MHB Finding Eigenvalues of Matrix A: Wrong Answer, What Am I Doing Wrong?

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The discussion revolves around finding the eigenvalues of the matrix A, specifically A = [[1, 1, k], [1, 1, k], [1, 1, k]]. The original poster consistently calculates the eigenvalues as 0, 1, and (k+1), while the expected answers are 0 and (k+2). A key suggestion is to correctly compute the determinant of (λI - A), which should yield a polynomial determinant of λ^2(λ - k - 2). This indicates a potential miscalculation in the determinant process. The conversation emphasizes the importance of accurately determining the characteristic polynomial to find the correct eigenvalues.
Yankel
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Hello all,

I have a matrix A and I am looking for it's eigenvalues. No matter what I do, I find that the eigenvalues are 0, 1 and (k+1), while the answer of both the book and Maple is 0 and (k+2). I tried two different technical approaches, both led to the same place.

The matrix is:

\[A=\begin{pmatrix} 1 &1 &k \\ 1 &1 &k \\ 1 &1 &k \end{pmatrix}\]

I have stated with calculating

\[\lambda I-A\]

which is

\[A=\begin{pmatrix} \lambda -1 &-1 &-k \\ -1 &\lambda -1 &-k \\ -1 &-1 &\lambda -k \end{pmatrix}\]

Now I calculate the determinant of this matrix. Whatever I do, I get the wrong answer. Can you please assist ?

Thank you.
 
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Hi Yankel,

I believe you actually want to find $\text{det }(A-\lambda I)=0$ in order to calculate the eigenvalues. What do you get when you try that?
 
Is your determinant [math]\lambda ^3 - \lambda ^2(k + 2)[/math]?

-Dan
 
Last edited by a moderator:
Yankel said:
Hello all,

I have a matrix A and I am looking for it's eigenvalues. No matter what I do, I find that the eigenvalues are 0, 1 and (k+1), while the answer of both the book and Maple is 0 and (k+2). I tried two different technical approaches, both led to the same place.

The matrix is:

\[A=\begin{pmatrix} 1 &1 &k \\ 1 &1 &k \\ 1 &1 &k \end{pmatrix}\]

I have stated with calculating

\[\lambda I-A\]

which is

\[A=\begin{pmatrix} \lambda -1 &-1 &-k \\ -1 &\lambda -1 &-k \\ -1 &-1 &\lambda -k \end{pmatrix}\]

Now I calculate the determinant of this matrix. Whatever I do, I get the wrong answer. Can you please assist ?

Thank you.
Check your calculations again! You should find that $\det(\lambda I - A) = \begin{vmatrix} \lambda -1 &-1 &-k \\ -1 &\lambda -1 &-k \\ -1 &-1 &\lambda -k \end{vmatrix} = \lambda^2(\lambda-k-2).$
 
Last edited:
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