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

[Yankel]

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.

 

[Jameson]

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?

 

[topsquark]

Is your determinant \(\displaystyle \lambda ^3 - \lambda ^2(k + 2)\)?

-Dan

 

[Opalg]

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 $\det(\lambda I - A) = \begin{vmatrix} \lambda -1 &-1 &-k \\ -1 &\lambda -1 &-k \\ -1 &-1 &\lambda -k \end{vmatrix} = \lambda^2(\lambda-k-2).$