Please check my Eigenvector solutions.

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The discussion revolves around finding the characteristic equations, eigenvalues, and eigenvectors of a given matrix. It is clarified that the initial eigenvector corresponding to λ=3 is correct, but the characteristic equation is incorrect, and λ=1 is not an eigenvalue. The correct characteristic equation is derived, leading to the conclusion that the eigenvector for λ=-1 is [0;1]. Participants emphasize the importance of verifying eigenvalue and eigenvector pairs by substituting them back into the original equation. The conversation highlights the need for careful calculations and understanding of eigenvalue problems in linear algebra.
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Homework Statement


Find the characteristic equations, eigenvalues and eigenvector of the following matrix

Homework Equations


Eigen 2.png

The Attempt at a Solution


Eigen 1.png


Somehow somewhere I think the solution is wrong, based on online Eigenvector calculator on the web. Please do provide me actual answers and solutions. Thanks in advance!
 
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Your first eigenvector corresponding to \lambda= 3 is correct.

However, you have the wrong characteristic equation and "1" is NOT an eigenvalue. You should have seen that when you wrote (A- \lambda I)x= 0:
\begin{bmatrix}2 & 0 \\ 8 & -2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}
which gives
\begin{bmatrix}2x_1 \\ 8x_1- 2x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0 \end{bmatrix}
which then requires that 2x_1= 0 and 8x_1- 2x_2= 0.
From the second x_2= 4x_1 but the first says x_1= 0 so x_2= 4(0)= 0. There is NO non-trivial vector satisfying this. 1 is NOT an eigenvalue
(I have no idea where you got "x_1+ x_2= 0".)

The characteristic equation is given by
\left|\begin{array}{cc}3- \lambda & 0 \\ 8 & -1- \lambda\end{array}\right|= 0
Which is, of course, simply (3-\lambda)(-1- \lambda)= 0.
 
Eigen 3.png

Take a look at this. I've corrected it. Please let this answer correct :)

Some mistake there;
Eigenvector for lambda = 1 is [0;1]

One more, on the second last line.
is that correct to state "Let x_2=t" or "Let x_1=t"?
 
Last edited:
If you compute what you think are an eigenvector and eigenvalue pair, stick them back in! They had better satisfy Ax = \lambda x since that is, after all, the equation whose solutions you were looking for in the first place.

hadizainud said:
One more, on the second last line.
is that correct to state "Let x_2=t" or "Let x_1=t"?

The variable x_2 should equal t, if that's what you're asking.
 
hadizainud said:
View attachment 37957
Take a look at this. I've corrected it. Please let this answer correct :)

Some mistake there;
Eigenvector for lambda = 1 is [0;1]
You mean "for lambda= -1".

One more, on the second last line.
is that correct to state "Let x_2=t" or "Let x_1=t"?
You had just shown that x_1= 0 so you can't say "let x_1=t".
 
Thanks Stringy and HallsofIvy :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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