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Let \(T: M_{22} \rightarrow \mathbb{R}\) be a linear transformation for which \\[\begin{array}{l}T\left[\begin{array}{ll}1 & 0 \\\0 & 0\end{array}\right]=1, \quad T\left[\begin{array}{ll} 1 & 1 \\\0 & 0\end{array}\right]=2 \\\T\left[\begin{array}{ll}1 & 1 \\\1 & 0\end{array}\right]=3, \quad T\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]=4\end{array}\\] Find \(T\left[\begin{array}{ll}1 & 3 \\ 4 & 2\end{array}\right]\) and \(T\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\)

Short Answer

Expert verified
\(T\left[\begin{array}{cc}1 & 3 \\ 4 & 2\end{array}\right] = 10\), \(T\left[\begin{array}{cc}a & b \\ c & d\end{array}\right] = a + b + c + d\)."

Step by step solution

01

Express matrix in terms of known matrices

To solve for \(T\left[\begin{array}{cc}1 & 3 \ 4 & 2\end{array}\right]\), express the matrix \(\begin{array}{cc}1 & 3 \ 4 & 2\end{array}\) as a linear combination of the given matrices. Use the basis matrices to create the target matrix:\[\begin{array}{cc}1 & 3 \ 4 & 2\end{array}\] can be represented as:\[\left[\begin{array}{cc}1 & 3 \ 4 & 2\end{array}\right] = \left[\begin{array}{cc}1 & 0 \ 0 & 0\end{array}\right] + 2\left[\begin{array}{cc}0 & 1 \ 0 & 0\end{array}\right] + 4\left[\begin{array}{cc}0 & 0 \ 1 & 0\end{array}\right] + 2\left[\begin{array}{cc}0 & 0 \ 0 & 1\end{array}\right]\]
02

Apply linearity of transformation

Now use the linearity property of the transformation \(T\):\[T\left(\left[\begin{array}{cc}1 & 3 \ 4 & 2\end{array}\right]\right) = T\left(\left[\begin{array}{cc}1 & 0 \ 0 & 0\end{array}\right]\right) + 3T\left(\left[\begin{array}{cc}0 & 1 \ 0 & 0\end{array}\right]\right) + 4T\left(\left[\begin{array}{cc}0 & 0 \ 1 & 0\end{array}\right]\right) + 2T\left(\left[\begin{array}{cc}0 & 0 \ 0 & 1\end{array}\right]\right)\]From the information provided, we generate a system of equations to find the transformation of these elementary matrices.
03

Identify transformations of basis matrices

We need to find the transformations of these basis matrices:- \(T\left(\left[\begin{array}{cc}0 & 1 \ 0 & 0\end{array}\right]\right) = T\left(\left[\begin{array}{cc}1 & 1 \ 0 & 0\end{array}\right]\right) - T\left(\left[\begin{array}{cc}1 & 0 \ 0 & 0\end{array}\right]\right) = 2 - 1 = 1\)- \(T\left(\left[\begin{array}{cc}0 & 0 \ 1 & 0\end{array}\right]\right) = T\left(\left[\begin{array}{cc}1 & 1 \ 1 & 0\end{array}\right]\right) - T\left(\left[\begin{array}{cc}1 & 1 \ 0 & 0\end{array}\right]\right) = 3 - 2 = 1\)- \(T\left(\left[\begin{array}{cc}0 & 0 \ 0 & 1\end{array}\right]\right) = T\left(\left[\begin{array}{cc}1 & 1 \ 1 & 1\end{array}\right]\right) - T\left(\left[\begin{array}{cc}1 & 1 \ 1 & 0\end{array}\right]\right) = 4 - 3 = 1\)
04

Solve for specific transformation

Using the results of Step 3 in the linearity equation:\[T\left(\left[\begin{array}{cc}1 & 3 \ 4 & 2\end{array}\right]\right) = 1 + 3 \cdot 1 + 4 \cdot 1 + 2 \cdot 1 = 1 + 3 + 4 + 2 = 10\]
05

Generalize for arbitrary matrix

For the arbitrary matrix \(\begin{array}{cc}a & b \ c & d\end{array}\), apply the linearity:\[T\left(\left[\begin{array}{cc}a & b \ c & d\end{array}\right]\right) = aT\left(\left[\begin{array}{cc}1 & 0 \ 0 & 0\end{array}\right]\right) + bT\left(\left[\begin{array}{cc}0 & 1 \ 0 & 0\end{array}\right]\right) + cT\left(\left[\begin{array}{cc}0 & 0 \ 1 & 0\end{array}\right]\right) + dT\left(\left[\begin{array}{cc}0 & 0 \ 0 & 1\end{array}\right]\right)\]Substitute the known transformations:\[= a \cdot 1 + b \cdot 1 + c \cdot 1 + d \cdot 1 = a + b + c + d\]
06

Final Result

Therefore, the transformation for the specific matrix is 10 and for the arbitrary matrix it is the sum of its elements, \(a + b + c + d\).

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Key Concepts

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

Basis Matrices
Understanding basis matrices is a core concept in linear transformations. In the context of the given problem, basis matrices serve as building blocks for constructing other matrices. They are often structured as elementary matrices where most elements are zero, except for a few designated ones.

In our exercise, the basis matrices come into play as matrices like \( \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \). This matrix has a non-zero element in the top-left corner. Such matrices help simplify the matrix transformation problem, as they can represent any given matrix as a linear combination of such basis matrices.

When solving for linear transformations, these basis matrices allow us to express any target matrix as a sum (or combination) of these simpler matrices. Thus they are crucial for simplifying complex problems into manageable calculations.

Key points about basis matrices include:
  • They simplify complex matrices into recognizable patterns.
  • Each basis matrix is almost entirely zeros, with a unique structure.
  • They help in constructing a matrix transformation by using linear combinations.
Linearity Property
The linearity property is an essential aspect of understanding linear transformations. This property allows us to decompose and compute transformations conveniently by ensuring that the transformation of a sum is the sum of the transformations.

More formally, for a linear transformation \( T \), and matrices \( A \) and \( B \), with a scalar \( c \), linearity implies that:
  • \( T(A + B) = T(A) + T(B) \)
  • \( T(cA) = cT(A) \)
In our exercise, this property allows us to find the transformation of a complex matrix by first breaking it down into simpler, basis matrices. We then apply the transformation to each basis matrix individually, ensuring each component’s transformation can be added together to achieve the transformation of the initial matrix.

The linearity property is powerful because it allows us to handle large and seemingly complex matrices by "piecewise" operating on simpler constituent parts. This simplifies computation tremendously and ensures accuracy at each step.
Matrix Transformation
Matrix transformation refers to mapping matrices from one space to another using a specific rule or function. In our example, you're transforming a matrix from a space of 2x2 matrices \( M_{22} \) to real numbers \( \mathbb{R} \).

This transformation is guided entirely by the rules set by the linearity property. The known outputs of specific basis matrices, such as \( T \left( \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \right) = 1 \), help set these rules.

Understanding matrix transformations involves knowing:
  • The domain and range (or output space) of your transformation.
  • The specific results for given input matrices, which act as guidelines or rules.
  • How to apply the linearity property to extend these rules to other matrices.
By calculating transformations like \( T \left( \begin{bmatrix} a & b \ c & d \end{bmatrix} \right) = a + b + c + d \), you generalize specific transformations to any matrix by extending the rules learned from known transformations.
Linear Combination
Linear combination is a mathematical expression whereby we construct new matrices by adding scalar multiples of other matrices. This concept is pivotal in breaking down complex matrices into simpler, solvable pieces using known matrices.

In the given problem, for the matrix \( \begin{bmatrix} 1 & 3 \ 4 & 2 \end{bmatrix} \), we express it as a linear combination of simpler matrices, each of which corresponds to a basis matrix:
  • \( 1 \cdot \begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix} \) - The first term contributes 1 to the top-left corner.
  • \( 3 \cdot \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} \) - The multiplier 3 equals the element in the top-right corner.
  • \( 4 \cdot \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix} \) - Produces the 4 seen in the bottom-left corner.
  • \( 2 \cdot \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix} \) - Equates to a 2 at the bottom-right corner.
By combining these, one constructs the initial matrix entirely in terms of these simpler matrices.

Linear combination focuses on this synthesis process, showcasing how any matrix can be built from simpler building blocks adjusted by scaling coefficients.

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