# What is the matrix of $A$?

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?

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 $$2$$)

• 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 $v_2$. – Widawensen Mar 12 at 13:25
• 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}} – Widawensen Mar 12 at 13:36
• 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! – Wallname Mar 12 at 14:01

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 $$A$$.