Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 26E from Chapter 1.8 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 26E
Chapter:
Problem:
Step-by-Step Solution
Given Information
We are given that the linearly independent vectors u and v are in $R^3$ and a plane which passes through the vectors . $u, v$ and $0$. The parametric equation of the plane is \[ \mathbf { x } = s \mathbf { u } + t \mathbf { v } \] We have to show that a linear transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ maps P onto a plane through 0 or onto a line through ), or just the origin in $R^3$.
Step-1:
Since T is a linear transformation, \[ \begin{aligned} T ( \mathbf { x } ) & = T ( s \mathbf { u } + t \mathbf { v } ) \\ & = T ( s \mathbf { u } ) + T ( t \mathbf { v } ) \\ & = s T ( \mathbf { u } ) + v T ( \mathbf { v } ) \end{aligned} \]
Step-2:
If T(u) and T(v) are linearly independent vectors in $R^3$, then the parametric equation of a plane in $R^3$ is: \[ T ( \mathbf { x } ) = s T ( \mathbf { u } ) + v T ( \mathbf { v } ) \] Given that the plane P passes through the vector 0 and T is a linear transformation, then $T(0)=0$. So that, the plane $T ( \mathbf { x } ) = s T ( \mathbf { u } ) + v T ( \mathbf { v } )$ passed through the origin. So, $T$ is a linear transformation from $R^3$ to $R^3$ that maps P onto a plane through 0 when $T(u)$ and $T(v)$ are linearly independent
Step-3:
In case, T(u) and T(v) are linearly dependent and both are nonzero, then one vector can be written as a linear combination of other two vectors. \[ T ( \mathbf { u } ) = c T ( \mathbf { v } ) \] A line in the direction of $T(v)$ is \[ T ( \mathbf { x } ) = \operatorname { sc } T ( \mathbf { v } ) + v T ( \mathbf { v } ) = ( s c + v ) T ( \mathbf { v } ) \] Since the line P passes through origin and T is a linear transformation, then $T ( \mathbf { x } ) = s T ( \mathbf { u } ) + v T ( \mathbf { v } )$ should pass through the origin. Therefore, there is a linear transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ that maps P onto a line passing through origin when the vectors T(u) and T(v) are linearly dependent and are nonzero.
Step-4:
If both the vectors are equal to 0, then $T(x)=0$. hence $x=0$. Thus, the parametric equation represents only origin.
Therefore, there is a linear transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ that maps P onto origin when the vectors T(u) and T(v) are both zero
https://imgur.com/uk4eeZH
We are given that the linearly independent vectors u and v are in $R^3$ and a plane which passes through the vectors . $u, v$ and $0$. The parametric equation of the plane is \[ \mathbf { x } = s \mathbf { u } + t \mathbf { v } \] We have to show that a linear transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ maps P onto a plane through 0 or onto a line through ), or just the origin in $R^3$.
Step-1:
Since T is a linear transformation, \[ \begin{aligned} T ( \mathbf { x } ) & = T ( s \mathbf { u } + t \mathbf { v } ) \\ & = T ( s \mathbf { u } ) + T ( t \mathbf { v } ) \\ & = s T ( \mathbf { u } ) + v T ( \mathbf { v } ) \end{aligned} \]
Step-2:
If T(u) and T(v) are linearly independent vectors in $R^3$, then the parametric equation of a plane in $R^3$ is: \[ T ( \mathbf { x } ) = s T ( \mathbf { u } ) + v T ( \mathbf { v } ) \] Given that the plane P passes through the vector 0 and T is a linear transformation, then $T(0)=0$. So that, the plane $T ( \mathbf { x } ) = s T ( \mathbf { u } ) + v T ( \mathbf { v } )$ passed through the origin. So, $T$ is a linear transformation from $R^3$ to $R^3$ that maps P onto a plane through 0 when $T(u)$ and $T(v)$ are linearly independent
Step-3:
In case, T(u) and T(v) are linearly dependent and both are nonzero, then one vector can be written as a linear combination of other two vectors. \[ T ( \mathbf { u } ) = c T ( \mathbf { v } ) \] A line in the direction of $T(v)$ is \[ T ( \mathbf { x } ) = \operatorname { sc } T ( \mathbf { v } ) + v T ( \mathbf { v } ) = ( s c + v ) T ( \mathbf { v } ) \] Since the line P passes through origin and T is a linear transformation, then $T ( \mathbf { x } ) = s T ( \mathbf { u } ) + v T ( \mathbf { v } )$ should pass through the origin. Therefore, there is a linear transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ that maps P onto a line passing through origin when the vectors T(u) and T(v) are linearly dependent and are nonzero.
Step-4:
If both the vectors are equal to 0, then $T(x)=0$. hence $x=0$. Thus, the parametric equation represents only origin.
Therefore, there is a linear transformation $T : \mathbb { R } ^ { 3 } \rightarrow \mathbb { R } ^ { 3 }$ that maps P onto origin when the vectors T(u) and T(v) are both zero
https://imgur.com/uk4eeZH