The problem is with the last part of the following question:
I will write my results to the first parts which are correct here :
Three Eigenvalues:
$$\lambda_1=1 , \lambda_2=2 , \lambda_3=3$$
Corresponding Eigenvectors:
$$e_1 = –2j + k$$
$$e_2 = i + j $$
$$e_3 = 2i + 2j + k$$
I proved the second part sucessfully:
$r$ is the parametric equation
$$r=te+sf$$
$$A(te+sf) = tAe+sAf= (t\lambda) e + (s \mu) f$$
Therefore we can conclude the statement
Now the problem is:
How to do the last part, what three planes?
I don't understand what's happening.
My book has taken two pairs of my three Eigenvectors and done the cross product to find the normal vector, since there are 3 combinations so they have got 3 equations, like this:
$e_1 × e_2 = –i + j + 2k ⇒ x – y – 2z = 0$
$e_1 × e_3 = 2i – j – 2k ⇒ 2x – y – 2z = 0$
$e_2 × e_3 = i – j ⇒ x – y = 0$
But I don't understand how this works, because I found the three Eigenvectors of A, so ALLL three Eigenvectors lie on the same plane, so then cross product of any vectors in one plane gives its normal vector which is the same for all. So there won't be any equations, Please help.
The book contradicts my belief that :
All Eigenvectors of $A$ lie on the plane which represents the vector space $Ax$
Hint : All 3 eigenvectors do not lie on the same plane. If they do, the matrix is nilpotent and this matrix is not, since the eigenvalues are distinct. The eigenvectors span a space but that does not mean that they are on a plane. The space, in this case, is 3-dimensional whereas the plane is 2-dimensional. You can see that the eigenvectors are linearly independent which means that they do not lie on the same plane but span $\mathbb{R}^3$.
You can visualize in 3 dimensions that a pair of eigenvectors will span a plane. Therefore you have $3 \choose 2 $ $=3$ planes.