Question

In each case, show that \(T\) is not a linear transformation. a. \(T: \mathbf{M}_{n n} \rightarrow \mathbb{R} ; T(A)=\operatorname{det} A\) b. \(T: \mathbf{M}_{n m} \rightarrow \mathbb{R} ; T(A)=\operatorname{rank} A\) c. \(T: \mathbb{R} \rightarrow \mathbb{R} ; T(x)=x^{2}\) d. \(T: V \rightarrow V ; T(\mathbf{v})=\mathbf{v}+\mathbf{u}\) where \(\mathbf{u} \neq \mathbf{0}\) is a fixed vector in \(V(T\) is called the translation by \(\mathbf{u})\)

Short answer

Each function fails to satisfy linearity conditions, so none are linear transformations.

Step-by-step solution

Understanding Linear Transformations

A transformation \( T \) is linear if it satisfies two conditions for all vectors \( \mathbf{u} \) and \( \mathbf{v} \) and scalar \( c \): 1) \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \), and 2) \( T(c\mathbf{u}) = cT(\mathbf{u}) \). If a transformation does not satisfy these conditions, it is not linear.

Verifying Linearity of Determinant Function

For the function \( T: \mathbf{M}_{n n} \rightarrow \mathbb{R} ; T(A)=\operatorname{det} A \), check the first property. Consider two matrices, \( A \) and \( B \). The determinant of \( A + B \) is not necessarily equal to \( \operatorname{det}(A) + \operatorname{det}(B) \). Thus, \( T \) does not satisfy the first condition and is not linear.

Verifying Linearity of Rank Function

For the function \( T: \mathbf{M}_{n m} \rightarrow \mathbb{R} ; T(A)=\operatorname{rank} A \), consider matrices \( A \) and \( B \). Generally, \( \operatorname{rank}(A + B) \leq \operatorname{rank}(A) + \operatorname{rank}(B) \), which doesn't guarantee the equality needed for linearity. Therefore, \( T \) is not linear.

Verifying Linearity of Squaring Function

For the function \( T: \mathbb{R} \rightarrow \mathbb{R} ; T(x)=x^{2} \), check the linearity conditions. If you take two numbers, \( x \) and \( y \), \( (x+y)^2 eq x^2 + y^2 \) in general. Hence, \( T \) does not satisfy the first condition of linearity and is not linear.

Verifying Linearity of Translation

For the function \( T: V \rightarrow V ; T(\mathbf{v}) = \mathbf{v} + \mathbf{u} \), where \( \mathbf{u} eq \mathbf{0} \), test with \( \mathbf{u} + \mathbf{v} + \mathbf{w} \). \( T(\mathbf{v} + \mathbf{w}) = (\mathbf{v} + \mathbf{w}) + \mathbf{u} \), but \( T(\mathbf{v}) + T(\mathbf{w}) = (\mathbf{v} + \mathbf{u}) + (\mathbf{w} + \mathbf{u}) \). The results are not equal, proving \( T \) is not linear since it doesn't satisfy the first condition.

Key concepts

Determinant Function

The determinant is a key property of square matrices, providing a scalar value reflecting certain geometric and algebraic properties. One important fact to note is that the determinant offer insights into matrix invertibility—only matrices with a non-zero determinant are invertible. In terms of linear algebra, any transformation involving determinants is likely to break the linearity rules due to how determinants behave with addition. If you consider two matrices, \( A \) and \( B \), the general relationship is \( \operatorname{det}(A + B) eq \operatorname{det}(A) + \operatorname{det}(B) \). This means that when a function based on determinants is checked for linearity, it often fails the additive property test. A transformation involving matrix determinants cannot be linear because it doesn't preserve simple addition.

Rank Function

The rank of a matrix is a crucial concept for understanding the solutions to matrix equations. It gives the dimension of the column space of the matrix, which translates to the number of linearly independent columns in the matrix. While analyzing a rank function for linearity properties, the additive property becomes the focal point. For matrices \( A \) and \( B \), typically \( \operatorname{rank}(A + B) \leq \operatorname{rank}(A) + \operatorname{rank}(B) \). This inequality tells us that adding two matrices concerns the linear independence of rows and columns, instead of simply adding up ranks. Hence, rank cannot satisfy the conditions needed for a linear transformation, primarily due to failing the additive property.

Squaring Function

Squaring is a familiar operation in mathematics, as it refers to multiplying a number by itself. In algebra, squaring possesses fascinating attributes but doesn't align with linear transformations' kindergarten rules. A function \( T(x) = x^2 \), for instance, flouts both fundamental tenets of linearity: the scalar multiplication and the additivity conditions. Consider two numbers, \( x \) and \( y \); it's generally true that \( (x+y)^2 eq x^2 + y^2 \). Thus, this squaring operation doesn't uphold the additivity requirement of linear transformations, showing that while powerful, squaring disrupts linearity.

Translation Vector

Translation in vector spaces involves shifting each vector by a fixed vector, \( \mathbf{u} \). Imagine in two or three-dimensional space, it's akin to moving every point by the same amount and direction. The translation expressed as \( T(\mathbf{v}) = \mathbf{v} + \mathbf{u} \) isn't linear because it introduces constant shifts, altering the zero vector's location which should remain stationary in genuinely linear operations. When verifying with \( \mathbf{v} + \mathbf{w} \), we find \( T(\mathbf{v} + \mathbf{w}) = (\mathbf{v} + \mathbf{w}) + \mathbf{u} eq (\mathbf{v} + \mathbf{u}) + (\mathbf{w} + \mathbf{u}) \), thus breaking the core additivity condition required for linearity. This constant addition by the translation vector deflects from maintaining a direct path through the origin point of vector space, marking it as a nonlinear transformation in a mathematical context.