# Thread: Eigenvalues and eigenvectors [Linear Algebra]

1. ## Eigenvalues and eigenvectors [Linear Algebra]

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

2. 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:
$\lambda_2,\lambda_3=1$

3. You're right. Thanks.

4. 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?

5. Yes, you are correct. x and y coefficients go in the column.

If you forget think about it this way.

$
\begin{bmatrix}{1}&{1}\end{bmatrix}\begin{bmatrix} {x}\\{y}\end{bmatrix} = \begin{bmatrix}x + y\end{bmatrix}
$