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It exist linear mapping $\phi_{1} : \mathbb R ^{3}\mapsto \mathbb R ^{3}$ which corresponds to a reflection about the plane $x_{1}-x_{2}=0 $.

It exist linear mapping $\phi_{2} : \mathbb R ^{3}\mapsto \mathbb R ^{3}$ which corresponds to a roation with angle 30$^{\circ}$ to the axis with direction vector $\vec{v} =\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} $.

Determine the matrices the linear mapping $\phi_{1}$ and $\phi_{2} $. That means, determine matrices A and B so that following is true:

$\phi_{1} \left(\vec{x} \right)=A\vec{x} $

$\phi_{2} \left(\vec{x} \right)=B\vec{x} $

I dont know how to solve this example. The only thing I can remember is formula for rotation matrix, but I am not sure how helpful it can be here.

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We need to find $A$ such that $\phi_{1} \left(\vec{x} \right)=A\vec{x}$

$\vec{x}$ can be written as $ \begin{pmatrix} x_{1}\\ x_{2}\\ x_{3} \end{pmatrix}$ since the mapping goes from $\mathbb R ^{3}$

The result of the transformation should be the vector: $ \begin{pmatrix} x_{2}\\ x_{1}\\ x_{3} \end{pmatrix}$ according to $x_{1}-x_{2}=0$

So: We know that $A$ should be a $3\times 3$ matrix from the fact that the mapping returns a vector. Reasoning as follows: We need the first component of the returned vectors to correspond to just $x_{2}$, so we need to get rid of $x_{1}$ and $x_{3}$. This means that the first line of the matrix $A$ is $\begin{pmatrix} 0 &1 &0 \end{pmatrix}$

Following the same idea, you should finally get $A$ is the matrix: $\begin{pmatrix} 0 &1 &0 \\ 1 &0 &0 \\ 0 &0 &1 \end{pmatrix}$

I will let you do the other mapping.

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  • $\begingroup$ I dont understand how do you get vector (x2,x1,x3) acording to the given equation. Can you write a little bit more detailed only that step? $\endgroup$ – Ana Matijanovic Jun 15 '16 at 20:04
  • $\begingroup$ $x_{1}-x_{2}=0$ tells you that the reflection plane is the plane containing the points $\begin{pmatrix} x &y &z \end{pmatrix}$ with $x=y$. Try to draw the coordinate axes and the line $y=x$ and reflect points according to the line. $\endgroup$ – John11 Jun 15 '16 at 20:19
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Every linear transformation is uniquely determined by a matrix given a basis of the domain and a basis of the vector space where it maps.

So we can choose the canonical base of R^3 i.e. e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1).

The first transformation maps the vector e1 into the vector e2, it maps the vector e2 into the vector e1 and doesn't move the vector e3.

So the first transformation is represented in the canonical basis of R^3 by the matrix A = ( 0 1 0; 1 0 0; 0 0 1 ).

The second transformation doesn't move the vector e1, it maps the vector e2 into the vector (0, sqrt(3)/2, 1/2) and it maps the vector e3 into the vector (0, -1/2, sqrt(3)/2).

Thus the second transformation is represented in the canonical basis of R^3 by the matrix B = ( 1 0 0; 0 sqrt(3)/2 -1/2; 0 1/2 sqrt(3)/2 ).

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