Value of an element of a symmetric matrix with given eigenvalue. Let \begin{bmatrix}
    a & b &  c \\
    b & d & e \\
    c & e & f
\end{bmatrix} be a real matrix with eigenvalues $1$,$0$ and $3$. If the eigenvectors corresponding to $1$ and $0$ are $(1,1,1)^{T}$ and$(1,-1,0)^{T}$ respectively, then what is the value of $3f?$  How to find value of $3f?$
From the given conditions on eigenvalues we have $$a+b+c=1\\b+d+e=1\\c+e+f=1\\a-b=0\\b-d=0\\c-e=0.$$ Is there any short way to find value of $f?$ Please suggest me. Thanks a lot.
 A: As Friedrich points out, we should make use of the assumption that the matrix is symmetric and therefore the eigenvectors are orthogonal each other. So the first task is to solve $$\left(\begin{array}{ccc}1 & 1 & 1\\1 & -1 & 0\end{array}\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=0$$ so that we obtain the remaining eigenvector (please check by yourself that the result is $(-1,-1,2)^T$). 
Now we know the image of the transformation for one set of basis $\{e_1=(1,1,1)^T,e_2=(1,-1,0)^T,e_3=(-1,-1,2)^T\}$ for $\mathbb{R}^3$; namely $$Ae_1=\left(\begin{array}{c}1\\1\\1\end{array}\right),\quad Ae_2=\left(\begin{array}{c}0\\0\\0\end{array}\right),\quad Ae_3=\left(\begin{array}{c}-3\\-3\\6\end{array}\right).$$
To obtain a value of $f$, easiest way I believe is to evaluate $$A\left(\begin{array}{c}0\\0\\1\end{array}\right)(=\left(\begin{array}{c}c\\e\\f\end{array}\right)).$$
So we want to know the set of coefficients $(c_1,c_2,c_3)$ such that $c_1e_1+c_2e_2+c_3e_3=(0,0,1)^T.$ Nice thing about the matrix which consists of orthogonal (if not orthonormal) vectors is that we can obtain diagonal matrix by multiplying its transpose. Thanks to this, we can solve the system easily (check by yourself that $(c_1,c_2,c_3)=(1/3,0,1/3)$). After all, we obtain $f=7/3.$
A: We know trace $a+d+f=4$
Take help from equations you have got from $(A-\lambda I)X=0$
$a=b=d$ and $c=e$
Now $a+d+f=2a+f=4-----(1)$
Find $a$ which is not very hard 
$a+b+c=2a+c=1------(2)$
$c+e+f=2c+f=1------(3)$
Solving $(2),(3)$ we have $4a-f=1----(4)$
Solving $1$ and $4$ we have $f=\dfrac{7}{3}$
$3f=7$
A: You forgot to take into account that $3$ is another eigenvalue. Moreover, recall that eigenvectors corresponding to different eigenvalues are orthogonal to each other. So, you can obtain an eigenvector corresponding to $3$.
