Eigenvalues and eigenvectors [Linear Algebra]

  • Thread starter Telemachus
  • Start date
  • #1
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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 [tex]R^3[/tex]

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:

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

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

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

[tex]\lambda_1=0[/tex]:

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

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 [tex]\lambda_2,\lambda_3=1[/tex]:

z=0,

Which gives: [tex]{(x,y,0)}[/tex], and implies: [tex]{(1,0,0),(0,1,0)}[/tex], 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
 
Last edited:

Answers and Replies

  • #2
296
15
(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. :tongue:
 
  • #4
832
30
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 [tex]T:\mathbb{R}^2\longrightarrow{\mathbb{R}^2}[/tex] its known that [tex]\vec{v}=(1,1)[/tex] its an eigenvector with the eigenvalue associated [tex]\lambda=2[/tex] and that [tex]T(0,1)=(1,2)[/tex].

Find [tex][T]_c[/tex]

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

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

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

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 [tex][T]_c=\begin{bmatrix}{1}&{1}\\{0}&{2}\end{bmatrix}[/tex] 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?
 
  • #5
455
121
Just like in a vector, the x and y go in the columns and not the rows.
 

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