$Q(\vec{0}) = 0$ for all quadratic forms?

Question

Let's define a quadratic form as $Q(\vec{x}) = \vec{x}^TA\vec{x}$.

If $\vec{x} = \vec{0}$, then is, $Q(\vec{x}) = 0$?

(As the title states, $Q(\vec{0}) = 0$ for all quadratic forms?)

I know this may be a simple question for others, but I just want to confirm that this is true (if it is).

Answer 1

Let $\mathbf{x}$ be the vector $\mathbf{x} = [x_1 \; x_2 \; \dots \; x_n]$ and $\mathbf{A}$ be the matrix :

$$\mathbf{A} = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{pmatrix}$$

Then, the general quadratic form $Q(\mathbf{x})$ is defined as :

$$Q(\mathbf{x}) = \mathbf{x}^\mathbf{T}\mathbf{A}\mathbf{x}$$

Now, for $\mathbf{x} = (0,0,\dots,0)$, thus the zero vector, one can easily yield that :

$$Q(\mathbf{0}) = \begin{bmatrix}0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n1} & a_{n2} & \cdots & a_{nn}\end{pmatrix} \begin{bmatrix} 0 & 0 & \cdots & 0 \end{bmatrix} \implies Q(\mathbf{0}) = 0$$

I'll leave the operations/calculations on you to exercise.