Similarity of Matrices: Finding the Invertible Matrix P

[boneill3]

Homework Statement

Show that the Matrices A and B are similar

$A=\[ \left( \begin{array}{cc}<br /> 1 & 1 \\<br /> -1 & 4 \end{array} \right)\] B=\[ \left( \begin{array}{cc}<br /> 2 & 1 \\<br /> 1 & 3 \end{array} \right)\]$

Homework Equations

B=PAP^-1

The Attempt at a Solution

I know they have the same trace of 5

I have found the charecteristic equations of both matrices which are the same

$\lambda^2-5\lambda+5$

So I solve for zero to get the eigenvalues.

$\lambda^2-5\lambda+5=0$ so the eigenvalues are

$\lambda_1=\frac{5+\sqrt{5}}{2}$ and $\lambda_2=\frac{5-\sqrt{5}}{2}$

I'm having trouble finding the invertable Matrix P

I know I have to use the matrix
$A=\[ \left( \begin{array}{cc}<br /> 1-\lambda & 1 \\<br /> -1 & 4-\lambda \end{array} \right)\]$

and solve for zero. then I believe I use the diagonals of matrix B to substitute for $\lambda$ eg 2 and 3

But not sure how to get there. Any help greatly appreciated

 

[tiny-tim]

Hi boneill3!

(have a lambda: λ and a square-root: √ and try using the X² tag just above the Reply box )

Show that the Matrices A and B are similar

$A=\[ \left( \begin{array}{cc}<br /> 1 & 1 \\<br /> -1 & 4 \end{array} \right)\] B=\[ \left( \begin{array}{cc}<br /> 2 & 1 \\<br /> 1 & 3 \end{array} \right)\]$
…
I have found the charecteristic equations of both matrices which are the same

$\lambda^2-5\lambda+5$

So I solve for zero to get the eigenvalues.

$\lambda^2-5\lambda+5=0$ so the eigenvalues are

$\lambda_1=\frac{5+\sqrt{5}}{2}$ and $\lambda_2=\frac{5-\sqrt{5}}{2}$

I'm having trouble finding the invertable Matrix P …

Do you need to find P, or even to find the eigenvalues?

If the characteristic equations are the same, don't the matrices have to be similar?

To find P, you could find the eigenvectors of each matrix first: then it's the matrix which transforms from one pair (as a basis) to the other pair.

 

[boneill3]

Hi Guy's
I'm not sure if I have to find P but I would like to know how to do it.
So I've found the eigenvectors of the two Matrices.

The eigenvectors for
$A=\[ \left( \begin{array}{cc}1 & 1 \\-\frac{1}{2}(\sqrt{5}-3) & \frac{1}{2}(\sqrt{5}+3) \end{array} \right)\]$

$B=\[ \left( \begin{array}{cc}1 & 1 \\-\frac{1}{2}(\sqrt{5}-1) & \frac{1}{2}(\sqrt{5}+1) \end{array} \right)\]$


I calculated D= P^-1AP and D= P^-1BP and both Matrices are diagonable.

How do you find the Matrix which transforms one pair to the other?

hence finding B=PAP-1

I have not seen this yet in my course
thanks

 

[tiny-tim]

I calculated D= P^-1AP and D= P^-1BP and both Matrices are diagonable.

Do you mean you calculated Q and R such that D= Q^-1AQ and D= R^-1BR with D diagonal?

Then B = RDR^-1 = RQ^-1AQR^-1 = (QR^-1)^-1A(QR^-1)

 

[beetle2]

Thanks Tiny_Tim, Your a Ledgend!

I found the change of basis Matrix P to be

$P=\[ \left( \begin{array}{cc}1 & 1 \\1 & 0 \end{array} \right)\]$

B = PAP^-1

Using form what you said that P = QR^-1

It makes heaps more sense to me know. Yippieeeee!