For what values of $k$ does the matrix
$$\mathbf{A} = \left[ \begin {array}{cc} 1&k\\ 1&1\end {array}
\right] $$
have eigenvalues $\lambda = 0.5, 1.5$?
I have tried to write out the determinant of $\lambda \mathbf{I}_2 – \mathbf{A}$: $(\lambda – 1)(\lambda – 1)-k$. To find the eigenvalues, this determinant should be equal to $0$. In other words, you solve the characteristic polynomial $(\lambda – 1)(\lambda – 1)-k = 0$. I just do not know 1) if this even is the correct/optimal approach to solve the question, 2) if it actually is, how to move forward?
Best Answer
As Alex R. said, your approach is perfectly reasonable.
The eigenvalues satisfy $(\lambda-1)(\lambda-1)-k=0$; i.e., $\lambda^2-2\lambda+(1-k)=0$.
If you want them to be $1.5$ and $0.5$, then $(\lambda-1.5)(\lambda-0.5)=\lambda^2-2\lambda+0.75=0$.
To make these the same, set $1-k=0.75$.
Can you take it from here?