Problem 39
Question
Suppose \(\mathbf{A}\) and \(\mathbf{B}\) are \(n \times n\) matrices and \(\mathbf{A}\) is nonsingular. Show that if \(\mathbf{A} \mathbf{B}=\mathbf{0}\), then \(\mathbf{B}=\mathbf{0}\).
Step-by-Step Solution
Verified Answer
If \(\mathbf{A}\mathbf{B} = \mathbf{0}\) and \(\mathbf{A}\) is nonsingular, then \(\mathbf{B} = \mathbf{0}\).
1Step 1: Understand the Problem
We are given two \(n \times n\) matrices \(\mathbf{A}\) and \(\mathbf{B}\), and it is known that \(\mathbf{A}\) is nonsingular, meaning \(\mathbf{A}\) has an inverse. We aim to show that if the product of \(\mathbf{A}\) and \(\mathbf{B}\) is the zero matrix, then \(\mathbf{B}\) itself must be the zero matrix.
2Step 2: Utilize Matrix Inverse
Since \(\mathbf{A}\) is nonsingular, it has an inverse denoted by \(\mathbf{A}^{-1}\). We will use this inverse to manipulate the equation \(\mathbf{A}\mathbf{B} = \mathbf{0}\) to solve for \(\mathbf{B}\).
3Step 3: Multiply by the Inverse
Multiply both sides of the equation \(\mathbf{A}\mathbf{B} = \mathbf{0}\) by \(\mathbf{A}^{-1}\) from the left:\[\mathbf{A}^{-1}(\mathbf{A}\mathbf{B}) = \mathbf{A}^{-1}\mathbf{0}.\]
4Step 4: Simplify Using Matrix Properties
Applying the property of the inverse, we know \(\mathbf{A}^{-1}\mathbf{A} = \mathbf{I}\), where \(\mathbf{I}\) is the identity matrix. Hence, we have:\[\mathbf{I}\mathbf{B} = \mathbf{0}.\]
5Step 5: Conclude the Proof
Since \(\mathbf{I}\mathbf{B} = \mathbf{B}\) for any matrix \(\mathbf{B}\), the equation \(\mathbf{I}\mathbf{B} = \mathbf{0}\) simplifies to \(\mathbf{B} = \mathbf{0}\). Thus, we have shown that \(\mathbf{B}\) must be the zero matrix.
Key Concepts
Nonsingular MatrixInverse of a MatrixZero MatrixMatrix Multiplication
Nonsingular Matrix
A nonsingular matrix, often referred to as an invertible or non-degenerate matrix, is one with a very important property: it has an inverse. In simpler terms, if you have a matrix \( \mathbf{A} \), it's considered nonsingular if you can find another matrix \( \mathbf{A}^{-1} \) such that multiplying them yields the identity matrix \( \mathbf{I} \).
- Mathematically, this can be represented as \( \mathbf{A} \mathbf{A}^{-1} = \mathbf{A}^{-1} \mathbf{A} = \mathbf{I} \).
- The identity matrix \( \mathbf{I} \) is like multiplying a number by 1; it leaves the original matrix unchanged.
Inverse of a Matrix
The inverse of a matrix is like the matrix equivalent of a reciprocal of a number. If you think about how multiplying a number by its reciprocal gives you 1, a matrix multiplied by its inverse gives you the identity matrix.
- The notation \( \mathbf{A}^{-1} \) is used to represent the inverse of \( \mathbf{A} \).
- Only nonsingular matrices have inverses, which is a key point to remember.
Zero Matrix
The zero matrix is akin to the number zero in regular arithmetic. It is a matrix in which every element is 0. Just as multiplying any number by zero results in zero, multiplying any matrix by a zero matrix results in a zero matrix of compatible dimensions.
- Notation for a zero matrix often simply uses \( \mathbf{0} \).
- Size matters: an \( n \times n \) zero matrix is different from any other dimension zero matrix.
Matrix Multiplication
Matrix multiplication is an operation where two matrices are combined to produce another matrix. It plays a foundational role in linear algebra due to its various applications. Unlike regular multiplication of numbers, matrix multiplication isn't commutative, meaning \( \mathbf{A} \mathbf{B} eq \mathbf{B} \mathbf{A} \) in general.
- The resulting matrix dimensions depend on the original matrices: for matrices \( \mathbf{A} \) of size \( m \times n \) and \( \mathbf{B} \) of size \( n \times p \), the result is \( m \times p \).
- During multiplication, the element at position \((i, j)\) in the result is calculated as the sum of the products of elements from row \(i\) of \( \mathbf{A} \) and column \(j\) of \( \mathbf{B} \).
Other exercises in this chapter
Problem 38
The \(m\) th power of a diagonal matrix $$ \mathbf{D}=\left(\begin{array}{cccc} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & & & \vdots \\ 0 &
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$$ (\mathbf{A}+\mathbf{B})^{2}=\mathbf{A}^{2}+2 \mathbf{A} \mathbf{B}+\mathbf{B}^{2} $$
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An \(n \times n\) matrix \(\mathbf{A}\) is said to be skew-symmetric if \(\mathbf{A}^{T}=-\mathbf{A}\). If \(\mathbf{A}\) is a \(5 \times 5\) skew-symmetric mat
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Solve the system \(2 x_{1}+3 x_{2}-x_{3}=6\) \(x_{1}-2 x_{2}=-3\) \(-2 x_{1}+\quad x_{3}=9\) by writing it as a matrix equation and finding the inverse of the c
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