Get the solution to "Show that mapping y↦reflLy is a linear transformation."

Caelan

Answered question

2021-01-19

Show that mapping

y \mapsto r e f l_{L} y

is a linear transformation.

Answer & Explanation

tabuordg

Skilled2021-01-20Added 99 answers

Given information:
$u \neq 0 \in R^{n} and L = S p a n {u} .$
For y in $R^{n}, the reflection of y in L is the point r e f l_{L} y defined by r e f l_{L} y = 2 \cdot p r o j_{L} y - y .$
$r e f l_{L} y is the sum of haty = p r o j_{L} y$ and hath - y.
Calculation:
Consider $T (y) = r e f l_{L} y .$
Substitute $2 \cdot p r o j_{L} y - y for r e f l_{L} y .$
$T (y) = 2 \cdot p r o j_{L} y - y (1)$
Apply Theorem 1 (b) as shown below.
Let u, v, and w be vectors in $R^{n}$ , and let c be a scalar. Then
$(u + v) \cdot w = u \cdot w + v \cdot w$
For any vectors y and z in $R^{n}$ , and any scalars c and d, the properties of inner product as shown below.
Substitute $c y + d z$ for y in Equation (1).
$T (c y + d z) = 2 \cdot p r o j_{L} (c y + d z) - (c y + d z)$
$= 2 (c p r o j_{L} y + d p r o j_{L} z) - (c y + d z)$
$= 2 c p r o j_{L} y - c y + 2 d p r o j_{L} z - d z$
$= c (2 p r o j_{L} y - y) + d (2 p r o j_{L} z - z)$
$= c T (y) + d T (z)$
Therefore, the mapping $y \mapsto r e f l_{L} y$ y is a linear transformation.

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