Thread: Finding eigenvectors and eigenvalues geometrically

1. Finding eigenvectors and eigenvalues geometrically

Find the eigenvalues and eigenvectors of A geometrically.

A=
[1 0
0 1]

(projection onto the x-axis)

2. the matrix for the projection on the x axis is
$\displaystyle \left( \begin{array}{ccc} 1 & 0 \\ 0 & 0 \end{array} \right)$

and since eigenvectors
are the vectors
$\displaystyle x$ which satisfy $\displaystyle Ax={\lambda}x$ with $\displaystyle A$ the matrix of your linear transformation and $\displaystyle {\lambda}$ an eigenvalue of $\displaystyle A$, the eigenvectors of the projection on the x axis must be all the vectors that are already on the x axis and have an eigenvalue of 1. you can also see this by observing the matrix which is already diagonnal and gives you right away its only eigenvalue 1.

3. There is also the eigenvalue 0, which corresponds to...

4. aaaaaaaaaaaaaah yess yes yes 0, of course! merci de me doublechecker bruno

5. Haha! Avec plaisir.

The eigenspace corresponding the the eigenvalue 0, in this case, is the vertical line through the origin; vectors on this line are projected to the zero vector.