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Find the dimension \(d\) of (a) \(\operatorname{Hom}\left(\mathbf{R}^{2}, \mathbf{R}^{8}\right)\), (b) \(\operatorname{Hom}\left(\mathbf{P}_{4}(t), \mathbf{R}^{3}\right),(\mathrm{c}) \operatorname{Hom}\left(\mathbf{M}_{2,4}, \mathbf{P}_{2}(t)\right)\).

Short Answer

Expert verified
The dimensions of the given Hom spaces are: (a) Hom(\(\mathbf{R}^2\), \(\mathbf{R}^8\)) has dimension 16, (b) Hom(\(\mathbf{P}_4(t)\), \(\mathbf{R}^3\)) has dimension 15, and (c) Hom(\(\mathbf{M}_{2,4}\), \(\mathbf{P}_{2}(t)\)) has dimension 12.

Step by step solution

01

(a) Finding the dimension of Hom(\(\mathbf{R}^2\), \(\mathbf{R}^8\))

A linear transformation from \(\mathbf{R}^2\) to \(\mathbf{R}^8\) can be considered as a matrix of size 8x2, where the input vectors from \(\mathbf{R}^2\) are multiplied on the right side. Each matrix entry represents a coefficient in the linear transformation. To find the dimension of the Hom space, we need to find the total number of entries in the 8x2 matrix, which is 8*2 = 16. Thus, the dimension of Hom(\(\mathbf{R}^2\), \(\mathbf{R}^8\)) is 16.
02

(b) Finding the dimension of Hom(\(\mathbf{P}_4(t)\), \(\mathbf{R}^3\))

A linear transformation from \(\mathbf{P}_4(t)\) to \(\mathbf{R}^3\) takes a polynomial of degree 4 or less and maps it to a vector in \(\mathbf{R}^3\). The transformation can be represented as a 3*(4+1) = 15-element matrix acting on a 5-element vector. Thus, the dimension of Hom(\(\mathbf{P}_4(t)\), \(\mathbf{R}^3\)) is 15.
03

(c) Finding the dimension of Hom(\(\mathbf{M}_{2,4}\), \(\mathbf{P}_{2}(t)\))

A linear transformation from \(\mathbf{M}_{2,4}\) to \(\mathbf{P}_{2}(t)\) takes a 2x4 matrix and maps it to a polynomial of degree 2 or less. The transformation can be represented as a \(4*(2+1) = 12\)-element matrix acting on a 8-element vector (each element represents a coefficient of a 2x4 matrix). Thus, the dimension of Hom(\(\mathbf{M}_{2,4}\), \(\mathbf{P}_{2}(t)\)) is 12.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dimension of Vector Spaces
The dimension of a vector space is a fundamental concept in linear algebra which reflects the number of independent directions within the space. Think of it as the number of coordinates needed to specify any point within the space. For example, the familiar three-dimensional space we live in is three-dimensional because we need three numbers (such as width, length, and height) to describe the position of any point.
In the case of our exercise, we are looking at the dimension of spaces of linear transformations, also known as Hom spaces. To find the dimension of a Hom space, you calculate the number of parameters it takes to describe a linear transformation within that space. Each linear transformation from an n-dimensional space to an m-dimensional space can be represented as an m x n matrix, which has m*n entries. Thus, the dimension of the Hom space is simply m*n.
This information is critical as it gives us a method to quantify the 'size' of the set of all linear transformations from one vector space to another, providing a solid ground for working with vector spaces and understanding their properties.
Linear Transformations
A linear transformation is a special type of function between two vector spaces that preserves the operations of vector addition and scalar multiplication. In simple terms, the transformation of the sum of two vectors is the same as the sum of the transformations, and the transformation of a scalar multiple of a vector is the same as the scalar multiple of the transformation of the vector.
The exercise introduces us to linear transformations via the Hom spaces - spaces consisting of all linear transformations from one vector space to another. To compare the 'size' of different Hom spaces, we looked at their dimensions, which directly relate to the dimensions of the matrices representing these transformations. Understanding linear transformations is pivotal as it is through these transformations that we can analyze and manipulate vector spaces in ways that are beneficial in various fields, such as computer graphics, engineering, and even theoretical physics.
Matrix Representation of Linear Transformations
The matrix representation of linear transformations is an essential tool that allows us to 'translate' the abstract concept of a function into a concrete set of numbers that we can compute with. When we represent a linear transformation as a matrix, we are essentially writing down how that transformation acts on a set of basis vectors. Each column of the matrix corresponds to the image of a basis vector under the transformation.
In our exercise, identifying the dimension of Hom spaces is simplified by using matrices. For instance, a linear transformation taking a vector from \(\mathbf{R}^2\) to \(\mathbf{R}^8\) can be visualized as a >8x2 matrix, while a transformation from a polynomial space \(\mathbf{P}_4(t)\) to \(\mathbf{R}^3\) is represented by a 3x5 matrix. These matrix representations make the underlying abstract concepts much more tangible and forms the crux of how we understand and compute with linear transformations.

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Most popular questions from this chapter

Consider the linear map \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) defined by \(G(x, y, z)=(x+y+z, y-2 z, y-3 z)\) and the unit sphere \(S_{2}\) in \(\mathbf{R}^{3}\), which consists of the points satisfying \(x^{2}+y^{2}+z^{2}=1 .\) Find (a) \(G\left(S_{2}\right)\), (b) \(G^{-1}\left(S_{2}\right)\).

Show that the following mappings are linear: (a) \(\quad F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=(x+2 y-3 z, 4 x-5 y+6 z)\) (b) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(a x+b y, c x+d y),\) where \(a, b, c, d\) belong to \(\mathbf{R}\)

Consider the mapping \(F: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y, z)=\left(y z, x^{2}\right) .\) Find (a) \(F(2,3,4)\) (b) \(F(5,-2,7)\) (c) \(F^{-1}(0,0),\) that is, all \(v \in \mathbf{R}^{3}\) such that \(F(v)=0\)

For each linear map \(G,\) find a basis and the dimension of the kernel and the image of \(G\) : (a) \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(G(x, y, z)=(x+y+z, 2 x+2 y+2 z)\) (b) \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{2}\) defined by \(G(x, y, z)=(x+y, y+z)\) (c) \(G: \mathbf{R}^{5} \rightarrow \mathbf{R}^{3}\) defined by \\[ G(x, y, z, s, t)=(x+2 y+2 z+s+t, \quad x+2 y+3 z+2 s-t, \quad 3 x+6 y+8 z+5 s-t) \\]

Let \(G: \mathbf{R}^{3} \rightarrow \mathbf{R}^{3}\) be the linear mapping defined by $$ G(x, y, z)=(x+2 y-z, \quad y+z, \quad x+y-2 z) $$ Find a basis and the dimension of (a) the image of \(G\), (b) the kernel of \(G\). (a) Find the images of the usual basis of \(\mathbf{R}^{3}\) : $$ \begin{array}{lll} G(1,0,0)=(1,0,1), & G(0,1,0)=(2,1,1), & G(0,0,1)=(-1,1,-2) \end{array} $$ By Proposition 5.4, the image vectors span Im \(G\). Hence, form the matrix \(M\) whose rows are these image vectors, and row reduce to echelon form: $$ M=\left[\begin{array}{rrr} 1 & 0 & 1 \\ 2 & 1 & 1 \\ -1 & 1 & -2 \end{array}\right] \sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 1 & -1 \end{array}\right] \sim\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right] $$ Thus, \((1,0,1)\) and \((0,1,-1)\) form a basis for \(\operatorname{Im} G\); hence, \(\operatorname{dim}(\operatorname{Im} G)=2\). (b) Set \(G(v)=0\), where \(v=(x, y, z)\); that is, $$ G(x, y, z)=(x+2 y-z, \quad y+z, \quad x+y-2 z)=(0,0,0) $$ Set corresponding entries equal to each other to form the following homogeneous system whose solution space is Ker \(G\) : \(\begin{array}{rlrl}x+2 y-z & =0 & x+2 y-z & =0 \\ y+z & =0 & \text { or } & y+z & =0 \\ x+y-2 z & =0 & & -y-z & =0\end{array} \quad\) or \(\quad x+2 y-z=0\) \(y+z=0\) The only free variable is \(z\); hence, \(\operatorname{dim}(\) Ker \(G)=1 .\) Set \(z=1\); then \(y=-1\) and \(x=3\). Thus, \((3,-1,1)\), forms a basis of \(\operatorname{Ker} G .\) [As expected, \(\operatorname{dim}(\operatorname{Im} G)+\operatorname{dim}(\) Ker \(G)=2+1=3=\operatorname{dim} \mathbf{R}^{3}\), the domain of \(G .]\)

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