So I want to find the eigen values and eigen vectors of a matrix with all 1's
\begin{bmatrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{bmatrix}
Only 1 independent would be left,
\begin{bmatrix}1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}
Now, Let's assume λ= 1.
A-λI would give me,
\begin{bmatrix}0&1&1&1\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}
With eigen values as, -1,-1,-1 and 3.
Ultimately, λ1=3 and λ2=-1.
But this is something I have assumed, how can I get eigen values and its vectors by a method? Or what steps should I take ahead? Thanks in advance
Best Answer
HINT
To begin with, obtain the characteristic polynomial:
\begin{align*} \det(A - \lambda I) & = \begin{vmatrix} 1 - \lambda & 1 & 1 & 1\\ 1 & 1 - \lambda & 1 & 1\\ 1 & 1 & 1 - \lambda & 1\\ 1 & 1 & 1 & 1 - \lambda \end{vmatrix}\\\\ & = \begin{vmatrix} 4 - \lambda & 1 & 1 & 1\\ 4 - \lambda & 1 - \lambda & 1 & 1\\ 4 - \lambda & 1 & 1 - \lambda & 1\\ 4 - \lambda & 1 & 1 & 1 - \lambda \end{vmatrix}\\\\ & = (4-\lambda)\begin{vmatrix} 1 & 1 & 1 & 1\\ 1 & 1 - \lambda & 1 & 1\\ 1 & 1 & 1 - \lambda & 1\\ 1 & 1 & 1 & 1 - \lambda \end{vmatrix}\\\\ & = (4-\lambda)\begin{vmatrix} 1 & 0 & 0 & 0\\ 1 & -\lambda & 0 & 0\\ 1 & 0 & -\lambda & 0\\ 1 & 0 & 0 & - \lambda \end{vmatrix}\\\\ & = (\lambda - 4)\lambda^{3} \end{align*}
whence we get the eigenvalues $\lambda\in\{0,4\}$.
Can you take it from here?