Diagonalization of Matrix A

Hi,

A is a 2 by 2 matrix:

a 1

ab b

i found the eigenvalues λ: 0 and a+b,

the corresponding (column)eigenvectors:

[1/√(1+a^2) -a/√(1+a^2)] & [1/b√(1+(1/b)^2) 1/√(1+(1/b)^2)]

Wich gives matrix Q:

1/√(1+a^2) 1/b√(1+(1/b)^2)

-a/√(1+a^2) 1/√(1+(1/b)^2)

Now i have to show that inverse(Q) * A * Q = Λ (diagonalization of A)

I get inverse(Q): (found det(Q)= 1+a)

1/(1+a)(1/√(1+a^2)) a/(1+a)√(1+a^2)

-1/(1+a)(b√(1+(1/b)^2)) 1/(1+a)√(1+(1/b)^2)

but i'm not getting the right anwser. What am i doing wrong? Is there a quicker way to do this?

Please help me out.

regards,

Moljka.

Re: Diagonalization of Matrix A

here is what i would do:

the eigenvector you have for a+b is a bit messy, in the denominator, √(1+(1/b)^2) = √((b/b)^2+(1/b)^2) = √[(1+b^2)/b^2],

so you could have chosen (1/√(1+b^2), b/√(1+b^2)) as the eigenvector instead (which matches the first eigenvector in style, at least).

now, re-calculate det(Q), i think your answer should have a's AND b's in it (when i did it, it was kinda ugly).

you might also wish to rationalize your denominators, by multiplying the first eigenvector by √(1 + a^2)/√(1 + a^2), and similarly for b.

Re: Diagonalization of Matrix A

is there a way to prove inverse(Q) * A * Q = Λ for non symmetric matrices, like one can prove transpose(Q) * A *Q= Λ for symmetrix matrixes by premultiplying Aq1=λq1 by transpose(q) (q= a column vector)?

This would also show that inverse(Q) * A * Q = Λ (diagonalization of A). Its impossible to premultiply Aq1=λq1 by inverse(q) since the inverse does not exist for a vector though.

Re: Diagonalization of Matrix A

for certain kinds of matrices, yes. but not every matrix is diagonalizable. for example,

[1 1]

[0 1]

is not diagonalizable. this matrix has the eigenvalue 0, but any eigenvector is a multiple of (1,0), which only gives us one column

for the change of basis matrix (and we need two).

Re: Diagonalization of Matrix A

Thank you for helping me out.

after computing inverse(Q) * A * Q = Λ i got the matrix Λ:

0 fcd(a+(a^2)b)+fce(1+ab)/(f^3)(d^3)(e^3)

0 (-afd+abfb)(f^2)(d^2)(e)-(f^2)(e^2)(d)(fd+bfe)/(f^3)(d^3)(e^3)

for convenience i substituted the following with the letters:

c=√1+a^2

e=b√1+(1/b)^2 ( or b√(b^2+1)/b^2)

d=√1+(1/b)^2 ( or √(b^2+1)/b^2)

f=1+a

either that can some how be rewritten to:

0 0

0 a+b

or once again i made an error somewere along the way.