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The 2-norm of a vector \(\mathbf{y}\), denoted as \(\|\mathbf{y}\|_2\), is defined as: \[ \|\mathbf{y}\|_2 = \sqrt{\sum_{i=1}^{n} y_i^2}, \] where \(y_i\) are the components of the vector \(\mathbf{y}\). The 2-norm of a matrix \(Y\), which is also called the spectral norm, is denoted as \(\|Y\|_2\). It can be defined as the maximum value of \(\frac{\|Y\mathbf{x}\|_2}{\|\mathbf{x}\|_2}\), where the maximum is taken over all nonzero vectors \(\mathbf{x}\). That means, \(\|Y\|_2 = \max_{\mathbf{x} \neq 0} \frac{\|Y\mathbf{x}\|_2}{\|\mathbf{x}\|_2}\). Now, let's move to the next step and prove the equality for (a).

We know that \(Y=\mathbf{y}\), and we want to show that \(\|Y\|_2 = \|\mathbf{y}\|_2\). We know that the 2-norm of a matrix is the maximum value of \(\frac{\|Y\mathbf{x}\|_2}{\|\mathbf{x}\|_2}\) for all nonzero \(\mathbf{x}\). In this case, \(Y\) is an \(n \times 1\) matrix (a column vector), and \(\mathbf{x}\) is an \(n\)-dimensional vector. Therefore, when we calculate \(Y\mathbf{x}\), we will get a scalar product of two vectors \(\mathbf{y}^T \mathbf{x}\). Considering the scalar product, we know that \(|\mathbf{y}^T \mathbf{x}| \leq \|\mathbf{y}\|_2 \cdot \|\mathbf{x}\|_2\) (by Cauchy-Schwarz inequality). Thus, \[\frac{\|Y\mathbf{x}\|_2}{\|\mathbf{x}\|_2} = \frac{|\mathbf{y}^T \mathbf{x}|}{\|\mathbf{x}\|_2} \leq \|\mathbf{y}\|_2. \] Taking the maximum over all nonzero \(\mathbf{x}\), we have \[\|Y\|_2 \leq \|\mathbf{y}\|_2. \] Now let's consider \(\mathbf{x} = \frac{\mathbf{y}}{\|\mathbf{y}\|_2}\), then: \[\frac{\|Y\mathbf{x}\|_2}{\|\mathbf{x}\|_2} = \frac{|\mathbf{y}^T \frac{\mathbf{y}}{\|\mathbf{y}\|_2}|}{\frac{\|\mathbf{y}\|_2}{\|\mathbf{y}\|_2}} = \|\mathbf{y}\|_2. \] Therefore, \[\|Y\|_2 = \|\mathbf{y}\|_2. \] So, we have proven part (a) of the exercise. For part (b), we need to prove that the 2-norm of the transpose of the matrix representation of the vector \(\mathbf{y}\) is equal to the 2-norm of the vector \(\mathbf{y}\).

In this step, we will prove that \(\|Y^T\|_2 = \|\mathbf{y}\|_2\). Recall that \(Y^T = \mathbf{y}^T\) is a \(1 \times n\) matrix (a row vector). We know that the 2-norm of a matrix is the maximum value of \(\frac{\|Y^T\mathbf{x}\|_2}{\|\mathbf{x}\|_2}\), for all nonzero \(\mathbf{x}\). In this case, when we calculate \(Y^T\mathbf{x}\), we get a scalar product of two vectors \(\mathbf{y}^T \mathbf{x}\). From our proof in Step 2, we have already shown that \(\frac{\|Y\mathbf{x}\|_2}{\|\mathbf{x}\|_2} \leq \|\mathbf{y}\|_2\). Since the scalar product \(\mathbf{y}^T \mathbf{x}\) is the same as in Step 2, the same inequality holds for the transpose: \[ \frac{\|Y^T\mathbf{x}\|_2}{\|\mathbf{x}\|_2} = \frac{|\mathbf{y}^T \mathbf{x}|}{\|\mathbf{x}\|_2} \leq \|\mathbf{y}\|_2. \] Taking the maximum over all nonzero \(\mathbf{x}\), we have \[\|Y^T\|_2 \leq \|\mathbf{y}\|_2. \] Now, using the same argument as in Step 2, we can choose \(\mathbf{x} = \frac{\mathbf{y}}{\|\mathbf{y}\|_2}\) and get: \[\frac{\|Y^T\mathbf{x}\|_2}{\|\mathbf{x}\|_2} = \frac{|\mathbf{y}^T \frac{\mathbf{y}}{\|\mathbf{y}\|_2}|}{\frac{\|\mathbf{y}\|_2}{\|\mathbf{y}\|_2}} = \|\mathbf{y}\|_2. \] Thus, we have \[\|Y^T\|_2 = \|\mathbf{y}\|_2. \] We have proven part (b) of the exercise. In conclusion, we have shown that for a vector \(\mathbf{y}\) and its matrix representations \(Y\) and \(Y^T\): (a) \(\|Y\|_2 = \|\mathbf{y}\|_2\) (b) \(\|Y^T\|_2 = \|\mathbf{y}\|_2\)