Are Similar Matrices' Eigenvalues the Same? Solving for Symmetric Matrices

[shreddinglicks]

Homework Statement

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.

Homework Equations

B = P^-1*A*P,
x = P*y

The Attempt at a Solution

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?

 

[mathwonk]

similar matrices represent the same operator on a space but with different coordinates. since eigenvalues do not depend on coordinates they have the same eigenvalues.

 

[shreddinglicks]

Right, which is why I have 0 and 5 as eigenvalues for both matrices.

How do I show that x = P*y are eigenvectors of A? I feel like I'm getting the wrong answer.

 

[Mark44]

Homework Statement

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.

Homework Equations

B = P^-1*A*P,
x = P*y

The Attempt at a Solution

I have

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

B = [0 0;2.333 4.999]

You should leave the numbers as fractions, not as rounded decimal values.
$P^{-1} =\begin{bmatrix} -2 & 1 \\ 1 & -\frac 1 3 \end{bmatrix}$ and $B =\begin{bmatrix} 0 & 0 \\ \frac 7 3 & 5 \end{bmatrix}$

eigenvalues for matrices A and B are 0 and 5.

eigenvectors, y are

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

The latter is really $\begin{bmatrix} -\frac{15} 7 \\ 1\end{bmatrix}$

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]

Hard to tell what you did here. When I multiply Py, I get a matrix whose columns are vectors that are scalar multiples of $\begin{bmatrix} -2 \\ 1\end{bmatrix}$ and $\begin{bmatrix} \frac 1 2 \\ 1\end{bmatrix}$

What am I doing wrong?

 

[Cutter Ketch]

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

Your matrix of y’s on the right appears to be transposed. Might have been better off leaving them as two vectors. That would avoid this error anyway

 

[Cutter Ketch]

How do I show that x = P*y are eigenvectors of A? I feel like I'm getting the wrong answer.

This might not be an issue once you construct the x’s properly, but on the off chance you are asking the more basic question of “how”, you prove the x’s are eigenvectors of A by using the definition of an eigenvector

 

[shreddinglicks]

You should leave the numbers as fractions, not as rounded decimal values.
$P^{-1} =\begin{bmatrix} -2 & 1 \\ 1 & -\frac 1 3 \end{bmatrix}$ and $B =\begin{bmatrix} 0 & 0 \\ \frac 7 3 & 5 \end{bmatrix}$
The latter is really $\begin{bmatrix} -\frac{15} 7 \\ 1\end{bmatrix}$
Hard to tell what you did here. When I multiply Py, I get a matrix whose columns are vectors that are scalar multiples of $\begin{bmatrix} -2 \\ 1\end{bmatrix}$ and $\begin{bmatrix} \frac 1 2 \\ 1\end{bmatrix}$

I made x2 = 1

Then I calculated P*y

This gave me

P*y = [1 3; 3 6] * [0 -15/7; 1 1] = [3 6/7; 6 -3/7]

 

[Mark44]

The columns of your last matrix, Py, are multiples of $\begin{bmatrix} \frac 1 2 \\ 1\end{bmatrix}$ and $\begin{bmatrix} -2 1\end{bmatrix}$, so they are also eigenvectors of A.

 

[shreddinglicks]

Yes, you're right. I didn't even notice that.

6*[.5 1]'
and
(3/7)*[-2 1]

 

[shreddinglicks]

Thanks! You have been a great help.