calculating eigen values from an equation

Question

I'm trying to use this equation(in yellow) to calculate the eigen values of B = \begin{pmatrix} 1&1&1\\ 1&1&1\\ 1&1&1\\ \end{pmatrix}

but I'm getting $$-λ^3+3λ^2-3λ$$

and the proper answer using the other method is $$det( \begin{pmatrix} 1-λ&1&1\\ 1&1-λ&1\\ 1&1&1-λ\\ \end{pmatrix})$$ $$=-λ^3+3λ^2$$

Anyone see where I could have gone wrong or if the yellow equation only works in certain situations?

Answer 1

$\DeclareMathOperator\tr{tr}$The full proper formula for $n=3$ is: $$\det(M-\lambda I_3)=(-1)^3\lambda^3 + (-1)^2\tr(M)\lambda^2 + (-1)\cdot \frac 12\big[(\tr M)^2-\tr(M^2)\big]\lambda + \det(M)$$ In this case: $$\tr\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}^2 =\tr \begin{pmatrix}3&3&3\\3&3&3\\3&3&3\end{pmatrix} =9$$ So the result is indeed $-\lambda^3+3\lambda^2$.

Answer 2

You should find that $\det(M)=0,$ so that the two methods agree.

Answer 3

$\det(M-\lambda I)=\det\begin{pmatrix}1-\lambda&1&1 \\ 1&1-\lambda&1\\ 1&1&1-\lambda\end{pmatrix}=$

add to the 1st row two others

=$\det\begin{pmatrix}3-\lambda&3-\lambda&3-\lambda \\ 1&1-\lambda&1\\ 1&1&1-\lambda\end{pmatrix}=(3-\lambda)\cdot\det\begin{pmatrix}1&1&1 \\ 1&1-\lambda&1\\ 1&1&1-\lambda\end{pmatrix}=$

Subtract the 1st row from the each of the others

=$(3-\lambda)\cdot\det\begin{pmatrix}1&1&1 \\ 0&-\lambda&0\\ 0&0&-\lambda\end{pmatrix}=\lambda^2(3-\lambda)$

The other approuch

Take $a=(1,1,1)^T$, then $Ma=3a$, so $\lambda=3$ is eigenvalue. Evidently $\lambda=0$ is a root of $\det(M-\lambda I)$, so $\lambda=0$ is eigenvalue too. As $\text{rank} M=1$ the multiplisity of $\lambda=0$ is 2. So $\det(M-\lambda I)=\lambda^2(3-\lambda).$