Yes, you are correct.
3 eigen values 0, 1, 1.
Projection collapses every vector parallel to the z axis to the zero vector. (0,0,a) is transformed to 0*(0,0,a) = (0,0,0), so its an eigen vector right?
And I think you mean:
Hi there. I must give the eigenvalues and the eigenvectors for the matrix transformation of the orthogonal projection over the plane XY on
So, at first I thought it should be the eigenvalue 1, and the eigenvectors (1,0,0) and (0,1,0), because they don't change. But I also tried doing the calculus, and then I've confused.
I have:
From the characteristic polynomial I get to:
Then, I have as eigenvalues 0, and 1 twice.
:
I get to:
and then the eigenvector:
So I thought, shouldn't it be zero? because of the projection. I have doubts with this, but I know that as it is a symmetrical matrix it should be diagonalizable, and then I should get a basis from the eigenvectors, which I wouldn't find with just the first reasoning, and then I need a linear independent vector, like this one, respect to the first I gave.
And then for :
z=0,
Which gives: , and implies: , I think that have sense.
Well, I need some help with this. Can anybody tell me if this is right, and if it isn't, what I did wrong?
Thank you!
Bye there.
PS: I have my final exam tomorrow :P
I have another doubt about this. Now its about the construction of a transformation matrix. The thing is I thought I was constructed in one way, and it seems to be that I was wrong.
I have this problem, which says: From a linear transformation its known that its an eigenvector with the eigenvalue associated and that .
Find
I thought of as
The thing is that when I constructed the matrix I did it this way:
But when I make the multiplication with the vectors that the problem data gives, I don't get the right results, but when I changed to it seems to be the matrix I'm looking for.
So, it seems to be that x, and y go in the columns, and not in the rows as I thought. I was wrong then?