Identifying the nature of the eigenvalues I wish somebody could help me in this one. We have to choose one of the $4$ options.
Let $a,b,c$ be positive real numbers such that $b^2+c^2<a<1$. Consider the $3 \times 3$ matrix
$$A=\begin{bmatrix}
        1 & b & c \\
        b & a & 0 \\
        c & 0 & 1 \\
        \end{bmatrix}.$$


*

*All the eigenvalues of $A$ are negative real numbers.

*All the eigenvalues of $A$ are positive real numbers.

*$A$ can have a positive as well as a negative eigenvalue.

*Eigenvalues of $A$ can be non-real complex numbers.
Now, $\det(A-\lambda I)=0$, so
$$\begin{vmatrix}
        1-\lambda & b & c \\
        b & a-\lambda & 0 \\
        c & 0 & 1-\lambda \\
        \end{vmatrix}=0$$
$\implies(1-\lambda)(a-a\lambda -\lambda +\lambda^2)-b(b-b\lambda)-c(ac-c\lambda)=0$
$\implies a-a\lambda -\lambda +\lambda^2-a\lambda+a\lambda^2 +\lambda^2 -\lambda^3-b^2+b^2\lambda-ac^2+c^2\lambda=0$
$\implies-\lambda^3+\lambda^2(2+a)+\lambda(-2a-1+b^2+c^2)+a-b^2-ac^2=0$
I am stuck here, don't know how to proceed. Need your help, please.

Let $\mathbf{x}=[x_1\quad x_2\quad x_3]^{T}$ is an arbitrary vector in $\mathbb{R}^3$. Then,\begin{align} \mathbf{x^T}A\mathbf{x}=&[x_1\quad x_2\quad x_3]\begin{bmatrix}
1 & b & c\\
b & a & 0\\
c & 0 & 1\\
\end{bmatrix}\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} \\
\ =& x_1^2+x_3^2+ax_2^2+2bx_1x_2+2cx_1x_3\\
\ =& (x_1+bx_2+cx_3)^2+ax_2^2+x_3^2-(bx_2+cx_3)^2\\
\ =& (x_1+bx_2+cx_3)^2+(a-b^2)x_2^2+(1-c^2)x_3^2-2bcx_2x_3\\
\ >& (x_1+bx_2+cx_3)^2+c^2x_2^2+b^2x_3^2-2bcx_2x_3\\
\ =& (x_1+bx_2+cx_3)^2+(cx_2-bx_3)^2\ge 0
\end{align}
Hence $A$ is positive definite and also it is symmetric $\Rightarrow $ all the eigenvalues of $A$ are positive and real. So option $(2)$ is the correct one.
