Eigenvalues and eigenvectors [Linear Algebra]

[Telemachus]

Homework Statement

Hi there. I must give the eigenvalues and the eigenvectors for the matrix transformation of the orthogonal projection over the plane XY on R^3

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:

[T]_c=\begin{bmatrix}{1}&{0}&{0}\\{0}&{1}&{0}\\{0}&{0}&{0}\end{bmatrix}

From the characteristic polynomial I get to: -\lambda(1-\lambda)^2

Then, I have as eigenvalues 0, and 1 twice.

\lambda_1=0:

I get to: \begin{Bmatrix}x=0\\y=0\end{matrix}
and then the eigenvector: {(0,0,1)}

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 \lambda_2,\lambda_3=1:

z=0,

Which gives: {(x,y,0)}, and implies: {(1,0,0),(0,1,0)}, 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

 

[Citan Uzuki]

(0, 0, 1) is indeed an eigenvector with eigenvalue 0. Remember, the definition of eigenvector doesn't require that the vector remain unchanged, but rather that it be sent to a scalar multiple of the original vector. Observe that 0(0, 0, 1) = (0, 0, 0) = T(0, 0, 1).

Also, you mention that you know that your matrix is diagonalizable because it is symmetric. This is true, but I would point out the much more obvious fact that it's diagonalizable on account of already being a diagonal matrix.

 

[Telemachus]

Thanks!

 

[Telemachus]

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 T:\mathbb{R}^2\longrightarrow{\mathbb{R}^2} its known that \vec{v}=(1,1) its an eigenvector with the eigenvalue associated \lambda=2 and that T(0,1)=(1,2).

Find [T]_c

I thought of [T]_c as T(x,y)=(x+y,2y)

The thing is that when I constructed the matrix I did it this way:

[T]_c=\begin{bmatrix}{1}&{0}\\{1}&{2}\end{bmatrix}

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 [T]_c=\begin{bmatrix}{1}&{1}\\{0}&{2}\end{bmatrix} 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?

 

[romsofia]

Just like in a vector, the x and y go in the columns and not the rows.