mathbf{p} = 0\).
• This condition tells us that the vector \(\mathbf{p}\) lies in the plane when the dot product equals zero.

In the problem, the normal vector undergoes transformation via a matrix \(B = (A^{-1})^T\), resulting in a new normal vector \(\mathbf{n}_1\). This new normal vector maintains the plane's orientation after the linear transformation applied by the invertible matrix \(A\).

Dot Product

The dot product is a mathematical operation that returns a scalar quantity. For two vectors \(\mathbf{v}\) and \(\mathbf{w}\), their dot product is \(\mathbf{v} \cdot \mathbf{w} = \mathbf{v}^T \mathbf{w}\). Here's why it's fundamental in this context:

• It provides a method for checking orthogonality. When the dot product is zero, the vectors involved are perpendicular.
• In the problem, using the dot product lets us check if the transformed normal vector \(\mathbf{n}_1\) is orthogonal to \(A \mathbf{p}\).

The steps outlined in the exercise show us how to use the property that \(\mathbf{v} \cdot \mathbf{w} = \mathbf{v}^T \mathbf{w}\) to demonstrate that even after the transformation by matrix \(A\), the orientation properties of the plane are preserved.

Image of a Plane

When a plane undergoes linear transformation via a matrix, the image of the plane refers to its new position and orientation in space. A key aspect of this transformation is that if \(A\) is invertible, the transformation is reversible:

• The image of the plane through the origin results from applying the matrix \(A\) to every point \(\mathbf{p}\) on the original plane.
• Because of the invertibility of \(A\), the transformed plane remains a plane through the origin, just with a new normal vector \(\mathbf{n}_1\).

This transformation doesn’t change the fundamental characteristics of the plane; it simply changes its orientation. This is a crucial insight when exploring how transformations affect geometric entities in linear algebra.