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We know the following about the linear map $A$: $\mathbb{R}$$^3$ -> $\mathbb{R}$$^3$:

$A$ is orthogonal

$A$(1,2,2) = (1,2,2)

The vector (2,0,-1) is eigenvector for eigenvalue -1

dim $E_1$ = 1

Determine the matrix of $A$

I'm not quite sure which properties to use, such that i can create a matrix $A$. Any help/tips? on proceeding this particular question?

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From the given conditions you have two equations $Av_1=v_1$ and $Av_2=-v_2$.

Notice that here additionally $v_1^Tv_2=0$ what means that both vectors are orthogonal.

You can find also transformed the third vector $v_3$ using as input vector cross product of $v_1$ and $v_2$, the result vector is orthogonal to both $= \pm (v_1 \times v_2)$ (transformation with orthogonal matrix preserves lengths and angles of vectors )

With this you have transformation $A[v_1 \ \ v_2 \ \ v_1 \times v_2] = [v_1 \ \ -v_2 \ \ \pm v_1 \times v_2]$ what leads to direct calculation of

$A= [v_1 \ \ -v_2 \ \ \pm v_1 \times v_2][v_1 \ \ v_2 \ \ v_1 \times v_2]^{-1} $

(two solutions).

Additionally if $\text{dim} \ E_1= 1$ means that $1$ is the eigenvalue with multiplicty $1$ then you can exclude one solution. (the $-1$ is then eigenvalue with multiplicity $2$)

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  • $\begingroup$ At Wolphram Alpha I have received as the result { {1,-2,2}, {2,0,-5} , {2,1,4} }. inverse { {1,2,-2}, {2,0,5} , {2,-1, -4} }={{-7/9, 4/9, 4/9}, {4/9, -1/9, 8/9}, {4/9, 8/9, -1/9}} =A which provides the required transformations of $v_1$ and $v_2$. $\endgroup$ – Widawensen Mar 12 at 13:25
  • $\begingroup$ Direct checking : first vector {{-7/9, 4/9, 4/9}, {4/9, -1/9, 8/9}, {4/9, 8/9, -1/9}} . {{1}, {2}, {2}}={{1}, {2}, {2}} , second vector {{-7/9, 4/9, 4/9}, {4/9, -1/9, 8/9}, {4/9, 8/9, -1/9}} . {{2}, {0}, {-1}} = {{-2}, {0}, {1}} $\endgroup$ – Widawensen Mar 12 at 13:36
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    $\begingroup$ Thank you very much, at first i understood your method but i couldn't get the same cross product for $v_3$ but then i saw your fix and everything clicked. I get it now thanks! $\endgroup$ – Wallname Mar 12 at 14:01
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As $A$ is orthogonal, it is orthogonally diagonalizable. You know two eigenvalues and you know that the determinant is $\pm 1$. With the information on the dimension of the eigenspace you get the third eigenvalue. The third eigenvector you can find by completing the orthonormal basis (e.g. Gram-Schmidt). Now you have all the information to compute $A$.

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