Let T ∈ℒ(ℝ^2) be defined byT( x y )=( x x+y ), for all ( x y ) ∈ℝ^2

step 1 For exercise 15, part A, we have that T sub 1, T sub 2, A, B, C, D is equal to A plus D. And for

Let T ∈ℒ(ℝ^2) be defined byT( x y )=( x ...

Let $T \in \mathcal{L}\left(\mathbb{R}^{2}\right)$ be defined by$$T\left(\begin{array}{l} x \\y\end{array}\right)=\left(\begin{array}{c}x \\ x+y\end{array}\right), \text { for all }\left(\begin{array}{l}x \\y \end{array}\right) \in \mathbb{R}^{2}$$ Define two real numbers $\lambda_{+}$ and $\lambda_{-}$ as follows: $$\lambda_{+}=\frac{1+\sqrt{5}}{2}, \lambda_{-}=\frac{1-\sqrt{5}}{2}$$ (a) Find the matrix of $T$ with respect to the canonical basis for $\mathbb{R}^{2}$ (both as the domain and the codomain of $T$; call this matrix $A$ ). (b) Verify that $\lambda_{+}$ and $\lambda_{-}$ are eigenvalues of $T$ by showing that $v_{+}$ and $v_{-}$ are eigenvectors, where $$v_{+}=\left(\begin{array}{c}1 \\\lambda_{+}\end{array}\right), v_{-}=\left(\begin{array}{c}1 \\\lambda_{-}\end{array}\right)$$ (c) Show that $\left(v_{+}, v_{-}\right)$ is a basis of $\mathbb{R}^{2}$. (d) Find the matrix of $T$ with respect to the basis $\left(v_{+}, v_{-}\right)$ for $\mathbb{R}^{2}$ (both as the domain and the codomain of $T$; call this matrix $B$ ).

Let $T \in \mathcal{L}\left(\mathbb{R}^{2}\right)$ be defined by$$T\left(\begin{array}{l}
x \\y\end{array}\right)=\left(\begin{array}{c}x \\
x+y\end{array}\right), \text { for all }\left(\begin{array}{l}x \\y
\end{array}\right) \in \mathbb{R}^{2}$$
Define two real numbers $\lambda_{+}$ and $\lambda_{-}$ as follows:
$$\lambda_{+}=\frac{1+\sqrt{5}}{2}, \lambda_{-}=\frac{1-\sqrt{5}}{2}$$
(a) Find the matrix of $T$ with respect to the canonical basis for $\mathbb{R}^{2}$ (both as the domain and the codomain of $T$; call this matrix
$A$ ).
(b) Verify that $\lambda_{+}$ and $\lambda_{-}$ are eigenvalues of $T$ by showing that $v_{+}$ and $v_{-}$ are eigenvectors, where
$$v_{+}=\left(\begin{array}{c}1 \\\lambda_{+}\end{array}\right), v_{-}=\left(\begin{array}{c}1 \\\lambda_{-}\end{array}\right)$$
(c) Show that $\left(v_{+}, v_{-}\right)$ is a basis of $\mathbb{R}^{2}$.
(d) Find the matrix of $T$ with respect to the basis $\left(v_{+}, v_{-}\right)$ for $\mathbb{R}^{2}$ (both as the domain and the codomain of $T$; call this matrix $B$ ).

Key Concepts

Matrix Representation of a Linear Transformation

This concept involves expressing a linear transformation as a matrix with respect to a chosen basis for both the domain and codomain. In finite-dimensional vector spaces, every linear transformation can be represented by a matrix whose entries are determined by the images of basis vectors. The canonical basis is a common choice because it often simplifies computation and understanding of the transformation's structure.

Eigenvalues and Eigenvectors

Eigenvalues are scalars associated with a linear transformation that characterize when certain nonzero vectors (eigenvectors) are scaled by the transformation rather than redirected. Verifying eigenvalues and eigenvectors involves showing that when the transformation is applied to an eigenvector, the result is equal to the eigenvalue multiplied by that eigenvector. This process is fundamental to understanding the geometry of linear maps and simplifies many matrix operations.

Basis and Diagonalization

A set of eigenvectors corresponding to distinct eigenvalues forms a basis of the vector space. Such a basis, known as an eigenbasis, allows the transformation to be represented by a diagonal matrix, a process known as diagonalization. Diagonalization simplifies the analysis of the transformation, making it easier to compute powers of the transformation, solve differential equations, and perform other operations.

Change of Basis

Changing the basis of a vector space, particularly to one composed of eigenvectors, is a powerful tool in linear algebra. It involves transforming the matrix representation of a linear transformation from one basis (such as the canonical basis) to another (such as an eigenbasis), which can simplify the structure of the matrix. This concept is central to many applications in mathematics and science, including simplifying the computation of transforms and understanding invariant subspaces.

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