Question

how to work out the eigenvectors of the identity matrix?

Answer 1

eigenvectors \(v\ne 0_v\) of a square matrix \(A\) satisfy \(Av=\lambda v\) for some scalar \(\lambda\)

Answer 2

by definition

Answer 3

im doing this in R2 and the wya i figure it is that nothing changes so ive got to make the basis for R2 so they would be (0,1) and (1,0)

Answer 4

the identity matrix \(I\) satisfies \(Iv=v=1\cdot v\) for all \(v\) so every vector in \(\mathbb{R}^2\setminus{(0,0)}\) is an eigenvalue

Answer 5

eigenvector* rather

Answer 6

you don't need to pick a basis or anything for this problem really

Answer 7

oh. thanks. what would happen if i had multiplied the identity by a scalor before trying to figre out the eigenvectors?

Answer 8

Nothing.

Answer 9

Remember, eigenvectors are scalar multiples to begin with.

Answer 10

in fact you could argue some basis \(v_1,v_2\) exists therefore \(v=c_1v_1+c_2v_2=[c_1,c_2]^T\) for some \(c_1,c_1\in\mathbb{R}\) then just let it work:$$k\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}c_1\\c_2\end{bmatrix}=k\begin{bmatrix}c_1\\c_2\end{bmatrix}$$therefore all \(v\) are eigenvectors with eigenvalue \(k\) for the matrix \(kI\)

Answer 11

^^ this is a way to bring bases in to it for explicit computation of the matrix multiplication and for premultiplying by an arbitrary scalar \(k\)

Answer 12

notice we don't need to PICK any particular basis -- there's no need

Answer 13

In fact, you could have written a matrix like this:

\[\left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\]

Answer 14

in fact, the fact that eigenvectors are basis-independent is exactly why \(k\) does not affect anything... eigenvalues may change but eigenvectors will not

Answer 15

It can be nice to make inverting a matrix of eigenvalues easier if they're normalized though. But yeah, doesn't really matter.