Find the basis for both eigenvalues

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    Basis Eigenvalues
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Homework Help Overview

The discussion centers around finding the basis for the eigenvalues of the matrix A = {[39/25, 48/25], [48/25, 11/25]}. The original poster attempts to find the eigenvalues and corresponding eigenvectors but encounters difficulties in verifying their correctness.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of row reduction and its appropriateness for finding eigenvalues. There is a focus on deriving equations from the eigenvalue definitions and determining eigenvectors from those equations.

Discussion Status

Some participants provide guidance on how to derive eigenvectors from the equations formed for each eigenvalue. There is an exploration of the relationships between the variables in the equations, but no consensus on the final basis has been reached.

Contextual Notes

The original poster's approach involved row reduction, which some participants question as a valid method for finding eigenvalues. The discussion highlights the need to derive eigenvectors from the characteristic equations instead.

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Homework Statement



Given matrix A= {[39/25,48/25],[48/25,11/25]} find the basis for both eigenvalues.


Homework Equations





The Attempt at a Solution



I row reduced the matrix and found both eigenvalues. I found λ = -1, and λ = 3. Then, I used diagonalization method [-1I2 - A 0] and [3I2 - A 0]. I got a basis of {[1,-3/4]} and {[1,4/3]}. However, I checked these but these were incorrect. I wasn't sure what I did wrong.
 
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Row reduce the matrix? You can't find eigenvalues by row reduction!

If [x, y] is an eigenvector corresponding eigenvalue -1, then
\begin{bmatrix}\frac{39}{25} & \frac{48}{25} \\ \frac{49}{25} & \frac{11}{25}\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}-x \\ -y\end{bmatrix}

\begin{bmatrix}\frac{39}{25}x+ \frac{48}{25}y \\ \frac{48}{25}x+ \frac{11}{25}y\end{bmatrix}= \begin{bmatrix}-x \\ -y \end{bmatrix}

So we must have 39x/25+ 48y/25= -x, which is the same as 64x+ 48y= 0 and 48x/25+ 11y/25= -y which is the same as 48x+ 36y= 0.

Both of those equations reduce to a single equation of the form ax+ by= 0 so that y= (b/a)x.

If [x, y] is an eigenvector corresponding eigenvalue 3, then
\begin{bmatrix}\frac{39}{25} & \frac{48}{25} \\ \frac{49}{25} & \frac{11}{25}\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}3x \\ 3y\end{bmatrix}

\begin{bmatrix}\frac{39}{25}x+ \frac{48}{25}y \\ \frac{48}{25}x+ \frac{11}{25}y\end{bmatrix}= \begin{bmatrix}-x \\ -y \end{bmatrix}

So we must have 39x/25+ 48y/25= 3x, which is the same as -36x+ 48y= 0 and 48x/25+ 11y/25= 3y which is the same as 48x- 64y= 0. Again, both of those equations reduce to a single equation.
 
Okay, so given those two solutions, do I row reduce it to determine the basis?
 
LosTacos said:
Okay, so given those two solutions, do I row reduce it to determine the basis?
No. The first equation in HallsOfIvy's post is 48x + 36y = 0, which is equivalent to 4x + 3y = 0. From that equation you can find the eigenvector that is associated with the eigenvalue λ = -1.

The second equation is 48x - 64y = 0, or 3x - 4y = 0. Use this equation to find the eigenvector that is associated with the eigenvalue λ = 3.
 
thank you
 
Last edited:

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