Linear Algebra Question

Question

Let A =

-17 -15

10 8

Find a matrix S, a diagonal matrix D, and S^-1 such that A=SDS^-1.

Answer 1

We first use Cayley Hamilton' theorem to determine eigenvalues by solving:

det(A- aI) = 0 where I is the identity matrix, we get that

det([-17-a -15

10 8-a]) = 0

So computing the determinant, we get

(a+17)(a-8) + 150 = 0

t

that is

a^2 +9a +14 = 0

which could be written as

(a+2)(a+7) = 0

so the roots are -2,-7 which also represent the eigenvalues, hence the components of matrix D

D = [-7 0;0 -2]

Now to compute S, we need to find two column vectors s1 and s2 such that S =[s1 s2]. To do so we start with s1 as

A.s1 = -7.s1

Denoting s1 = [a b] as unknowns we get

-17a - 15b = -7a

10a + 8b = -7b

which are both

10a + 15b = 0, so if we were to solve for a unit norm vector a^2 + b^2 = 1 we get a=-0.8321 and b = 0.5547

Next solve s2 as

A.s2 = -2.s2

Denoting s1 = [a b] as unknowns we get

-17a - 15b = -2a

10a + 8b = -2b

Both give a = -b, so if we were to solve for a unit norm vector a^2 + b^2 = 1 we get a=0.7071 and b = -0.7071. Now, we have S as

S =

-0.8321 0.7071

0.5547 -0.7071

Answer 2

This is asking you to find the eigenvalues (diagonal entries of S) and the eigenvectors (the columns of S), so we first solve for the characteristic equation

det(A - xI) = 0 that is

det([-17-x -15

10 8-x]) = (x-8)(x+17) + 150 = 0

which could be written as

(x + 2)(x + 7) = 0

Hence the eigenvalues are -2 and -7. So

D = [ -7 0

0 -2]

Now the first column of S should satisfy

Av = -7v

solving in v =[v1 v2] we get

v1 = -0.8321 and v2 = 0.5547

Also the second column of S should satisfy

Av = -2v

solving in v =[v1 v2] we get

v1 = 0.7071 and v2 = -0.7071

So S should be

S = -0.8321 0.7071

0.5547 -0.7071