Example 3

We start with a new scalar but where now the vector $\boldsymbol{y}$ is replaced by a vector $\boldsymbol{x}$ and the matrix $\boldsymbol{A}$ is a square matrix with dimension $n\times n$.

$$\alpha = \boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x},$$

with $\boldsymbol{x}$ a vector of length $n$.

We write out the specific sums involved in the calculation of $\alpha$

$$\alpha = \sum_{i=0}^{n-1}\sum_{j=0}^{n-1}x_i a_{ij}x_j,$$

taking the derivative of $\alpha$ with respect to a given component $x_k$ we get the two sums

$$\frac{\partial \alpha}{\partial x_k} = \sum_{i=0}^{n-1}a_{ik}x_i+\sum_{j=0}^{n-1}a_{kj}x_j,$$

for $\forall k =0,1,2,\dots,n-1$. We identify these sums as

$$\frac{\partial \alpha}{\partial \boldsymbol{x}} = \boldsymbol{x}^T\left(\boldsymbol{A}^T+\boldsymbol{A}\right).$$

If the matrix $\boldsymbol{A}$ is symmetric, that is $\boldsymbol{A}=\boldsymbol{A}^T$, we have

$$\frac{\partial \alpha}{\partial \boldsymbol{x}} = 2\boldsymbol{x}^T\boldsymbol{A}.$$

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