Prove eigenvalues of a symmetric matrix are in a certain interval I am given a matrix $A=\begin{bmatrix}1&2&0\\0&1&2\\0&0&1\end{bmatrix}$. I am asked to compute $A^tA=\begin{bmatrix}1&2&0\\2&5&2\\0&2&5\end{bmatrix}$ 
and then to prove that the eigenvalues of $A^tA$ are all in $]0,8[$.
I have no idea how to prove this. I think it has something to do with matrix norms ?
I don't know whether the previous questions are relevant here, but I was asked to compute 
$e^{tA}=\begin{bmatrix}e^t&2te^t&2t^2e^t\\0&e^t&2te^t\\0&0&e^t\end{bmatrix}$ and $A^{-1}=\begin{bmatrix}1&-2&4\\0&1&-2\\0&0&1\end{bmatrix}$.

If $A^TAx = \lambda x$, then $x^TA^TAx = \lambda x^Tx$, so $\lambda ||x||^2 = ||Ax||^2$, where $||y|| = \sqrt{\sum y_i^2}$ is the norm of the vector $y$. It follows that $\lambda > 0$, since $A$ is invertible, so $||Ax|| \neq 0$ for $x \neq 0$.
To see that $\lambda < 8$, we need to show that for all $x$, we have $||Ax||^2 < 8 ||x||^2$. For this, note that $x = (a,b,c) \implies Ax = (a+2b,b+2c,c)$, in which case $$||Ax||^2 = a^2 + 4b^2 + 4ab  + b^2 + 4c^2 + 4bc + c^2 = a^2 + 5b^2 + 5c^2 + 4ab+4bc$$
Take the difference $8||x||^2 - ||Ax||^2$, it is equal to $7a^2+3b^2+3c^2 -4ab-4bc$. Can we prove this is greater than zero for all $(a,b,c)$ non-zero?
Well, we can, by combining the $ab$ and $bc$ nicely into squares.Like this:
$$
(2b^2 -4bc+2c^2) + (4a^2 - 4ab+b^2) + c^2 + 3a^2 = 2(b-c)^2 + (2a-b)^2 + c^2 + 3a^2
$$
which is a positive linear combination of squares. Note that if the RHS equals zero, this forces $c=a=0$ and $b-c = 0$ so $b = 0$. In other words, if $(a,b,c) \neq (0,0,0)$ then the difference is positive, giving $\lambda < 8$.
