Lower triangular matrix eigenvectors problem

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Homework Help Overview

The discussion revolves around finding the eigenvectors and eigenvalues of a given lower triangular matrix. The matrix is defined with specific entries, and the original poster attempts to derive the eigenvalues and corresponding eigenvectors, leading to some confusion regarding the signs and calculations involved.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the calculation of eigenvalues and eigenvectors, with some questioning the validity of the signs of the eigenvalues. There is also a discussion about the formulation of the eigenvector equations and the clarity of the steps taken to derive them.

Discussion Status

Some participants have provided clarifications on the calculations and pointed out potential errors in the original poster's reasoning. There is acknowledgment of a typo regarding the sign of one eigenvalue, and the conversation reflects a collaborative effort to clarify the steps involved in finding the eigenvectors.

Contextual Notes

Participants note the confusion stemming from the signs of the eigenvalues and the structure of the eigenvector equations. There is an indication that the original poster had a moment of realization regarding their calculations, suggesting that the discussion has been productive in addressing misunderstandings.

eherrtelle59
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Ok, this is starting to come back to me, but I'm stuck again

Homework Statement




M=\begin{bmatrix}
(1-\frac{4}{3}) & 0 \\
-\frac{c}{3} & -c \\
\end{bmatrix}


Find eigenvectors and eigenvalues.

Homework Equations





The Attempt at a Solution



Eigenvalues are [itex]λ_1= (1-\frac{4}{3})>0[/itex] and [itex]λ_2=-c>0[/itex]

Eigenvector for [itex]λ_2[/itex] is <0 1>
For [itex]λ_1[/itex], I should get [itex]<(\frac{c}{3}-1) (\frac{4}{3})>[/itex]

However, I end up with (without writing out the matrix again, just giving the equation mind you)

[itex]-c*e_2 = e_1 - \frac{4}{3}*e_1 +\frac{c}{3}*e_1[/itex]

This is [itex]-c*e_2 = (1-c)*e_1[/itex]which gets an eigenvector of something like <(1-c) c>

Anyone see my error? Thanks
 
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The clearly negative eigenvalue being allegedly positive is the first thing that struck me as odd, but not really a big deal in the grand scheme of things

I don't understand what you did to get that equation (the left hand side looks like Me2, and the right hand side is completely unclear to me where it came from, but the fact that you have a scaling of e1= a scaling of e2 seems pretty suspicious), or what you are claiming the eigenvector for the first eigenvalue is since it only has one entry as written - is it supposed to be ( (c/3-1), 4/3)?
 
"The clearly negative eigenvalue being allegedly positive is the first thing that struck me as odd, but not really a big deal in the grand scheme of things"

That's a typo on my part. λ_2 should be negative i.e. -c<0

"is it supposed to be ( (c/3-1), 4/3)?"

Yes

To show explicitly what I did, using the standard eigenvector equation with v_i instead of e_i


\begin{bmatrix}
(1-(4/3)c) & 0 \\
-c/3 & -c \\
\end{bmatrix} *\begin{bmatrix}
v_1 \\
v_2 \\
\end{bmatrix}

= \begin{bmatrix}
(1-4c/3)*v_1 \\
(1-4c/3)*v_1\\
\end{bmatrix}

When you multiply this matrices (as sort of shown...latex skills withstanding) then you get the equation I've written previously.
 
"but the fact that you have a scaling of e1= a scaling of e2 seems pretty suspicious"

When finding eigenvectors I always get equations like this.
 
I'm sorry everyone, that should be multiply by

\begin{pmatrix}
(1-4c/3)v_1\\
(1-4c/3)v_2\\
\end{pmatrix}
which does indeed give the right answer,

(c/3)*v_1 = (c/3 -1)*v_2

Sorry for wasting everyone's time, I've finally got it now. Thank you!
 

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