# fast way to find eigen vector and eigen value by inspection

$\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1& 1 & 0 \\ \end{bmatrix}$

trying to understand how to find Eigen value and vector by inspection for symmetry matrix

$\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1& 1 & 0 \\ \end{bmatrix}$
as the matrix has trace $0$, which mean sum of eigen values is zero. and sum up each number in row produce 2 which mean 2 is one of Eigen value. beside how can I find another Eigen vector by inspection? and also the other one Eigen value is $-1$ but with geometry multiplicity two. how can I find it using inspection?

• Since the matrix is symmetric, it will be led to the guess that all the three elements of the eigenvector are equal, by simple trial and error method, it would be $\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}$ for eigenvalue of 2. – user376921 Feb 18 '18 at 12:34
• For eigenvalue of $-1$, it will be led to think about $\begin{bmatrix} 1 \\ 0 \\ 1 \\ \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}$, or another set $\begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix}$, or $\begin{bmatrix} -1 \\ 0 \\ 1 \\ \end{bmatrix}$ and $\begin{bmatrix} -1 \\ 1 \\ 0 \\ \end{bmatrix}$by symmetry. – user376921 Feb 18 '18 at 12:38
• You can spot that the matrix becomes singular if you add the identiy matrix, so -1 is an eigenvalue. In fact, it's singular with rank 1, so that eigenvalue must have multiplicity 2. – Max Freiburghaus Feb 18 '18 at 12:38
• @LKM thanks! but for this $\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}$and $\begin{bmatrix} 1 \\ 1 \\ 0 \\ \end{bmatrix}$, I multiplied it but wont get $-1$ as $\lambda$ . I multiplied and got $\begin{bmatrix} 1 \\ 1 \\ 2 \\ \end{bmatrix}$. – fiksx Feb 18 '18 at 13:07
• @MaxFreiburghaus but add identity matrix and make it singular only works for symmetry matrices? I mean for other matrix, I need to compute the Eigen value and vector manually? – fiksx Feb 18 '18 at 13:09