Symmetric matrices: for A = [1 2;2 4], P = [1 3;3 6]; using B = P^-1*A*P...

[shreddinglicks]

Consider matrices A = [1 2;2 4] and P = [1 3;3 6]. Using B = P^-1*A*P, verify that similar matrices have the same eigenvalues. Find the eigenvectors y for B and show that x = P*y are eigenvectors of A.


I have

P^-1 = [-2 1;1 -.333]

B = [0 0;2.333 4.999]

eigenvalues for matrices A and B are 0 and 5.

eigenvectors, y are

x2*[0 1]
x2*[-2.14 1]

eigenvectors for matrix A are

x2*[-2 1]
x2*[.5 1]

P*y = [1 3;3 6]*[0 1 ; -2.14 1]

I get x = [3 .86;6 -.42]

What am I doing wrong?

 

[HallsofIvy]

Consider matrices A = [1 2;2 4] and P = [1 3;3 6]. Using B = P^-1*A*P, verify that similar matrices have the same eigenvalues. Find the eigenvectors y for B and show that x = P*y are eigenvectors of A.


I have

P^-1 = [-2 1;1 -.333]

B = [0 0;2.333 4.999]

More precisely, B= [0 0, 7/3 5]

eigenvalues for matrices A and B are 0 and 5.

Yes, that is correct.

eigenvectors, y are

x2*[0 1]
x2*[-2.14 1]

Do you mean the eigenvectors of B?
Yes, an eigenvector of B corresponding to eigenvalue 5 is any multiple of [0, 1]. But for eigenvalue 0, we must have \(\displaystyle \begin{bmatrix}0 & 0 \\ 7/3 & 5 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}0 \\ 7x/3+ 5y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}\).

So 7x/3+ 5y= 0, y= (-7/15)x. Taking x= 15, y= -7 so any eigenvalue corresponding to eigenvalue 0 is a multiple of \(\displaystyle \begin{bmatrix}15 & -7\end{bmatrix}\). Or taking y= 1, x= -15/7 which is approximately 2.14.

eigenvectors for matrix A are
x2*[-2 1]
x2*[.5 1]

Yes, those are correct.

P*y = [1 3;3 6]*[0 1 ; -2.14 1]

I get x = [3 .86;6 -.42]

What am I doing wrong?

 

[shreddinglicks]

I'm confused, so my values of x = [3 .86;6 -.42] is correct?

How does this show that x = P*y are eigenvectors of A?