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Chapter 6: Problem 47

Find the change-of-basis matrix \(P\) from the usual basis \(E\) of \(\mathbf{R}^{2}\) to a basis \(S\), the change-of-basis matrix \(Q\) from \(S\) back to \(E,\) and the coordinates of \(v=(a, b)\) relative to \(S,\) for the following bases \(S\) : (a) \(\quad S=\\{(1,2),(3,5)\\}\) (c) \(S=\\{(2,5),(3,7)\\}\) (b) \(\quad S=\\{(1,-3),(3,-8)\\}\) (d) \(\quad S=\\{(2,3), \quad(4,5)\\}\)

Short Answer

Expert verified
For the given bases, the change-of-basis matrices and the coordinates of the vector \(v=(a, b)\) relative to each basis are: (a) \(S= \{(1,2), (3,5)\}\) P = \(\begin{bmatrix}1 & 3\\ 2 & 5\end{bmatrix}\), Q = \(\begin{bmatrix}5 & -3\\ -2 & 1\end{bmatrix}\), v' = \((5a - 3b, -2a + b)\) (c) \(S= \{(2,5), (3,7)\}\) P = \(\begin{bmatrix}2 & 3\\ 5 & 7\end{bmatrix}\), Q = \(\begin{bmatrix}7 & -3\\ -5 & 2\end{bmatrix}\), v' = \((7a - 3b, -5a + 2b)\) (b) \(S= \{(1,-3), (3,-8)\}\) P = \(\begin{bmatrix}1 & 3\\ -3 & -8\end{bmatrix}\), Q = \(\begin{bmatrix}-8 & -3\\ 3 & 1\end{bmatrix}\), v' = \((-8a - 3b, 3a + b)\) (d) \(S= \{(2,3), (4,5)\}\) P = \(\begin{bmatrix}2 & 4\\ 3 & 5\end{bmatrix}\), Q = \(\begin{bmatrix}-5 & 4\\ 3 & -2\end{bmatrix}\), v' = \((-5a + 4b, 3a - 2b)\)

Step by step solution

01

Compute the change-of-basis matrix P from the standard basis to the given basis S

To compute the change-of-basis matrix P, we need to express each vector of the standard basis E = {(1,0), (0,1)} in terms of the basis S. For this exercise: P = \(\begin{bmatrix}1 & 3\\ 2 & 5\end{bmatrix}\)
02

Compute the change-of-basis matrix Q from the given basis S back to the standard basis E

To compute the matrix Q, we need to find the inverse of the matrix P: Q = \(P^{-1}\) We calculate the inverse of P: \(\begin{bmatrix}1 & 3\\ 2 & 5\end{bmatrix}^{-1}\) = \(\begin{bmatrix}5 & -3\\ -2 & 1\end{bmatrix}\) So, Q = \(\begin{bmatrix}5 & -3\\ -2 & 1\end{bmatrix}\)
03

Compute the coordinates of the vector v relative to the given basis S

To compute the coordinates of the vector v=(a, b) relative to S, we need to multiply the matrix Q by the vector v: v' = Q * v = \(\begin{bmatrix}5 & -3\\ -2 & 1\end{bmatrix}\) * \(\begin{bmatrix}a\\ b\end{bmatrix}\) = \(\begin{bmatrix}5a - 3b\\ -2a + b\end{bmatrix}\) So, the coordinates of v relative to S are: v' = \((5a - 3b, -2a + b)\) This process can be repeated for the remaining bases: #exercise#: (c)\(S= \{(2,5), (3,7)\}\) P = \(\begin{bmatrix}2 & 3\\ 5 & 7\end{bmatrix}\) Q = \(\begin{bmatrix}7 & -3\\ -5 & 2\end{bmatrix}\) v' = \((7a - 3b, -5a + 2b)\) #exercise#: (b) \(S= \{(1,-3), (3,-8)\}\) P = \(\begin{bmatrix}1 & 3\\ -3 & -8\end{bmatrix}\) Q = \(\begin{bmatrix}-8 & -3\\ 3 & 1\end{bmatrix}\) v' = \((-8a - 3b, 3a + b)\) #exercise#: (d) \(S= \{(2,3), (4,5)\}\) P = \(\begin{bmatrix}2 & 4\\ 3 & 5\end{bmatrix}\) Q = \(\begin{bmatrix}-5 & 4\\ 3 & -2\end{bmatrix}\) v' = \((-5a + 4b, 3a - 2b)\)

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Key Concepts

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

Linear Transformations
Linear transformations are functions between vector spaces that respect the operations of vector addition and scalar multiplication. They are the backbone of linear algebra and have many practical applications.

Imagine a linear transformation as a rule that takes a vector and alters it, often changing its direction or magnitude.

For example, you might have a transformation that rotates every vector by 90 degrees or doubles their length. The transformation is linear if for any vectors \(u\) and \(v\), and scalars \(c\) and \(d\), you have:
  • \(T(u + v) = T(u) + T(v)\)
  • \(T(cu) = cT(u)\)
Generally, linear transformations can be represented by matrices. The output of a linear transformation on a vector can be calculated by multiplying the matrix of the transformation by the vector.
Matrix Inversion
Matrix inversion is a crucial concept when dealing with systems of linear equations and transformations. The inverse of a matrix essentially "undoes" the transformation. If you have a transformation matrix \(P\), its inverse \(P^{-1}\) will reverse what \(P\) does.

To find an inverse of a 2x2 matrix \(\begin{bmatrix}a & b \ c & d\end{bmatrix}\), you can use the formula:
\[ P^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d & -b \ -c & a\end{bmatrix}\]
However, not every matrix has an inverse. A matrix will have an inverse only if its determinant (\(ad-bc\) for a 2x2 matrix) is non-zero.

In problems involving change of basis, once you have the change-of-basis matrix \(P\), finding its inverse allows you to convert coordinates back to the standard basis.
Coordinate Systems
Changing coordinate systems might sound complicated, but it's simply like looking at a graph from a different angle. Each basis gives a new perspective.

The standard basis in \(\mathbf{R}^{2}\) is \(\{(1, 0), (0, 1)\}\). It鈥檚 like the default view. Changing the basis means choosing different vectors that span the space. For example, instead of the standard basis, you might use \(\{(1, 2), (3, 5)\}\) to be your basis.

The process of finding the coordinates of a vector \(v = (a, b)\) in a new basis \(S\) involves calculating how \(v\) can be represented as a linear combination of the vectors in \(S\). This involves multiplying the inverse of the change-of-basis matrix by the vector \(v\).

Each choice of basis offers a unique perspective on the same space, making tasks like solving a system of equations simpler in some contexts.
Vector Spaces
Vector spaces are collections of vectors that can be scaled and added together. They're foundational in understanding linear algebra and transformations.

Each vector space has a dimension, which tells you how many vectors are needed to span the entire space without any redundancy. For example, \(\mathbf{R}^{2}\) is a 2-dimensional space.

To change the basis of a vector space means selecting a new set of vectors that can describe every possible vector in that space, just like the original basis.

This new set must be linearly independent, meaning no vector in the set can be written as a combination of the others, and it must span the entire space, meaning any vector in the space can be expressed as a linear combination of the basis vectors.
  • Understanding vector spaces helps in various applications, such as physics and computer graphics, where dimensions and movements are crucial.

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Most popular questions from this chapter

Let \(A: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) be defined by the matrix \(A=\left[\begin{array}{rr}5 & -1 \\ 2 & 4\end{array}\right]\). (a) Find the matrix \(B\) representing \(A\) relative to the basis \(S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,3),(2,8)\\}\). (Recall that \(A\). represents the mapping \(A\) relative to the usual basis \(E\).) (b) For \(v=(a, b)\), find \([v]_{S}\) and \([A(v)]_{s}\).

Two linear operators \(F\) and \(G\) on \(V\) are said to be similar if there exists an invertible linear operator \(T\) on \(V\), such that \(G=T^{-1} \circ F \circ T .\) Prove (a) \(F\) and \(G\) are similar if and only if, for any basis \(S\) of \(V,[F]_{S}\) and \([G]_{S}\) are similar matrices. (b) If \(F\) is diagonalizable (similar to a diagonal matrix), then any similar matrix \(G\) is also diagonalizable.

The set \(S=\left\\{e^{3, t}, t^{3 t}, t^{2} e^{3 \gamma}\right\\}\) is a basis of a vector space \(V\) of functions \(f: \mathbf{R} \rightarrow \mathbf{R}\). Let \(\mathbf{D}\) be the differential operator on \(V ;\) that is, \(\mathbf{D}(f)=d f / d t .\) Find the matrix representation of \(\mathbf{D}\) relative to the basis \(S\).

Show that all matrices similar to an invertible matrix are invertible. More generally, show that similar matrices have the same rank.

Let \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) be defined by \(F(x, y)=(4 x+5 y, 2 x-y)\). (a) Find the matrix \(A\) representing \(F\) in the usual basis \(E\). (b) Find the matrix \(B\) representing \(F\) in the basis \(S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,4),(2,9)\\}\). (c) Find \(P\) such that \(B=P^{-1} A P\). (d) For \(v=(a, b)\), find \([v]_{S}\) and \([F(v)]_{s}\). Verify that \([F]_{S}[v]_{S}=[F(v)]_{s}\).
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