Question

Show that \(\left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right]=L U\) is impossible where \(L\) is lower triangular and \(U\) is upper triangular.

Short answer

It's impossible; the equations lead to contradictions.

Step-by-step solution

Define the Matrices

For a matrix \(A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}\), we need to verify if there exist matrices \(L\) and \(U\) such that \(A = LU\), where \(L\) is lower triangular and \(U\) is upper triangular. Express \(L\) and \(U\) in their general forms: \[ L = \begin{bmatrix} a & 0 \ b & c \end{bmatrix} \qquad U = \begin{bmatrix} d & e \ 0 & f \end{bmatrix} \]

Multiply the Matrices

Compute the product \(LU\) using the matrices \(L\) and \(U\) defined previously:\[ LU = \begin{bmatrix} a & 0 \ b & c \end{bmatrix} \begin{bmatrix} d & e \ 0 & f \end{bmatrix} = \begin{bmatrix} ad & af \ bd & bf + cf \end{bmatrix} \]

Compare with Matrix A

Compare the product \(LU\) with the given matrix \(A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}\): - We have \(ad = 0\), \(af = 1\), \(bd = 1\), \(bf + cf = 0\).

Analyze the Equations

Discuss the equations obtained:- From \(ad = 0\), either \(a = 0\) or \(d = 0\).- From \(af = 1\), if \(a = 0\) then there is a contradiction because \(1eq 0\).- Similarly, from \(bd = 1\), if \(d = 0\) there is a contradiction because \(1eq 0\).These account for impossibilities when \(a = 0\) or \(d = 0\), as they contradict both \(af = 1\) and \(bd = 1\).

Conclusion

Since any choice that resolves \(ad = 0\) leads to contradictions with the other equations \(af = 1\) and \(bd = 1\), it is impossible for such matrices \(L\) and \(U\) to exist that satisfy these conditions. Therefore, \(A = LU\) is not possible for lower triangular \(L\) and upper triangular \(U\).

Key concepts

Lower Triangular Matrix

A lower triangular matrix is a square matrix where all entries above the main diagonal are zero. For example, in a 2x2 matrix, it looks like this:

• The main diagonal is from the top left to the bottom right.
• All elements above this diagonal are zero.

In the context of our problem, we represent the lower triangular matrix \( L \) as \( \begin{bmatrix} a & 0 \ b & c \end{bmatrix} \).
Here, \( a \) and \( c \) are on the main diagonal, while \( b \) is below it, with zero above the diagonal. Understanding this helps us see how triangular matrices play into matrix decomposition.

Upper Triangular Matrix

An upper triangular matrix is the opposite of a lower triangular matrix. In these matrices, all entries below the main diagonal are zero. The example form for a 2x2 matrix is:

• The main diagonal includes elements from the top left moving to the bottom right.
• Below this diagonal, all values are zero.

For the exercise, our upper triangular matrix \( U \) is \( \begin{bmatrix} d & e \ 0 & f \end{bmatrix} \).
Here, the non-zero values can appear anywhere along the diagonal and upper part of the matrix, helping us understand how an upper triangular matrix keeps its structure during operations like matrix multiplication.

Linear Algebra

Linear algebra is a branch of mathematics concerning linear equations and their representations in vector spaces and matrices. It is foundational for understanding concepts such as vector spaces, transformations, and especially matrix decomposition.
In our context, matrix decomposition aims to express a matrix as a product of simpler matrices, often triangular for easier computations. Linear algebra provides the tools:

• To analyze systems of linear equations.
• To perform operations on matrices such as decomposition, multiplication, and transformations.

Grasping linear algebra concepts is essential for higher-level mathematical studies, engineering, computer science, and more.

Matrix Multiplication

Matrix multiplication is a way to combine two matrices to produce another matrix. It follows specific rules:

• The number of columns in the first matrix must match the number of rows in the second matrix.
• Each element of the resulting matrix is computed as a dot product of the corresponding row and column.

In our problem, to check if \( L \) and \( U \) could multiply to form matrix \( A \), matrix multiplication was used.
We multiplied \( L = \begin{bmatrix} a & 0 \ b & c \end{bmatrix} \) and \( U = \begin{bmatrix} d & e \ 0 & f \end{bmatrix} \), resulting in a new matrix.
However, comparing it to \( A = \begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix} \) showed a contradiction due to the structure of triangular matrices, demonstrating the importance of understanding these operations.