Well all variants of A have the form (a k-a) (K-a a) So rows and columns add to k Det(A-lambda*I)=0 means the eigenvalue lambda should be k-2a to result in zero determinant Let V(X) (Y) Be an eigenvector of A A*V=lambda*V ( a. k-a). (X). (X) (k-a. a). *. (Y). =. (K-2a)* (Y) aX+(k-a)Y=(k-2a)X (k-a)X+ay =(k-2a)y (a-k)X+(k-a)y=0 => X=Y for eigenvectors So only column vector (1,1) works
Well all variants of A have the form (a k-a)
(K-a a) So rows and columns add to k
Det(A-lambda*I)=0 means the eigenvalue lambda should be k-2a to result in zero determinant
Let V(X)
(Y) Be an eigenvector of A
A*V=lambda*V
( a. k-a). (X). (X)
(k-a. a). *. (Y). =. (K-2a)* (Y)
aX+(k-a)Y=(k-2a)X
(k-a)X+ay =(k-2a)y
(a-k)X+(k-a)y=0
=> X=Y for eigenvectors
So only column vector (1,1) works
Isn't such a matrix always of the form
( a b )
( b a )
? That makes it even more obvious.
Yeah, this question wasn’t too complicated, but it was a pretty early question in this particular exam, so a nice warmup.