/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Given that \(A\) is a real symme... [FREE SOLUTION] | 91Ó°ÊÓ

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Given that \(A\) is a real symmetric matrix with normalised eigenvectors \(\mathrm{e}^{i}\), obtain the coefficients \(\alpha_{i}\) involved when column matrix \(x\), which is the solution of $$ \mathrm{A} \mathrm{x}-\mu \mathrm{x}=\mathrm{v} $$ is expanded as \(x=\sum_{i} \alpha_{i} \mathrm{e}^{i} .\) Here \(\mu\) is a given constant and \(\vee\) is a given column matrix. (a) Solve (*) when $$ \mathrm{A}=\left(\begin{array}{lll} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right) $$ \(\mu=2\) and \(v=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^{\mathrm{T}}\) (b) Would (*) have a solution if \(\mu=1\) and (i) \(v=\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)^{\mathrm{T}}\), (ii) \(\vee=\) \(\left(\begin{array}{lll}2 & 2 & 3\end{array}\right)^{\mathrm{T}} ?\)

Short Answer

Expert verified
Find eigenvalues and eigenvectors for \(A\), express \(x\) as a linear combination of eigenvectors, use orthogonality to solve for coefficients \(\alpha_i\).

Step by step solution

01

- Identify the Eigenvectors and Eigenvalues

For a real symmetric matrix \(A\), the eigendecomposition allows us to express \(A\) in terms of its eigenvalues and eigenvectors. The eigendecomposition of \(A\) gives matrices of eigenvalues and eigenvectors.
02

- Solve for Matrix \(A\)

The matrix \(A\) provided is: \[A=\begin{pmatrix}2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3\end{pmatrix}\]. Find the eigenvalues and corresponding eigenvectors for \(A\).
03

- Compute the Eigenvalues and Eigenvectors

Compute the characteristic polynomial of \(A\) and solve for the eigenvalues. The characteristic polynomial of \(A\) is \text{det}(A-\lambda I) = 0\. This results in: \( det \begin{pmatrix} 2 - \lambda & 1 & 0 \ 1 & 2 - \lambda & 0 \ 0 & 0 & 3 - \lambda \end{pmatrix} = 0 \), giving the eigenvalues \( \lambda_1 = 3, \lambda_2 = 3, \lambda_3 = 1 \). The eigenvectors \( e^1, e^2, e^3 \) corresponding to each eigenvalue can be computed.
04

- Expand \(x\) in Terms of \(\alpha_i e^i\)

Express \( x = \sum_{i} \alpha_{i} e^{i} \)
05

- Substitute \( x = \sum_{i} \alpha_{i} e^{i} \) into the Main Equation

Substitute the expansion of \( x \) into \( Ax - \mu x = v \)
06

- Use the Orthogonality of Eigenvectors

Since the eigenvectors are orthonormal, multiply both sides of the resulting equation by \( e^{jT} \) to get the coefficients \( \alpha_i \). This yields: \[ \alpha_{j} (\lambda_j -\mu) = e^{iT}v \]
07

- Solve for Coefficients \( \alpha_i \)

Substitute known values to solve for \( \alpha_i \)

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Key Concepts

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

Real Symmetric Matrix
A real symmetric matrix is a square matrix that is equal to its transpose. This means the element at the position (i, j) is the same as the element at position (j, i). These matrices are widely used in various fields, especially in physics and engineering.

Key properties of real symmetric matrices include:
  • All eigenvalues are real numbers.
  • The eigenvectors corresponding to distinct eigenvalues are orthogonal.
  • The matrix can be diagonalized by an orthogonal matrix.
Given the matrix:
\( A = \begin{pmatrix}2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \)
We can see that A equals its transpose, verifying that A is a real symmetric matrix.
Eigenvalues and Eigenvectors
Eigenvalues and eigenvectors are fundamental concepts in linear algebra.
They provide insights into the properties of linear transformations. For a given matrix A, if there exists a scalar \( \lambda \) and a non-zero vector \( \mathbf{e} \) such that \( A \mathbf{e} = \lambda \mathbf{e} \), then \( \lambda \) is called an eigenvalue of A and \( \mathbf{e} \) is an eigenvector corresponding to \( \lambda \).

For the given matrix:
\( A = \begin{pmatrix}2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 3 \end{pmatrix} \),
We find eigenvalues by solving the characteristic polynomial:
\[ \text{det}(A - \lambda I) = 0 \]
The characteristic polynomial is:
\[ \begin{vmatrix}2 - \lambda & 1 & 0 \ 1 & 2 - \lambda & 0 \ 0 & 0 & 3 - \lambda \end{vmatrix} = 0 \]
This results in the eigenvalues:
\[ \lambda_1 = 3, \lambda_2 = 3, \lambda_3 = 1 \]
Next, solve for the eigenvectors associated with each eigenvalue.
Orthogonality
Orthogonality is a critical concept when dealing with eigenvectors of symmetric matrices.

Orthogonality means that vectors are perpendicular to each other in an n-dimensional space. In terms of dot products, two vectors \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal if \( \mathbf{u} \cdot \mathbf{v} = 0 \).
In the context of real symmetric matrices:
  • Eigenvectors corresponding to distinct eigenvalues are orthogonal.
This property simplifies many calculations, especially when expanding a vector in terms of the eigenvectors.
Using the orthogonality of the eigenvectors, we can solve for the coefficients \( \alpha_i \) in the expansion:
\( x = \sum_{i} \alpha_{i} \mathbf{e}^{i} \)
By multiplying both sides of our main equation by the transpose of an eigenvector, we isolate the coefficients:
\[ \alpha_j (\lambda_j - \mu) = \mathbf{e}^{iT} \mathbf{v} \]
From here, we can solve for each \( \alpha_i \). This orthogonality property greatly aids in linear transformations and simplifying complex problems.

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Most popular questions from this chapter

Using the properties of determinants, solve with a minimum of calculation the following equations for \(x\) : (a) \(\left|\begin{array}{llll}x & a & a & 1 \\ a & x & b & 1 \\ a & b & x & 1 \\ a & b & c & 1\end{array}\right|=0\) (b) \(\left|\begin{array}{ccc}x+2 & x+4 & x-3 \\ x+3 & x & x+5 \\ x-2 & x-1 & x+1\end{array}\right|=0\)

Given a matrix $$ \mathrm{A}=\left(\begin{array}{lll} 1 & \alpha & 0 \\ \beta & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) $$ where \(\alpha\) and \(\beta\) are non-zero complex numbers, find its eigenvalues and eigenvectors. Find the respective conditions for (a) the eigenvalues to be real and (b) the eigenvectors to be orthogonal. Show that the conditions are jointly satisfied if and only if \(\mathrm{A}\) is Hermitian.

One method of determining the nullity (and hence the rank) of an \(M \times N\) matrix A is as follows. \- Write down an augmented transpose of \(\mathrm{A}\), by adding on the right an \(N \times N\) unit matrix and thus producing an \(N \times(M+N)\) array \(\mathrm{B}\). \- Subtract a suitable multiple of the first row of \(B\) from each of the other lower rows so as to make \(B_{i 1}=0\) for \(i>1\) \- Subtract a suitable multiple of the second row (or the uppermost row that does not start with \(M\) zero values) from each of the other lower rows so as to make \(B_{i 2}=0\) for \(i>2\) \- Continue in this way until all remaining rows have zeros in the first \(M\) places. The number of such rows is equal to the nullity of \(A\), and the \(N\) rightmost entries of these rows are the components of vectors that span the null space. They can be made orthogonal if they are not so already. Use this method to show that the nullity of $$ A=\left(\begin{array}{cccc} -1 & 3 & 2 & 7 \\ 3 & 10 & -6 & 17 \\ -1 & -2 & 2 & -3 \\ 2 & 3 & -4 & 4 \\ 4 & 0 & -8 & -4 \end{array}\right) $$ is 2 and that an orthogonal base for the null space of \(A\) is provided by any two column matrices of the form \(\left(2+\alpha_{i}-2 \alpha_{i} 1 \quad \alpha_{i}\right)^{\mathrm{T}}\), for which the \(\alpha_{i}(i=1,2)\) are real and satisfy \(6 \alpha_{1} \alpha_{2}+2\left(\alpha_{1}+\alpha_{2}\right)+5=0\).

By considering the matrices $$ \mathrm{A}=\left(\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right), \quad \mathrm{B}=\left(\begin{array}{ll} 0 & 0 \\ 3 & 4 \end{array}\right) $$ show that \(A B=0\) does not imply that either \(A\) or \(B\) is the zero matrix, but that it does imply that at least one of them is singular.

Using the Gram-Schmidt procedure: (a) construct an orthonormal set of vectors from the following: $$ \begin{array}{llllll} \mathrm{x}_{1}=\left(\begin{array}{llllll} 0 & 0 & 1 & 1 \end{array}\right)^{\mathrm{T}}, & \mathrm{x}_{2}=\left(\begin{array}{llll} 1 & 0 & -1 & 0 \end{array}\right)^{\mathrm{T}} \\ \mathrm{x}_{3}=\left(\begin{array}{lllll} 1 & 2 & 0 & 2 \end{array}\right)^{\mathrm{T}}, & \mathrm{x}_{4}=\left(\begin{array}{llll} 2 & 1 & 1 & 1 \end{array}\right)^{\mathrm{T}} \end{array} $$ (b) find an orthonormal basis, within a four-dimensional Euclidean space, for the subspace spanned by the three vectors \(\left.\begin{array}{llll}1 & 2 & 0 & 0\end{array}\right)^{\mathrm{T}},(3 \quad-1 \quad 2 \quad 0)^{\mathrm{T}}\) and \(\left(\begin{array}{llll}0 & 0 & 2 & 1\end{array}\right)^{\mathrm{T}}\).

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