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Chapter 5: Problem 6

Show that the only matrix similar to a scalar matrix \(A=r I, r\) in \(\mathbb{R},\) is \(A\) itself.

Short Answer

Expert verified
The only matrix similar to a scalar matrix \( A = rI \) is \( A \) itself.

Step by step solution

01

Define Scalar Matrix and Similar Matrices

A scalar matrix is a special type of diagonal matrix where all the diagonal elements are equal, denoted as \( A = rI \), where \( r \) is a scalar and \( I \) is the identity matrix. Matrix \( B \) is similar to \( A \) if there exists an invertible matrix \( P \) such that \( B = P^{-1}AP \). Our aim is to show that \( B = A = rI \).
02

Simplify P^{-1}AP for Scalar Matrix

Since \( A = rI \), we have \( P^{-1}AP = P^{-1}(rI)P = rP^{-1}IP = rP^{-1}P \). The identity matrix \( I \) does not change the matrix \( P \), so \( r P^{-1}IP = rI \). Thus, \( P^{-1}AP = rI \).
03

Establish that Similar Matrices Have the Same Eigenvalues

Eigenvalues are invariant under similarity transformations. The eigenvalues of \( A \) (which is \( rI \)) are \( r \) repeated for the dimension of \( I \). Thus, any matrix similar to \( A \) must also have only \( r \) as its eigenvalue, with the same multiplicity.
04

Prove Matrix Equality

Since \( P^{-1}AP = rI = A \) from calculation and shares identical eigenvalues, \( B = rI \). Therefore, the only matrix similar to \( A = rI \) is \( A \) itself, i.e., \( B \) cannot differ from \( A \).
05

Conclude with Matrix Equality Proof

For any invertible matrix \( P \), \( P^{-1}AP = A \) proves that \( A \) is similar only to itself because no other matrices satisfy both the algebraic and eigenvalue conditions of \( A \). Therefore, the proof shows that the only matrix similar to a scalar matrix is the matrix itself.

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Key Concepts

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

Scalar Matrix
A scalar matrix is a particular type of diagonal matrix where all the elements on the main diagonal are the same constant value, denoted by a scalar.
It is written in the form of \( A = rI \), where \( r \) is a real number termed as the scalar, and \( I \) is the identity matrix.
  • Each diagonal element is equal to \( r \).
  • All off-diagonal elements are zero.
  • An example of a scalar matrix for a 3x3 matrix would be \[ r \cdot\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} r & 0 & 0 \ 0 & r & 0 \ 0 & 0 & r \end{pmatrix} \]
In essence, a scalar matrix is just a multiple of the identity matrix, scaling it by multiplying each diagonal by \( r \). This uniformity simplifies many matrix operations involving scalar matrices.
Eigenvalues
Eigenvalues are specific values which show how a linear transformation described by a matrix can scale vectors without changing their direction.
For scalar matrices, determining eigenvalues becomes straightforward. If you have a scalar matrix \( A = rI \), its eigenvalues are simply \( r \) repeated \( n \) times, where \( n \) is the size/dimension of the matrix.
  • These eigenvalues are computed by solving the characteristic equation \( \det(A - \lambda I) = 0 \).
  • For a scalar matrix \( A \), it transforms into \( \det(rI - \lambda I) = (r - \lambda)^n = 0 \).
  • So, \( \lambda = r \) is the eigenvalue, repeated \( n \) times.
Eigenvalues being invariant under similarity transformations ensures that any matrix similar to our scalar matrix \( A = rI \) must have these same eigenvalues. This property plays a crucial role in proving that scalar matrices are only similar to themselves.
Identity Matrix
An identity matrix, typically denoted as \( I \) or \( I_n \) for an \( n imes n \) matrix, is a square matrix with ones on the main diagonal and zeros elsewhere.
This matrix functions as the equivalent of "1" in matrix algebra.
  • Any matrix \( A \) multiplied by the identity matrix results in \( A \) itself. ( \( AI = IA = A \))
  • For any real number \( r \) and an identity matrix \( I \), \( rI \) turns the identity matrix into a scalar matrix by multiplying each diagonal element by \( r \).
The role of the identity matrix in our problem is pivotal because it preserves the matrix form when matrices multiply, specifically helping to simplify expressions like \( P^{-1}AP \) to \( rI \). This preservation demonstrates that the transformation does not alter the scalar matrix \( A \), affirming its uniqueness in similarity.

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Most popular questions from this chapter

If \(A\) is \(m \times n\) show that $$ \operatorname{col}(A)=\left\\{A \mathbf{x} \mid \mathbf{x} \text { in } \mathbb{R}^{n}\right\\} $$

}\( satisfying the following four conditions: \)A A^{\\#} A=A ; A^{\鈥� # If \(A\) is an \(m \times n\) matrix, it can be proved that there exists a unique \(n \times m\) matrix \(A^{\\#}\) satisfying the following four conditions: \(A A^{\\#} A=A ; A^{\\#} A A^{\\#}=A^{\\#} ; A A^{\\#}\) and \(A^{\\#} A\) are symmetric. The matrix \(A^{\\#}\) is called the generalized inverse of \(A\), or the Moore-Penrose inverse. a. If \(A\) is square and invertible, show that \(A^{\\#}=A^{-1}\). b. If \(\operatorname{rank} A=m,\) show that \(A^{\\#}=A^{T}\left(A A^{T}\right)^{-1}\). c. If \(\operatorname{rank} A=n,\) show that \(A^{\\#}=\left(A^{T} A\right)^{-1} A^{T}\).

Let \(\\{\mathbf{x}, \mathbf{y}, \mathbf{z}\\}\) be a linearly independent set in \(\mathbb{R}^{4}\). Show that \(\left\\{\mathbf{x}, \mathbf{y}, \mathbf{z}, \mathbf{e}_{k}\right\\}\) is a basis of \(\mathbb{R}^{4}\) for some \(\mathbf{e}_{k}\) in the standard basis \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}, \mathbf{e}_{4}\right\\}\).

a. Show that \(\mathbf{x}\) and \(\mathbf{y}\) are orthogonal in \(\mathbb{R}^{n}\) if and only if \(\|\mathbf{x}+\mathbf{y}\|=\|\mathbf{x}-\mathbf{y}\|\) b. Show that \(\mathbf{x}+\mathbf{y}\) and \(\mathbf{x}-\mathbf{y}\) are orthogonal in \(\mathbb{R}^{n}\) if and only if \(\|\mathbf{x}\|=\|\mathbf{y}\|\).

We often write vectors in \(\mathbb{R}^{n}\) as rows. If \(a \neq 0\) is a scalar, show that \(\operatorname{span}\\{a \mathbf{x}\\}=\operatorname{span}\\{\mathbf{x}\\}\) for every vector \(\mathbf{x}\) in \(\mathbb{R}^{n}\)
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