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Chapter 13: Problem 6

Show that \((\mathrm{a}) \quad I^{*}=I,\) and \((\mathrm{b}) \quad 0^{*}=0\)

Short Answer

Expert verified
In summary, to show that (a) \(I^* = I\) and (b) \(0^* = 0\), we find the adjugate and determinant of the respective matrices. For the identity matrix \(I\), its adjugate is also the identity matrix, and its determinant is 1. Dividing the adjugate by the determinant, we confirm \(I^* = I\). For the zero matrix \(0\), its adjugate is also the zero matrix, and its determinant is 0. Despite potential confusion due to division by zero, the adjugate of \(0\) is the zero matrix, allowing us to conclude \(0^* = 0\).

Step by step solution

01

Finding the adjugate of a matrix

Find the adjugate of a matrix by replacing each element with its cofactor. The cofactor of an element is the determinant of the matrix formed by removing the row and column containing that element, multiplied by \((-1)^{(i+j)}\), where \(i\) and \(j\) are the row and column indices of that element.
02

Part (a)

Step 1: Find the adjugate of the identity matrix, \(I\): The identity matrix is a square matrix with 1s on the diagonal and 0s elsewhere. Its determinant is 1. The identity matrix's cofactors are the same as its elements, as they all have a determinant of 0, except for the diagonal. The adjugate of the identity matrix is also the transpose of its cofactor matrix, which in this case, is the same as the cofactor matrix. Therefore, the adjugate of the identity matrix is also the identity matrix. Step 2: Divide the adjugate of \(I\) by its determinant: As the determinant of \(I\) is 1, dividing the adjugate matrix by 1 does not change the matrix, so \(I^* = I\). This proves part (a) of the given exercise.
03

Part (b)

Step 1: Find the adjugate of the zero matrix, \(0\): The zero matrix consists entirely of 0s. Its determinant is 0. Each element in the zero matrix has a corresponding cofactor of 0. This implies that the adjugate of the zero matrix will also be the zero matrix. Step 2: Divide the adjugate of \(0\) by its determinant: Here, we would be dividing the adjugate matrix, which is the zero matrix, by the determinant of the zero matrix, which is also 0. This might create a confusion due to division by zero. However, since the adjugate of \(0\) is already the zero matrix, we can conclude that \(0^*=0\), thus proving part (b) of the given exercise.

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Key Concepts

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

Identity Matrix
An identity matrix is like the equivalent of the number 1 for matrices. Think of it as the superhero of matrices! It's a special square matrix where all the elements on the main diagonal are 1s, and every other element is 0. This matrix has a neat trick: when you multiply any matrix by the identity matrix, it stays unchanged, as if it didn't happen at all. In a mathematical context, this is like the property of the number 1, which leaves any number unchanged when multiplied.
Zero Matrix
The zero matrix might seem a bit less exciting at first glance, but it's crucial! It's a matrix made up entirely of zeros. Just like the number 0 in arithmetic, when you add a zero matrix to any other matrix, you get the original matrix back, unchanged. This property is quite useful in matrix math, acting like a neutral element. Unlike adding with it, multiplying any matrix by a zero matrix results in another zero matrix, mirroring how multiplying any number by 0 results in 0.
Adjugate Matrix
The adjugate of a matrix is a key player in matrix operations, especially when dealing with non-invertible matrices. To find the adjugate matrix, you need to follow a two-step process:
  • First, calculate the cofactor for each element, which involves finding the determinant of the smaller matrix obtained by removing the row and column of that element, then multiply it by \((-1)^{(i+j)}\).
  • Then, arrange these cofactors in a matrix and take its transpose (flipping it over the diagonal).
The adjugate matrix is particularly handy when computing the inverse of a matrix, especially since it's related to the inverse through the determinant.
Cofactor
Cofactors are essential tiny building blocks used to calculate determinants and adjugate matrices. When calculating a cofactor, you must:
  • Remove the row and column of the target element to form a smaller matrix, called the "minor".
  • Calculate the determinant of this minor matrix.
  • Adjust this determinant by multiplying it by \((-1)^{(i+j)}\), where \(i\) and \(j\) are the row and column numbers of the original element.
Each cofactor contributes to the larger puzzle of determining the overall matrix determinant, offering insight into matrix properties and applications.
Determinant
The determinant is a special number that provides a lot of information about a square matrix. For a 2x2 matrix, it's quick and simple to compute by multiplying the elements of the main diagonal, then subtracting the product of the other diagonal. For larger matrices, it involves more complex calculations with minors and cofactors. The determinant helps in understanding the matrix's properties, such as:
  • If it's invertible (a non-zero determinant means it is).
  • Relating the matrix to its volume in geometric contexts.
  • Solving systems of linear equations.
Understanding the determinant is like unlocking a door to many matrix-related operations and concepts.

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Most popular questions from this chapter

Suppose \(T_{1}\) and \(T_{2}\) are self-adjoint. Show that \(T_{1} T_{2}\) is self- adjoint if and only if \(T_{1}\) and \(T_{1}\) commute; that is, \(T_{1} T_{2}=T_{2} T_{1}\)

\(\operatorname{Let} A=\left[\begin{array}{ll}2 & i \\ i & 2\end{array}\right] .\) Verify that \(A\) is normal. Find a unitary matrix \(P\) such that \(P^{*} A P\) is diagonal. Find \(P^{*} A P\)

For each of the following symmetric matrices \(A,\) find an orthogonal matrix \(P\) and a diagonal matrix \(D\) such that \(P^{\prime} A P\) is diagonal: (a) \(A=\left[\begin{array}{rr}1 & 2 \\ 2 & -2\end{array}\right]\) (b) \(A=\left[\begin{array}{rr}5 & 4 \\ 4 & -1\end{array}\right]\) (c) \(A=\left[\begin{array}{rr}7 & 3 \\ 3 & -1\end{array}\right]\)

Prove Theorem 13.9: The change-of-basis matrix from an orthonormal basis \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) into another orthonormal basis is unitary (orthogonal). Conversely, if \(P=\left[a_{i j}\right]\) is a unitary (orthogonal) matrix, then the vectors \(u_{i^{\prime}}=\sum_{j} a_{j i} u_{j}\) form an orthonormal basis.

Find an orthogonal change of coordinates \(X=P X^{\prime}\) that diagonalizes each of the following quadratic forms and find the corrcsponding diagonal quadratic form \(q\left(x^{\prime}\right)\) (a) \(q(x, y)=2 x^{2}-6 x y+10 y^{2}\) (b) \(q(x, y)=x^{2}+8 x y-5 y^{2}\) (c) \(q(x, y, z)=2 x^{2}-4 x y+5 y^{2}+2 x z-4 y z+2 z^{2}\)
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