Eigenvector different to expected using eig

Hello! The problem is as the title suggests.

A=[1.5 -0.5; -1 1];

[Vec,Val]=eig(A)

The outputted eigenvalues are 2 and 0.5

The outputted eigenvectors are: [0.7071 0.4472; -0.7071 0.8944]

The expected eigenvectors from calculations are (1 2) and (1 -1)

Anyway I can get the expected values ?

Answer 1

[Steven Lord]

A=[1.5 -0.5; -1 1];

[Vec,Val]=eig(A)

Vec = 2×2

0.7071 0.4472 -0.7071 0.8944

Val = 2×2

2.0000 0 0 0.5000

What does the eig function return? From its documentation: "[V,D] = eig(A) returns diagonal matrix D of eigenvalues and matrix V whose columns are the corresponding right eigenvectors, so that A*V = V*D." So is A*V close to V*D?

format longg
A*Vec-Vec*Val

ans = 2×2

0 2.77555756156289e-17 0 0

Yeah, those are pretty close to 0. How about for your matrix of eigenvectors?

Vec2 = [1 1; 2 -1]

Vec2 = 2×2

1 1 2 -1

A*Vec2-Vec2*Val

ans = 2×2

-1.5 1.5 -3 -1.5

Perhaps, from the orientation of your eigenvectors, you're trying to use the left eigenvectors? "[V,D,W] = eig(A) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'."

[Vec, Val, VecLeft] = eig(A);
VecLeft

VecLeft = 2×2

0.894427190999916 0.707106781186547 -0.447213595499958 0.707106781186547

VecLeft'*A - Val*VecLeft'

ans = 2×2

0 0 0 0

Vec2'*A-Val*Vec2'

ans = 2×2

-2.5 -2.5 2 -1

Still no. So please show me why you believe that the columns of Vec2 are eigenvectors for this matrix.

If we scaled VecLeft a bit it looks a little similar to your Vec2 matrix but not exactly.

VecLeftScaled = VecLeft ./ VecLeft(2, :)

VecLeftScaled = 2×2

-2 1 1 1

VecLeftScaled'*A-Val*VecLeftScaled'

ans = 2×2

0 0 0 0

[Joshua Tsui]

Thankyou Steven. It appears, I read my textbook incorrectly. I mistook the arbitary relationships as vectors.