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Let \(T: M_{22} \rightarrow \mathbb{R}\) be a linear transformation. Show that there are scalars \(a, b, c,\) and \(d\) such that \\[T\left[\begin{array}{ll}w & x \\\y & z\end{array}\right]=a w+b x+c y+d z\\] for all \(\left[\begin{array}{ll}w & x \\ y & z\end{array}\right]\) in \(M_{22}\)

Short Answer

Expert verified
Any linear transformation from 2x2 matrices to real numbers can be expressed as a linear combination of matrix elements.

Step by step solution

01

Understand Linear Transformations

A linear transformation from one vector space to another is a function that respects the addition and scalar multiplication of vectors. Here, we have a transformation \( T: M_{22} \to \mathbb{R} \), where \( M_{22} \) denotes the vector space consisting of all 2x2 matrices over the real numbers.
02

Expressing Matrix Elements

Any 2x2 matrix \( A \in M_{22} \) can be written in terms of its components as \( \left[\begin{array}{ll}w & x \ y & z\end{array}\right] \), where \( w, x, y, \) and \( z \) are real numbers.
03

Write Transformation as Matrix Expression

Since \( T \) is linear, we can express it in terms of a combination of the basis elements of \( M_{22} \): \[\left[\begin{array}{ll}1 & 0 \ 0 & 0\end{array}\right], \left[\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right], \left[\begin{array}{ll}0 & 0 \ 1 & 0\end{array}\right], \left[\begin{array}{ll}0 & 0 \ 0 & 1\end{array}\right].\] This implies \( T \) can be expressed in terms of the linearity over these basis elements.
04

Define Scalars for Basis Elements

Introduce scalars \( a, b, c, \) and \( d \) such that \( T \) can be expressed for each basis element as:\[T\left[\begin{array}{ll}1 & 0 \ 0 & 0\end{array}\right] = a, T\left[\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right] = b, T\left[\begin{array}{ll}0 & 0 \ 1 & 0\end{array}\right] = c, T\left[\begin{array}{ll}0 & 0 \ 0 & 1\end{array}\right] = d.\]
05

Compute Linear Combination of Scalars

For a matrix \( \left[\begin{array}{ll}w & x \ y & z\end{array}\right] \), the linearity property allows us to write:\[T\left[\begin{array}{ll}w & x \ y & z\end{array}\right] = wT\left[\begin{array}{ll}1 & 0 \ 0 & 0\end{array}\right] + xT\left[\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right] + yT\left[\begin{array}{ll}0 & 0 \ 1 & 0\end{array}\right] + zT\left[\begin{array}{ll}0 & 0 \ 0 & 1\end{array}\right]\]\[= aw + bx + cy + dz.\]
06

Conclusion

It is shown that the transformation \( T \) can be expressed as \( T\left[\begin{array}{ll}w & x \ y & z\end{array}\right] = aw + bx + cy + dz \) for scalars \( a, b, c, \) and \( d \). This demonstrates that any linear transformation of a 2x2 matrix to a real number can be boiled down to such a scalar multiplication form.

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Key Concepts

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

Vector Spaces
In linear algebra, a vector space is a collection of vectors that can be added together and multiplied by scalars to produce another vector within the same space. In simple terms, vectors are objects that can be scaled and moved in a certain way in the space they inhabit.
For example, the set of all 2x2 matrices, like those in the exercise, forms a vector space. This is denoted by \( M_{22} \), where each element of the space is a matrix of the form \( \left[\begin{array}{ll}w & x \ y & z\end{array}\right] \). Here, \( w, x, y, \) and \( z \) are real numbers.
This space is vast and contains all possible 2x2 matrices with real number entries. It must uphold certain properties, such as closure under addition and scalar multiplication. This means if you add or scale any matrices in the space, you end up with another matrix that also lies within \( M_{22} \).
These properties make \( M_{22} \) a perfect candidate for performing linear transformations, like the one seen in the exercise, mapping to the real number space \( \mathbb{R} \).
Matrix Representation
Matrices are a powerful way of representing information, particularly when discussing linear transformations between vector spaces. In many applications, a matrix can be thought of as an object that helps transform another object, like a vector, into something else.
For instance, in the exercise, the transformation \( T: M_{22} \to \mathbb{R} \) utilizes matrices to map a 2x2 matrix into a single scalar value. The key here is how we use the 2x2 matrix in the domain to achieve a real number in the codomain.
Every element of the matrix \( \left[\begin{array}{ll}w & x \ y & z\end{array}\right] \) can be manipulated using the transformation to yield the desired scalar result. Through linearity, each matrix element is independently weighted by a scalar as seen from the effects each component plays in contributing to the real number output.
Such representation allows for managing multi-dimensional data and breaking it down to simpler, manageable forms through operations like transformations.
Basis Elements
In a vector space, basis elements are the foundation since they allow every vector within the space to be expressed as a unique linear combination of these elements. Think of them as the building blocks or the "coordinates" system of the space.
For the vector space \( M_{22} \), basis elements are matrices themselves. Recall these specific matrices:
  • \( \left[\begin{array}{ll}1 & 0 \ 0 & 0\end{array}\right] \)
  • \( \left[\begin{array}{ll}0 & 1 \ 0 & 0\end{array}\right] \)
  • \( \left[\begin{array}{ll}0 & 0 \ 1 & 0\end{array}\right] \)
  • \( \left[\begin{array}{ll}0 & 0 \ 0 & 1\end{array}\right] \)

These four matrices are sufficient to "build" any other 2x2 matrix by combining them with suitable scalars \( w, x, y, \) and \( z \). This implies you can generate any matrix in \( M_{22} \) just by adjusting the scalars.
The basis elements are vital for understanding and performing any transformation or operation within the vector space because they provide the framework through which everything else can be interpreted and constructed.
Scalars
Scalars in linear algebra are numbers that are used to scale vectors or matrices, which means they alter the size or magnitude without affecting the direction.
In the exercise, scalars \( a, b, c, \) and \( d \) play crucial roles in defining how the linear transformation \( T \) maps a matrix to a real number. They are used to weight the contribution of each matrix component \( w, x, y, \) and \( z \) respectively.
When we say a matrix transformation is linear, it fundamentally means using these scalars to define how different parts of the matrix are scaled, combined, and transformed. The expression \( T\left[\begin{array}{ll}w & x \ y & z\end{array}\right] = aw + bx + cy + dz \) illustrates exactly how each matrix element is scaled individually.
Thus, each scalar \( a, b, c, \) and \( d \) represents the degree of impact or "weight" corresponding to the unit matrix elements in the basis. Without scalars, we would have no means to define how much influence or contribution each part of the matrix has in resultant transformations.

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Most popular questions from this chapter

Determine whether the linear transformation T is invertible by considering its matrix with respect to the standard bases. If \(T\) is invertible, use Theorem 6.28 and the method of Example 6.82 to find \(T^{-1}\). $$T: \mathscr{P}_{2} \rightarrow \mathscr{P}_{2} \text { defined by } T(p(x))=p^{\prime}(x)$$

Find the solution of the differential equation that satisfies the given boundary condition\((s)\) $$h^{\prime \prime}-4 h^{\prime}+5 h=0, h(0)=0, h^{\prime}(0)=-1$$

Let \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) be a basis for a vector space \(V\) and let \(T: V \rightarrow V\) be a linear transformation. Prove that if \(T\left(\mathbf{v}_{1}\right)=\mathbf{v}_{1}, T\left(\mathbf{v}_{2}\right)=\mathbf{v}_{2} \ldots, T\left(\mathbf{v}_{n}\right)=\mathbf{v}_{n},\) then \(T\) is the identity transformation on \(V\)

A pendulum consists of a mass, called a bob, that is affixed to the end of a string of length \(L\) (see Figure 6.26 . When the bob is moved from its rest position and released, it swings back and forth. The time it takes the pendulum to swing from its farthest right position to its farthest left position and back to its next farthest right position is called the period of the pendulum. (Figure can't copy) Let \(\theta=\theta(t)\) be the angle of the pendulum from the vertical. It can be shown that if there is no resistance, then when \(\theta\) is small it satisfies the differential equation \\[ \theta^{\prime \prime}+\frac{g}{L} \theta=0 \\] where \(g\) is the constant of acceleration due to gravity, approximately \(9.7 \mathrm{m} / \mathrm{s}^{2} .\) Suppose that \(L=1 \mathrm{m}\) and that the pendulum is at rest (i.e., \(\theta=0\) ) at time \(t=0\) second. The bob is then drawn to the right at an angle of \(\theta_{1}\) radians and released. (a) Find the period of the pendulum. (b) Does the period depend on the angle \(\theta_{1}\) at which the pendulum is released? This question was posed and answered by Galileo in \(1638 .\) [Galileo Galilei \((1564-1642)\) studied medicine as a student at the University of Pisa, but his real interest was always mathematics. In \(1592,\) Galileo was appointed professor of mathematics at the University of Padua in Venice, where he taught primarily geometry and astronomy. He was the first to use a telescope to look at the stars and planets, and in so doing, he produced experimental data in support of the Copernican view that the planets revolve around the sun and not the earth. For this, Galileo was summoned before the Inquisition, placed under house arrest, and forbidden to publish his results. While under house arrest, he was able to write up his research on falling objects and pendulums. His notes were smuggled out of Italy and published as Discourses on Two New Sciences in \(1638 .\)

Find the solution of the differential equation that satisfies the given boundary condition\((s)\) $$\text { : } x^{\prime}+x=0, x(1)=1$$

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