Question

Let T: R^2-> R^2 be the linear transformation that projects points onto the line y= 3x. Write
down all the eigenvalues and eigenvectors of T. n explain how do u get them?

Answer 1

Well intuitively (1,3) is an eigenvector.

But to get at this formally, write down the matrix of T.
\[T : i \to i + 3j\] and
\[T : j \to \frac{1}{3} i + j\]

Hence \[T = \left[\begin{matrix}1 & 1/3 \\ 3 & 1 \end{matrix}\right]\]

Now solve for eigenvalues and -vectors as usual

\[\det(A - \lambda I) = 0 \implies ...\]

Answer 2

n again thank you

Answer 3

The other vector that is intuitively an eigenvector is the 1-d basis of the subspace that gets sent to the null vector; that is every vector lying along the line perpendicular to y = 3x; that is, the line y = -x/3.

Hence the basis for that subspace is (1,-1/3)

Answer 4

So hopefully something like that comes out of the formal calculation. (and hopefully I haven't made a sign error!)