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Chapter 14: Problem 23

Show that \(A \otimes B=0\) if and only if \(A=0\) or \(B=0\).

Short Answer

Expert verified
In summary, we have proved that \(A \otimes B = 0\) if and only if either \(A = 0\) or \(B = 0\). We used the definition of tensor product to show that if \(A \otimes B = 0\), then every element of \(A\) or \(B\) must be 0. Additionally, we demonstrated the converse: if either \(A = 0\) or \(B = 0\), then their tensor product, \(A \otimes B\), will be the zero matrix.

Step by step solution

01

Understand the problem

In this proof, we're given an equation \(A \otimes B=0\) and we are asked to prove that if this is true, then either \(A=0\) or \(B=0\). So the first step is to understand the meaning of \(A \otimes B\) in terms of matrices \(A\) and \(B\).
02

Apply the definition of the tensor product

To unravel the tensor product \(A \otimes B\), remember that it is calculated by multiplying each element of \(A\) with whole matrix \(B\). The result is a block matrix where each block corresponds to an element of \(A\) times \(B\).
03

Infer from \(A \otimes B = 0\)

Now, consider the given \(A \otimes B = 0\). Given the definition of tensor product, if \(A \otimes B = 0\), it means that each block in the resulting block matrix is the zero matrix. This can only happen if every element in \(A\) is 0 (resulting in \(A = 0\)) or the whole of matrix \(B\) is the zero matrix (resulting in \(B = 0\)).
04

Proving the converse

For the other direction of the proof, we need to show that if either \(A = 0\) or \(B = 0\), then \(A \otimes B = 0\). This is straightforward, because if either \(A\) or \(B\) is the zero matrix, then by the definition of the tensor product, \(A \otimes B\) will indeed be the zero matrix.
05

Final statement

Therefore, we have proved that \(A \otimes B = 0\) if and only if either \(A = 0\) or \(B = 0\). Putting Steps 3 and 4 together completes the proof.

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Key Concepts

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

Zero Matrix
A zero matrix is a matrix in which all elements are zero. These matrices hold significant importance in matrix algebra, especially when dealing with operations like addition and multiplication. In essence, if we multiply any matrix by a zero matrix, the result is always a zero matrix because:
  • Each element in the resulting matrix is obtained by multiplying elements involving zero.
  • This applies to both rows and columns in any given matrix.
Zero matrices serve as the additive identity in the set of all matrices. This means that adding a zero matrix to any matrix leaves the original matrix unchanged. Moreover, in the context of exploratory problems like our exercise, understanding how zero matrices interact with other matrices, such as tensor products, is pivotal.
Zero matrices are also used as a simplification tool in proofs, often indicating that certain conditions within matrix equations must hold true, as seen with the tensor product example.
Block Matrix
Block matrices play a critical role in understanding complex matrix operations. A block matrix is essentially a matrix partitioned into smaller matrices or "blocks." This structure simplifies many operations, making it easier to work with large or complex datasets. When dealing with the tensor product, the matrix formed can directly be regarded as a block matrix. Here鈥檚 why:
  • Each element of one matrix interacts with an entire matrix on the other side of the tensor product operation.
  • The resulting blocks are matrices themselves, which correspond to element-wise operations.
This approach allows for easier visualization and computation of large tensor products. Block matrix representations provide a clearer way of understanding how different sections of matrices contribute to the outcome. They allow us to see patterns in computations and make seemingly intricate tasks more manageable by focusing on smaller, easier-to-manage matrix segments.
Matrix Multiplication
Understanding matrix multiplication is essential when dealing with advanced concepts like the tensor product. Matrix multiplication involves taking the rows of the first matrix and the columns of the second matrix to compute each element of the resulting matrix. Here鈥檚 a breakdown of what this entails:
  • Each element in the resulting matrix is a sum of products. Specifically, each position \(i, j\) in the product matrix is formed by summing up the products of corresponding elements from the row of the first and the column of the second matrix.
  • This operation requires that the number of columns in the first matrix match the number of rows in the second matrix.
In our specific exercise context, understanding the basics of matrix multiplication helps clarify why zero matrices result from tensor products when either participating matrix is zero. This matrix interaction forms the bedrock of further operations in linear algebra and is pivotal in solving more advanced matrix equations.
Mastery of matrix multiplication also opens doors to more advanced concepts and applications in mathematics, physics, and even computer science, illustrating its wide-ranging applications.

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Most popular questions from this chapter

Prove that \(\operatorname{tr}(A \otimes B)=\operatorname{tr}(A) \cdot \operatorname{tr}(B)\).

Show that a) \((A \otimes B)^{t}=A^{t} \otimes B^{t}\) b) \((A \otimes B)^{*}=A^{*} \otimes B^{*}(\) when \(F=\mathbb{C})\)

Prove that \(U \otimes V \approx V \otimes U\).

Let \(A=\left(a_{i, j}\right)\) be the matrix of a linear operator \(\tau \in \mathcal{L}(V)\) with respect to the ordered basis \(\mathcal{A}=\left(u_{1}, \ldots, u_{n}\right)\). Let \(B=\left(b_{i, j}\right)\) be the matrix of a linear operator \(\sigma \in \mathcal{L}(V)\) with respect to the ordered basis \(\mathcal{B}=\left(v_{1}, \ldots, v_{m}\right)\). Consider the ordered basis \(\mathcal{C}=\left(u_{i} \otimes v_{j}\right)\) ordered lexicographically; that is \(u_{i} \otimes v_{j}

Let \(\mathcal{B}=\left\\{b_{i}\right\\}\) be a basis for \(U\) and \(\mathcal{C}=\left\\{c_{i}\right\\}\) be a basis for \(V\). Show that any function \(f: \mathcal{B} \times \mathcal{C} \rightarrow W\) can be extended to a linear function \(\bar{f}: U \otimes V \rightarrow W\). Deduce that the function \(f\) can be extended in a unique way to a bilinear map \(\widehat{f}: U \times V \rightarrow W\). Show that all bilinear maps are obtained in this way.
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