Let $f : R^n \to R_+$ with $f(x) = x^TAx + 2b^Tx + c$ , where $A \in R^{n\times n} , b \in R^n , c \in R$. Show that the block matrix
$G =\pmatrix{A & b \\ b^T & c}$
is positive semidefinite.
I know that positive semidefinite means $x^TGx \geq 0$ for all x in vector space. For this question I think I need to somehow compute the expression f(x) from block matrix and from that conclude that it is positive semidefinite. However, I lost in dimensions while computing $x^TGx$ since G is matrix of matrix. So, I thought to break vector x in $R^n$ by vector u and v and compute $\pmatrix{u^T & v^T}{G}$ $\pmatrix{u & v}$. But that will not give f(x) expression. How can I approach this question ?
Let $y\in \mathbb R^{n+1}$, denote by $u$ the last coordinate and by $x$ the first $n$ coordinate as a vector of $\mathbb R^n$. Observe that \begin{align*} y^T G y&=x^T A x+ub^Tx+x^Tbu+u^2 c\\ &=x^TAx+2b^Txu+cu^2 \end{align*} Now if $u=0$, then this is exactly $x^TAx$ and if this is negative, then for some $a\in \mathbb R_+$ big enough we can have $f(ax)=x^TAxa^2+2b^T xa+c <0$, indeed it is a second order polynomial in $a$ with negative leading coefficient. This is a contradiction therefore the case $u=0$ is solved, suppose therefore that $u\neq 0$, then \begin{align*} y^T G y &= u^2\left(\frac{x^TAx}{u^2}+\frac{2b^Tx}{u} + c \right)\\ &= u^2 f\left( \frac{x}{u} \right)\\ &\geq 0 \end{align*}
Therefore $G$ is positive semidefinite.