Prove $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$ and determine those $\mathbf{x}$ such that $\mathbf{x}^{\trans}A\mathbf{x}=0$

Problem 559

For each of the following matrix $A$, prove that $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$ for all vectors $\mathbf{x}$ in $\R^2$. Also, determine those vectors $\mathbf{x}\in \R^2$ such that $\mathbf{x}^{\trans}A\mathbf{x}=0$.

(a) $A=\begin{bmatrix}
4 & 2\\
2& 1
\end{bmatrix}$.

 
(b) $A=\begin{bmatrix}
2 & 1\\
1& 3
\end{bmatrix}$.

 
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Proof.

(a) $A=\begin{bmatrix}
4 & 2\\
2& 1
\end{bmatrix}$.

Let $\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}$ be a vector in $\R^2$. Then we have
\begin{align*}
\mathbf{x}^{\trans}A\mathbf{x}&=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
4 & 2\\
2& 1
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
x & y
\end{bmatrix}\begin{bmatrix}
4x+2y \\
2x+y
\end{bmatrix}\\[6pt] &=x(4x+2y)+y(2x+y)=4x^2+2xy+2xy+y^2\\
&=4x^2+4xy+y^2\\
&=(2x+y)^2 \geq 0.
\end{align*}
Thus we have $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$.
Note that $\mathbf{x}^{\trans}A\mathbf{x}=0$ if and only if $2x+y=0$.
Thus those vectors $\mathbf{x}$ such that $\mathbf{x}^{\trans}A\mathbf{x}=0$ are
\[\mathbf{x}=\begin{bmatrix}
x \\
-2x
\end{bmatrix}=x\begin{bmatrix}
1 \\
-2
\end{bmatrix}\] for any real number $x$.

(b) $A=\begin{bmatrix}
2 & 1\\
1& 3
\end{bmatrix}$.

Let $\mathbf{x}=\begin{bmatrix}
x \\
y
\end{bmatrix}$ be a vector in $\R^2$.
We compute
\begin{align*}
\mathbf{x}^{\trans}A\mathbf{x}&=\begin{bmatrix}
x & y
\end{bmatrix}
\begin{bmatrix}
2 & 1\\
1& 3
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}=
\begin{bmatrix}
x & y
\end{bmatrix}
\begin{bmatrix}
2x+y \\
x+3y
\end{bmatrix}\\[6pt] &=x(2x+y)+y(x+3y)=2x^2+xy+xy+3y^2\\
&=2x^2+2xy+3y^2\\
&=x^2+(x+y)^2+2y^2 \geq 0. \tag{*}
\end{align*}
Thus we obtain $\mathbf{x}^{\trans}A\mathbf{x} \geq 0$.
It follows from (*) that $\mathbf{x}^{\trans}A\mathbf{x}=0$ if and only if
\[x^2+(x+y)^2+2y^2=0.\] Since each of $x^2, (x+y)^2, 2y^2$ is nonnegative, we must have $x=y=0$.
Therefore $\mathbf{x}^{\trans}A\mathbf{x}=0$ if and only if $\mathbf{x}=\mathbf{0}$.

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