Show that the largest eigenvalue of $A$ lies in the given interval 
Show that if the given matrix $A$ is positive semi-definite then the largest eigen value of $A$ lies in the interval $(6,7)$.

$$A=\begin{bmatrix} 
5&1&1&1&1&1\\
1&2&1&0&0&0\\
1&1&2&0&0&0\\
1&0&0&1&0&0\\
1&0&0&0&1&0\\
1&0&0&0&0&1
\end{bmatrix}$$
My try:
Since the matrix is given to be positive semidefinite so the spectral radius $\rho(A)$ must be an eigen value of $A$.
Also $\rho (A)=\max_{||x||=1}x^TAx$
I considered the vector $x=(1,0,0,0,0,0)$ then $\rho(A)\ge 5$
So I tried various $x$ such that $||x||=1$ but I find largest lower bound to be $5$.
Is there any way I can show that $\rho\ge 6$
 A: There exist $2\times 2$ orthogonal matrix $Q_2$ and $3\times 3$ orthogonal matrix $Q_3$ such that
$$Q_2\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}\sqrt{2}\\0\end{bmatrix}\quad\text{and}\quad Q_3\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}\sqrt{3}\\0\\0\end{bmatrix}\,,$$
e.g. Householder reflections. Define $6\times6$ orthogonal block diagonal matrix $Q$ with $Q=\operatorname{diag}(1,Q_2,Q_3)$ and note that matrix $A$ is similar to the matrix
$$QAQ^T=\begin{bmatrix}5&\sqrt{2}&0&\sqrt{3}&0&0\\\sqrt{2}&3&0&0&0&0\\0&0&1&0&0&0\\\sqrt{3}&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1\end{bmatrix}\,.$$
Now we know that eigenvalues of $A$ different from $1$ must be eigenvalues of the matrix $B$ defined as
$$B=\begin{bmatrix}5&\sqrt{2}&\sqrt{3}\\\sqrt{2}&3&0\\\sqrt{3}&0&1\end{bmatrix}\,.$$
Also,
$\rho(A)=\max\{1,\rho(B)\}\,.$
Characteristic polynomial of matrix $B$ is
$$k(\lambda) = \lambda^3-9\lambda^2+18\lambda-4\,.$$
Its derivative is a quadratic polynomial whose zeros are $3\pm\sqrt{3}$. From here we conclude that $k'(\lambda)>0$ for $\lambda\in(3+\sqrt{3},+\infty)$. This means that $k(\lambda)$ is strictly increasing on that interval and has at most one zero in it.
Since $k(6)=-4<0$ and $k(7)=24>0$, we know that zero is actualy in $(6,7)$.
This shows that the largest eigenvalue of $B$, $\rho(B)$, satisfies $6<\rho(B)<7$, from where $6<\rho(A)<7$ follows.
