Expected value of Normal distributed vectors

Let $$A_{k \times n}$$ be a matrix where each element is a independent random normal distribution defined by $$\mathcal{N}(0, \frac{1}{k})$$, and $$x \in \mathbb{R}^n$$.

Each $$j$$-th element of resulting vector $$Ax \in \mathbb{R}^k$$ is given by the distribution $$\mathcal{N}(0, \frac{\sum_{i=1}^d x_i^2}{k})$$. Since $$$$((Ax)_j)^2 = \left( \sum_{i=1}^d A_{ji} x_i \right)^2$$$$

Then, in particular for $$j=1$$ we have. $$$$\mathbb{E} \left[ \| A x \|_2^2 \right] = k \;\mathbb{E} \left[ \left( (A x)_1 \right)^2 \right] = k \frac{\| x \|_2^2}{k} = \| x \|_2^2$$$$

In the last equality, I could not understand how

$$$$\mathbb{E} \left[ \left( (A x)_1 \right)^2 \right] = \frac{\| x \|_2^2}{k}$$$$

This claim was taken from MIT 6.854 Spring 2016 Lecture 5: Johnson Lindenstrauss Lemma and Extensions.

I understand that for a $$X$$ random variable:

$$$$\mathbb{E}(X) = \sum_{i=1}^{n} \lambda_i x_i\quad\text{and}\quad\sum_{i=1}^n \lambda_i = 1$$$$

The matrix $$A_{k \times d}$$, vector $$x_{d \times 1}$$ product is given by $$$$Ax = \begin{bmatrix} \langle A_1, x \rangle \\ \langle A_2, x \rangle \\ \vdots \\ \langle A_k, x \rangle \\ \end{bmatrix}_{k \times 1}$$$$

And the vector norm $$$$\| X \|_2^2 = \sum_{i=1}^n x_i^2$$$$

After working a little with definitions I got the answer.

The definition of a variance is given by:

$$$$Var(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2$$$$

Hence

$$$$\mathbb{E}(X^2) = Var(X) +(\mathbb{E}(X))^2$$$$

For centered gaussian distribution we have $$\mathbb{E}(X) = 0 \implies (\mathbb{E}(X))^2 = 0$$. Leading to

$$$$\mathbb{E}(X^2) = Var(X)$$$$

This last statement implies that square of a centered gaussian distribution is equal to it's variance.

From OP as variance is given by

$$$$\frac{\sum_{i=1}^d x_i^2}{k}$$$$ and by definition $$$$\| X\|_2^2 = \sum x_i^2$$$$ that is exactly the numerator part of variance we can substitute, giving $$$$\frac{\sum_{i=1}^d x_i^2}{k} = \frac{\| X\|_2^2}{k}$$$$

Q.E.D.