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Every straight line remains straight after a linear transformation. If \(z\) is halfway between \(x\) and \(y\), show that \(A z\) is halfway between \(A x\) and \(A y\).

Short Answer

Expert verified
Az is halfway between Ax and Ay due to the linearity of transformation A.

Step by step solution

01

Express z in Terms of x and y

Since \(z\) is halfway between \(x\) and \(y\), we can write \(z\) as \(z = \frac{x + y}{2}\). This expresses \(z\) as the average of \(x\) and \(y\).
02

Apply the Linear Transformation A

Apply the linear transformation \(A\) to the expression for \(z\): \(Az = A\left(\frac{x+y}{2}\right)\). Due to the properties of linearity, we can distribute \(A\) over the addition and scalar multiplication.
03

Use Linearity Property

Use the linearity property of transformation \(A\), which allows us to rewrite \(Az = A\left(\frac{x+y}{2}\right) = \frac{1}{2}(Ax + Ay)\). This shows that \(A z\) can be expressed as the average of \(Ax\) and \(Ay\).
04

Conclude the Proof

Since \(A z = \frac{1}{2}(Ax + Ay)\), it confirms that \(A z\) is halfway between \(A x\) and \(A y\). Therefore, \(A z\) maintains the halfway property, solidifying the property of linear transformations preserving straight lines as midpoint.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linearity Property
The linearity property is a fundamental characteristic of linear transformations. It allows the transformation to be applied separately to each vector and for the results to be added together. This power of linearity underpins the behavior and predictability of linear transformations.
  • Vector Addition: If a transformation is linear, like our transformation \( A \), it satisfies \( A(x + y) = Ax + Ay \). This means you can add vectors first, then transform, or transform each vector and then add the results.
  • Scalar Multiplication: Linearity also applies when multiplying vectors by scalars. This is shown as \( A(kx) = kAx \), where \( k \) is a scalar. This means the transformation \( A \) distributes over the scalar multiplication, maintaining its form.
These properties guarantee that if you took a combination of vectors and scalars before a linear transformation, the result would remain true after the transformation is applied.
Midpoint
The concept of the midpoint is crucial in understanding linear transformations and their effects. A midpoint is a point that is exactly halfway between two points. Mathematically, the midpoint \( z \) between two vectors \( x \) and \( y \) is defined as \( z = \frac{x + y}{2} \). This formula indicates that \( z \) is the average of \( x \) and \( y \), equally spaced from both.
In the context of linear transformations, maintaining the position of midpoints is vital. As shown in our exercise, applying a linear transformation to a midpoint results in a new midpoint among the transformed vectors. This confirms that linear transformations preserve the relative distances and positions among vectors, preserving linear structures.
Vector Addition
Vector addition is a key operation in linear algebra and works by combining corresponding components of two vectors. Let's say you have two vectors, \( x = (x_1, x_2) \) and \( y = (y_1, y_2) \). Their sum \( z \) is calculated as \( z = (x_1 + y_1, x_2 + y_2) \).
  • Application: Vector addition allows for defining new vectors like the midpoint, which is the average of vectors \( x \) and \( y \).
  • Linearity: When dealing with linear transformations, vector addition's role becomes pivotal. In transformations, like \( A \), performed on sums of vectors, each component can be treated individually, and results can be aggregated for an overall effect.
Through vector addition, we maintain structured manipulation of data in geometrical and algebraic contexts.
Scalar Multiplication
Scalar multiplication involves multiplying each component of a vector by a constant or scalar. If you have a vector \( x = (x_1, x_2) \) and a scalar \( k \), then the scalar multiplication \( kx \) gives \( (kx_1, kx_2) \).
This operation is fundamental in linear transformations. In our context, applying a transformation like \( A \) to a scalar multiple of a vector \( x \) results in \( A(kx) = kAx \). This property demonstrates that scaling a vector before or after applying a linear transformation will yield consistent results.
Scalar multiplication is crucial for ensuring linear transformations correctly adjust distances and directions, retaining proportional relationships within and across vectors.

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Most popular questions from this chapter

Find a basis for the plane \(x-2 y+3 z=0\) in \(\mathbf{R}^{3}\). Then find a basis for the intersection of that plane with the \(x y\)-plane. Then find a basis for all vectors perpendicular to the plane.

The product \((A B) C\) of linear transformations starts with a vector \(x\) and produces \(u=C x\). Then rule \(2 \mathrm{~V}\) applies \(A B\) to \(u\) and reaches \((A B) C x\). (a) Is this result the same as separately applying \(C\) then \(B\) then \(A\) ? (b) Is the result the same as applying \(B C\) followed by \(A\) ? Parentheses are unnecessary and the associative law \((A B) C=A(B C)\) holds for linear transformations. This is the best proof of the same law for matrices.

The space of all 2 by 2 matrices has the four basis "vectors" $$ \left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right], \quad\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right], \quad\left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right], \quad\left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] $$ For the linear transformation of transposing, find its matrix \(A\) with respect to this basis. Why is \(A^{2}=1 ?\)

A nonlinear transformation is invertible if \(T(x)=b\) has exactly one solution for every \(b\). The example \(T(x)=x^{2}\) is not invertible because \(x^{2}=b\) has two solutions for positive \(b\) and no solution for negative \(b\). Which of the following transformations (from the real numbers \(\mathbf{R}^{1}\) to the real numbers \(\mathbf{R}^{1}\) ) are invertible? None are linear, not even (c). (a) \(T(x)=x^{3}\). (b) \(T(x)=e^{x}\). (c) \(T(x)=x+11\). (d) \(T(x)=\cos x\).

For these transformations of \(\mathbf{V}=\mathbf{R}^{2}\) to \(\mathbf{W}=\mathbf{R}^{2}\), find \(T(T(v))\). (a) \(T(v) \Leftrightarrow=-v\). (b) \(T(v)=v+(1,1)\). (c) \(T(v)=90^{\circ}\) rotation \(=\left(-v_{2}, v_{1}\right)\). (d) \(T(v)=\) projection \(=\left(\frac{v_{1}+v_{2}}{2}, \frac{v_{1}+v_{2}}{2}\right)\).

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