Thread: determinant and eigenvalues

For an nxn matrix A

det(A) = ∏(i=1, .., n) λi ...... [1],

where the λi are the eigen values (appearing with the correct multiplicities)

the characteristic equation is det(λI-A)=0, and this is an nth polynomial. Now i can see that statment [1] is true, but not sure how to demontrate this fact. We know that the matrix will be singular if an only if atleast one of the eigenvalues are zero, is there a way i can use this to show that [1] is true? need some advice on how to get started (∏ is from i=1 to n if it was not clear)

Originally Posted by action259

For an nxn matrix A

det(A) = ∏(i=1, .., n) λi ...... [1],

where the λi are the eigen values (appearing with the correct multiplicities)

the characteristic equation is det(λI-A)=0, and this is an nth polynomial. Now i can see that statment [1] is true, but not sure how to demontrate this fact. We know that the matrix will be singular if an only if atleast one of the eigenvalues are zero, is there a way i can use this to show that [1] is true? need some advice on how to get started (∏ is from i=1 to n if it was not clear)

If you show that:

A=PDP^-1,

where P is the matrix composed of the eigen vectors of A, and D is
a diaginal matrix with the eigen values of A on the diagonal, then the
result follows from:

det(A)=det(P).det(P).det(P^-1),

and det(P^-1)=1/det(P).

RonL