Why does $f_A(X,X)>0$ for every $X$ (Question 8 of section 8.2 of Hoffman and Kunze's Linear Algebra) I'm studying Hoffman and Kunze's linear algebra book and I'm having troubles how to prove this exercise on page 276:

The only part of this exercise I couldn't prove was $f_A(X,X)>0$ for every real column matrix $X$.
 A: Let's compute:
$$
X^t A X = \begin{pmatrix} x & y \end{pmatrix} 
\begin{pmatrix} a & c \\ c & b \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = x(ax+cy)+y(cx+by) = ax^2+2cxy + by^2
$$
Now use the additional hypothesis that $a, b>0$ and $ab > c^2$ (which comes from $\det A > 0$). So we get
$$
ax^2+2cxy+by^2 \geq ax^2 + 2cxy+\frac{c^2}{a} y^2 
= \frac{1}{a}(a^2x^2+2acxy+c^2 y^2) = \frac{1}{a}(ax+cy)^2
$$
which will be positive except when $ax=-cy$. Similarly,
$$
ax^2+2cxy+by^2 \geq \frac{c^2}{b} x^2 + 2cxy+ b y^2 
= \frac{1}{b}(c^2x^2+2bcxy+b^2 y^2) = \frac{1}{a}(cx+by)^2
$$
which will be positive except when $cx=-by$. 
Now in case when $ax=-cy$ and $cx=-by$ both hold, we can multiply them to get $abxy = c^2xy$. Since $ab>c^2$, this forces $xy=0$ $\Rightarrow$ either $x=0$ or $y=0$. However, if $x=0$ but $y\neq 0$, then $cx=-by$ cannot hold (since $b>0$). Similarly, if $y=0$ but $x\neq 0$, then $ax=-cy$ cannot hold (since $a>0$). We conclude that if the equality in both cases above hold, then $x=y=0$, that is $X$ is the zero vector. 
So the contrapositive says that if $X$ is a non-zero vector, then the quantity $ax^2+2cxy+by^2$ is positive.
A: The first thing to notice here is that the eigenvalues of $A$ are real. This is because $A$ is real symmetric. The proof of this is standard, I think you can find it easily on the web. 
Then by spectral theorem, there is an orthogonal matrix $O$ such that 
$$
A=O^t D O$$
with $D$ diagonal. 
This gives
$$
X^t AX = (OX)^t D (OX)
$$
Let $Y=OX$. This change of variable gives
$$
X^t AX = Y^t DY = y_{1 }^2 \lambda_1 + y_{2 }^2 \lambda_2
$$
where $\lambda_1, \lambda_2$ are eigenvalues of $A$. 
Now, by $det A>0$, we see that $\lambda_1,  \lambda_2$ are nonzero and they have the same sign.
Finally, by $A_{11}>0$ and $A_{22}>0$, we have $tr A >0$. Since $tr A = \lambda_1+\lambda_2$, we must have 
$$
\lambda_1>0, \ \ \lambda_2 >0.
$$
Therefore, we have $f_A(X,X)>0$ for any nonzero column matrix $X$. 
A: Slightly simpler: With $a>0,c>0,ac>b^2$, write $a=b^2/c+r$ where $r>0$:
$$ \left(\begin{matrix} x & y \end{matrix}\right) 
\left(\begin{matrix} a  & b\\b & c  \end{matrix}\right) 
\left(\begin{matrix} x \\ y \end{matrix}\right) = ax^2 + 2bxy + cy^2 =  rx^2 + c\left(\frac{b}{c} x +y\right)^2\geq 0 $$
with equality iff $x=y=0$
