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.
 A: Notice that the matrix is symmetric (so, we can exclude option 4 immediately), and the question basically asks if the matrix is negative definite (all eigenvalues are negative), positive definite (all eigenvalues are positive), or indefinite (we have both negative and positive eigenvalues).
Let's check the leading principal minoras:
\begin{align*}
\det A_{11} &= \det \begin{bmatrix} 1 \end{bmatrix} = 1 > 0, \\
\det A_{12} &= \det \begin{bmatrix} 1 & b \\ b & a \end{bmatrix} = a - b^2 > c^2 > 0, \\
\det A_{33} &= \det A = \det \begin{bmatrix} 1 & b & c \\ b & a & 0 \\ c & 0 & 1 \end{bmatrix} = a - ac^2 - b^2 > c^2 - ac^2 = c^2 (1-a) > 0,
\end{align*}
So, all of these are positive and we can conclude that the matrix $A$ is positive definite, i.e., option 2 is correct.
A: 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.
