I am trying to find the characteristic polynomial for the following matrix:
$$
A = \begin{pmatrix}7&1&2&2\\
1&4&-1&-1\\
-2&1&5&-1\\
1&1&2&8 \end{pmatrix}
$$
We are only given one eigenvalue, $\lambda=6$.
Now, I realize that I can find the characteristic polynomial using: $\det(A-\lambda I)$, but this looks very long, so I assume there is much easier way to do so but I can't see it.
I also tried thinking about trace and the sum of columns (but the sum of columns isn't all the same so this doesn't really help me)
Could anyone help me see what I'm missing ?
Thanks a lot 🙂
Best Answer
I don't see any shortcut. The characteristic polynomial of that matrix is$$\lambda ^4-24 \lambda ^3+216 \lambda ^2-864 \lambda+1296,$$which turns out to be equal to $(\lambda-6)^4$. So, $6$ is not just an eigenvalue of $A$. It's the only eigenvalue.
You can simplify your computations a lot finding the eigenvectors with eigenvalue $6$ (it is given that they exist). It turns out that the eigenspace corresponding to the eigenvalue $6$ is $2$-dimensional. A basis of it is $\{(-1,-1,1,0),(-1,-1,0,1)\}$. You can add to vectors to these two so that you get a basis of $\Bbb R^4$. For instance, you can add $(1,0,0,0)$ and $(0,1,0,0)$. So, you know that $A.(-1,-1,1,0)=6(-1,-1,1,0)$ and that $A.(-1,-1,0,1)=6(-1,-1,0,1)$. It's not hard to get that:\begin{align}A.(1,0,0,0)&=(7,1,-2,1)\\&=(-1,-1,1,0)-2(-1,-1,0,1)+6(1,0,0,0)\end{align}and that\begin{align}A.(0,1,0,0)&=(1,4,1,1)\\&=(-1,-1,1,0)+(-1,-1,0,1)+3(1,0,0,0)+6(1,0,0,0).\end{align}Therefore, the matrix of $v\mapsto A.v$ with respect to this basis is$$\begin{bmatrix}6 & 0 & 1 & 1 \\ 0 & 6 & -2 & 1 \\ 0 & 0 & 6 & 3 \\ 0 & 0 & 0 & 6\end{bmatrix}.$$But this matrix is similar to $A$ and it is clear that its characteristic polynomial is $(\lambda-6)^4$.