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The 1-norm of a vector \(x \in \mathbf{R}^n\) is given by \[\|x\|_1 = |x_1| + |x_2| + \cdots + |x_n|\] Since each coordinate \(x_i\) is a real number, its absolute value \(|x_i|\) is nonnegative. Therefore, the sum of absolute values is also nonnegative, which shows that \(\|x\|_1 \geq 0\). Moreover, if \(\|x\|_1 = 0\), it means that the absolute value of each coordinate is 0, implying that \(x = 0\).

Consider a scalar \(\alpha \in \mathbf{R}\) and a vector \(x \in \mathbf{R}^n\). The 1-norm of the scaled vector \(\alpha x\) is: \[\|\alpha x\|_1 = |\alpha x_1| + |\alpha x_2| + \cdots + |\alpha x_n|\] Since the absolute value has the property that \(|\alpha x_i| = |\alpha| |x_i|\), we can rewrite this sum as: \[\|\alpha x\|_1 = |\alpha| (|x_1| + |x_2| + \cdots + |x_n|) = |\alpha| \|x\|_1\] This shows that the 1-norm satisfies the homogeneity property.

Given two vectors \(x, y \in \mathbf{R}^n\), we have: \[\|x + y\|_1 = |x_1 + y_1| + |x_2 + y_2| + \cdots + |x_n + y_n|\] Using the triangle inequality for absolute values, we have \(|x_i + y_i| \leq |x_i| + |y_i|\) for each \(i \in \{1, 2, \ldots, n\}\). Summing these inequalities, we get: \[\|x + y\|_1 \leq (|x_1| + |x_2| + \cdots + |x_n|) + (|y_1| + |y_2| + \cdots + |y_n|) = \|x\|_1 + \|y\|_1\] Thus, the 1-norm satisfies the triangle inequality. #Case 2#: Prove that the infinity norm is a norm on \(\mathbf{R}^{n}\)

The infinity norm of a vector \(x \in \mathbf{R}^n\) is given by \[\|x\|_\infty = \max(|x_1|, |x_2|, \dots, |x_n|)\] The maximum of nonnegative values is nonnegative. Therefore, \(\|x\|_\infty \geq 0\). If \(\|x\|_\infty = 0\), it means that the maximum absolute value of the coordinates is 0, implying that all coordinates are 0 and \(x = 0\).

Consider a scalar \(\alpha \in \mathbf{R}\) and a vector \(x \in \mathbf{R}^n\). The infinity norm of the scaled vector \(\alpha x\) is: \[\|\alpha x\|_\infty = \max(|\alpha x_1|, |\alpha x_2|, \dots, |\alpha x_n|)\] Recall that for any real number \(z\), we have \(|\alpha z| = |\alpha| |z|\). Thus, we can rewrite the maximum as: \[\|\alpha x\|_\infty = \max(|\alpha| |x_1|, |\alpha| |x_2|, \dots, |\alpha| |x_n|) = |\alpha| \max(|x_1|, |x_2|, \dots, |x_n|) = |\alpha| \|x\|_\infty\] This shows that the infinity norm satisfies the homogeneity property.

Given two vectors \(x, y \in \mathbf{R}^n\), we have: \[\|x + y\|_\infty = \max(|x_1 + y_1|, |x_2 + y_2|, \dots, |x_n + y_n|)\] Using the triangle inequality for absolute values, \(|x_i + y_i| \leq |x_i| + |y_i|\) for each \(i \in \{1, 2, \ldots, n\}\). Therefore, \[\|x + y\|_\infty \leq \max(|x_1| + |y_1|, |x_2| + |y_2|, \dots, |x_n| + |y_n|)\] Since the max function is monotonic, we know that \(\max(a_1, a_2, \dots, a_n) + \max(b_1, b_2, \dots, b_n) \geq \max(a_1 + b_1, a_2 + b_2, \dots, a_n + b_n)\). Applying this to the expression above, we get: \[\|x + y\|_\infty \leq \max(|x_1|, |x_2|, \dots, |x_n|) + \max(|y_1|, |y_2|, \dots, |y_n|) = \|x\|_\infty + \|y\|_\infty\] Thus, the infinity norm satisfies the triangle inequality. In conclusion, both the 1-norm and the infinity norm satisfy all three norm properties, and hence they are both norms on \(\mathbf{R}^{n}\).