Assume that \(M\) is symmetric, \(C\) is orthogonal, and \(D=C^{-1} M C\) is diagonal. Show that the sum of the squares of the elements of \(M\) cquals the sum of the squares of its eigenvalues. Hint: Consider \(\operatorname{Tr}\left(D^{2}\right)\)

Short Answer

Expert verified
\(\sum m_{ij}^2 = \sum \lambda_i^2 \)

Step by step solution

01

Understand the problem

Given a symmetric matrix \(M\) and an orthogonal matrix \(C\), we know that \(D = C^{-1} M C\) is a diagonal matrix. We need to show that the sum of the squares of the elements of \(M\) equals the sum of the squares of its eigenvalues.
02

Recall properties of matrices

Recall that for an orthogonal matrix \(C\), \(C^{-1} = C^T\) and for a symmetric matrix \(M\), all eigenvalues are real. Since \(D\) is diagonal, it consists of the eigenvalues of \(M\).
03

Express \(M\) using eigenvalues

Since \(M\) is symmetric, it can be diagonalized by an orthogonal matrix \(C\): \(M = C D C^T\). Here, \(D\) is the diagonal matrix whose diagonal elements are the eigenvalues of \(M\).
04

Trace relationship

The trace of a matrix is the sum of its diagonal elements. The trace of a matrix is invariant under similarity transformations. Therefore, \(\operatorname{Tr}(M^2) = \operatorname{Tr}((C D C^T)^2)\).
05

Simplify using orthogonality

Using the orthogonality of \(C\), \((C D C^T)^2 = C D^2 C^T\). Hence, \(\operatorname{Tr}(M^2) = \operatorname{Tr}(C D^2 C^T)\). Since trace is invariant under such transformations, \(\operatorname{Tr}(C D^2 C^T) = \operatorname{Tr}(D^2)\).
06

Sum of squares

The sum of the squares of the elements of \(M\) is \(\operatorname{Tr}(M^2)\). On the other hand, the sum of the squares of the eigenvalues of \(M\) is \(\operatorname{Tr}(D^2)\). Since \(\operatorname{Tr}(M^2) = \operatorname{Tr}(D^2)\), the sum of the squares of the elements of \(M\) equals the sum of the squares of its eigenvalues.

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Key Concepts

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

symmetric matrix
A symmetric matrix is a square matrix that is equal to its transpose. This simply means that the matrix remains unchanged when mirrored along its main diagonal.
For example, if you have a 2x2 symmetric matrix:
a =
orthogonal matrix
A matrix is orthogonal if its transpose is also its inverse. This property is very useful for computations because it preserves lengths and angles. In simpler terms, for an orthogonal matrix C: C^T = C^{-1} where C^T is the transpose of C and C^{-1} is the inverse of C.
  • This means multiplying an orthogonal matrix by its transpose yields the identity matrix, I (I is a matrix with 1s on the diagonal and 0s elsewhere).
  • Orthogonal matrices are very important in transformations and preserving the geometry of data.
eigenvalues
Eigenvalues are the special numbers associated with a square matrix that help in understanding many of its properties. If you have a matrix M, an eigenvalue \(\lambda\) and corresponding eigenvector v satisfy the equation: M*V = \lambdaV
  • This means that when we apply matrix M to vector v, the result is a scalar multiple of v (no change in direction).
  • The eigenvalues of a symmetric matrix are always real numbers.
trace of a matrix
The trace of a matrix is the sum of its diagonal elements. The trace is denoted as Tr(M) for a matrix M. An important property of trace is that it is invariant under similarity transformations. In other words, for matrices A and B, if A and B are similar (i.e., there exists a matrix C such that A = C^{-1}BC), then Tr(A) = Tr(B).
This property is used in the exercise to show that the sum of the squares of the elements of M is equal to the sum of the squares of its eigenvalues. Specifically:
  • Tr(M) = Tr(C^{-1}MC)
  • Thus, Tr(M^2) = Tr((C^{-1}MC)^2) = Tr(C^{-1}MCM'C^{-1}) = Tr(D^2).
In conclusion, the trace helps tie together the characteristics of a matrix and its eigenvalues.

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Most popular questions from this chapter

(a) If a body is rotating about a fixed axis, then its angular momentum \(L\) and its angular velocity \(\boldsymbol{o}\) are parallel vectors and \(\mathrm{L}=\mathrm{le} \mathrm{s}\), where \(I\) is the (scalar) moment of inertia of the body about the axis. However, in general, \(L\) and \(\omega\) are not parallel and \(I\) in the equation must be a second-order tensor; let us call it I. Find I in dyadic form in the following way: For simplicity, first consider a point mass \(m\) at the point \(r\). The angular momentum of \(m\) about the origin is by definition \(m r \times v\), where \(v\) is the linear velocity.From Chapter \(6, v=\omega \times r .\) Write out the triple vector product for \(L\) and from it write each component of \(\mathrm{L}\) in terms of the three components of 6 . Write your results in matrix form and in dyadic form $$ \mathbf{L}=\mathbf{I} \cdot \mathbf{\omega}=\left(\mathrm{ii} l_{x x}+\mathrm{ij} l_{x y}+\cdots\right)+\omega $$ You should have $$ I_{x z}=m\left(y^{2}+z^{2}\right), \quad I_{x y}=-m x y, \quad \text { etc. } $$ For a set of masses \(m_{i}\) or an extended body, replace the cxpressions for \(I_{z x+}, \operatorname{ctc}_{4}\), by the corresponding sums or integrals: $$ \begin{aligned} &I_{x x}=\sum m_{i}\left(y_{i}^{2}+z_{i}^{2}\right) \quad \text { or } \quad \int\left(y^{2}+z^{2}\right) d m_{4} \\ &I_{x y}=-\sum m_{j} x_{i} y_{i} \quad \text { or }-\int x y d m_{4} \quad \text { etc. } \end{aligned} $$ (b) Show that 1 is a second-order (Cartesian) tensor by expressing its components relative to a rotated system \(\left[I_{s^{\prime} x^{\prime}}=m\left(y^{2}+z^{\prime 2}\right)\right.\), etc. \(]\) in terms of \(x, y, z\) using \((11.7)\) or \((11.11)\), and hence in terms of \(I_{x \pi}\), etc., to show that I obeys the transformation equations \((11,13)\). (c) Show that if \(\mathrm{n}\) is a unit vector, the expression \(\mathbf{n}+\mathbf{1} \cdot \mathbf{n}\) gives the moment of inertia about an axis through the origin parallel to n. Hint: Consider I rotated to a system in which one of the axes is along \(n\). (d) Observe that the I matrix is symmetric and recall that a symmetric matrix may be diagonalized by a rotation of axes. The eigenvalues of the I matrix are called the principal moments of inertia. Show by part (c) that they are moments of inertia abour the new axes \(\left(x^{\prime}, y^{\prime}, z^{\prime}\right)\) relative to which \(\mathbf{I}\) is diagonal. These new axes are called the principal axes of incrtia. For the mass distribution consisting of point masses \(m\) at \((1,1,1)\) and \((1,1,-1)\), find the nine components of 1, and find the principal moments of inertia and the principal axes.

Find the inverse of the transformation $$ \begin{aligned} &x^{\prime}=2 x-3 y \\ &y^{\prime}=x+y \end{aligned} $$ that is, find \(x, y\) in terms of \(x^{\prime}, y^{\prime}\). (Hint : Use matrices.) Is the transformation orthogonal?

Let each of the following matrices \(M\) describe a deformation of the \((x, y)\) plane. For each given \(M\) find: the eigenvalues and eigenvectors of the transformation, the matrix \(C\) which diagonalizes \(M\) and specifies the rotation to new axes \(\left(x^{\prime}, y^{\prime}\right)\) along the eigenvectors, and the matrix \(D\) which gives the deformation relative to the new axes. Describe the deformation relative to the new axes.\(\left(\begin{array}{ll}3 & 1 \\ 1 & 3\end{array}\right)\)

The trace of a matrix is the sum of the elements on the main diagonal. Show that the trace is not changed by cyclic permutation of the matrices, that is, \(\operatorname{Tr}(A B C)=\operatorname{Tr}(C A B)=\) Tr \((B C A)\). [Caution: \(\operatorname{Tr}(A B C) \neq \operatorname{Tr}(A C B)\) in general. ]

Let each of the following matrices \(M\) describe a deformation of the \((x, y)\) plane. For each given \(M\) find: the eigenvalues and eigenvectors of the transformation, the matrix \(C\) which diagonalizes \(M\) and specifies the rotation to new axes \(\left(x^{\prime}, y^{\prime}\right)\) along the eigenvectors, and the matrix \(D\) which gives the deformation relative to the new axes. Describe the deformation relative to the new axes.\(\left(\begin{array}{ll}3 & 2 \\ 2 & 3\end{array}\right)\)

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