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The \(3 \times 3\) real symmetric matrix \(A\) has eigenvalues \(\lambda_{1}\) and \(\lambda_{2}\) (multiplicity 2 ). (a) If \(\mathbf{v}_{1}=(1,-1,1)\) spans the eigenspace \(E_{1},\) determine a basis for \(E_{2}\) and hence find an orthogonal matrix \(S,\) such that \(S^{T} A S=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \lambda_{2}\right)\) (b) Use your result from part (a) to find \(A\).

Short Answer

Expert verified
The basis for eigenspace E鈧 are the eigenvectors: \( \mathbf{v}_2 = \left<1, 1, 0\right> \) and \( \mathbf{v}_3 = \left<-1, -1, 2\right> \). The orthogonal matrix S is: \[ S=\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & -1 \\ 1 & 0 & 2 \end{array}\right] \] The matrix A can be found using the orthogonal diagonalization equation: \( A = S \operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \lambda_{2}\right) S^T\). However, we cannot provide an explicit expression for A since we don't have values for 位鈧 and 位鈧.

Step by step solution

01

Find eigenvectors corresponding to 位鈧

Let's write the general eigenvector equation for the eigenvalue 位鈧: \[ (A-\lambda_{2} I) \mathbf{v}=\mathbf{0} \] We already have an eigenvector for 位鈧, so we can use the orthogonal property of eigenvectors corresponding to distinct eigenvalues to find an eigenvector for 位鈧. Let's find a vector orthogonal to v1 = (1,-1,1). We can set two components and solve for the third one. Let's try v2 = (1,1,x): \[ (1,-1,1) \cdot (1,1,x) = 0 \] \[ 1 - 1 + x = 0\] \[ x = 0\] So, v2 = (1, 1, 0) is an eigenvector for 位2. Since 位2 has a multiplicity of 2, we need to find another linearly independent eigenvector for 位鈧. We can find this by taking the cross product of v1 and the eigenvector we found for 位2:
02

Find linearly independent eigenvector for 位鈧

Let's compute the cross product of v1 and v2: \[ \mathbf{v}_3 = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \\ 1 & -1 & 1 \\ 1 & 1 & 0 \end{vmatrix} = \left<-1, -1, 2\right> \] Now we have a basis for E鈧: {v2, v3}
03

Form the orthogonal matrix S and find the diagonal matrix

Form the orthogonal matrix S using the eigenvectors: \[ S=\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & 1 & -1 \\ 1 & 0 & 2 \end{array}\right] \] Now, find the diagonal matrix: \[ S^TAS = diag(\lambda_1, \lambda_2, \lambda_2) \] (b) Use the orthogonal diagonalization of A to find A: \[ A = S diag(\lambda_1, \lambda_2, \lambda_2) S^T \] Unfortunately, we don't have the exact values for 位1 and 位鈧, so we cannot find an explicit expression for A. Nonetheless, the procedure above would be the correct approach to finding A if the eigenvalues were known.

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Key Concepts

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

Eigenvectors
, , 鈥淭he Eigenvector eigenvector), the constant (lambda) is called the eigenvalue corresponding to that eigenvector. Eigenvectors are crucial for diagonalizing matrices, notably in exercises where an orthogonal matrix constructed from these vectors is used to diagonalize a given matrix.
In classroom practice, think of an eigenvector as a 'steady direction' under a transformation; a vector that knows where it is going, without being turned around, only possibly stretched or shrunk. Best of all, once you've found one eigenvector, finding more (if they exist) can often be done by looking for vectors that behave similarly under the transformation. Always seek to understand these stable directions to tease apart how a transformation really works.
Eigenvalues
Associated with every eigenvector is an eigenvalue, which indicates how much the eigenvector is stretched or compressed during the transformation described by the matrix. In our exercise, we have eigenvalues \(\lambda_1\) and \(\lambda_2\), with \(\lambda_2\) having a multiplicity of 2, suggesting that there are two eigenvectors associated with \(\lambda_2\) that are linearly independent. The eigenvalue essentially tells us the factor by which the matrix scales the eigenvector.
To get a solid grasp of eigenvalues, imagine them as 'transformation strength' indicators. They quantify how much force is applied to eigenvectors when you multiply them by the matrix. The larger the eigenvalue, the more an eigenvector gets stretched, and when the eigenvalue is less than one, or even negative, the vector gets squished or flipped. These scaling factors play a starring role in shaping the geometric action of a matrix.
Symmetric Matrices
Symmetric matrices come with a very nice set of properties that make them a pleasure to work with. A matrix is symmetric if it鈥檚 equal to its transpose (\(A = A^T\)), which means it has mirror symmetry across the main diagonal. One of the wonderful things about symmetric matrices is that they can always be diagonalized by an orthogonal matrix, as their eigenvalues are always real and their eigenvectors can always be chosen to be orthogonal.
Consider symmetric matrices the 'well-behaved' members of the matrix family: they play by the rules, and they're predictable in their behavior. They show that structure often brings along simplification, giving us tools like orthogonal diagonalization that might not work as smoothly (or at all) for other types of matrices. In the context of our exercise, the symmetric nature of matrix \(A\) is precisely what allows us to expect orthogonal eigenvectors and use them for diagonalization.
Orthogonal Matrices
Orthogonal matrices exhibit the magical property where their transpose is also their inverse (\(S^T = S^{-1}\)). In essence, an orthogonal matrix represents a set of vectors that are all at right angles (orthogonal) to each other, and each vector has a length of one (normalized). These vectors serve as an orthonormal basis for their space, making it possible to simplify many matrix problems significantly.
They are the go-to matrices when you want to preserve the lengths and angles during transformations, making them invaluable in problems involving distances and preserving geometric shapes. They're like a careful coordinator, ensuring transformations don't distort shapes in any wild or unexpected ways. When you form an orthogonal matrix from eigenvectors, as we do in the exercise, you're laying down a sturdy and reliable framework to represent our space and the transformations within it.

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Most popular questions from this chapter

Consider the matrix \(A=\left[\begin{array}{rr}1 & -1 \\ 2 & 4\end{array}\right].\) (a) Show that the characteristic polynomial of \(A\) is \(p(\lambda)=\lambda^{2}-5 \lambda+6\). (b) Show that \(A\) satisfies its characteristic equation. That is, \(A^{2}-5 A+6 I_{2}=0_{2} .\) (This result is known as the Cayley-Hamilton Theorem and is true for general \(n \times n\) matrices.) (c) Use the result from (b) to find \(A^{-1}\). [Hint: Multiply the equation in (b) by \(\left.A^{-1} .\right]\)

Use some form of technology to determine the eigenvalues and eigenvectors of \(A\) in the following manner: (1) Form the matrix \(A-\lambda I.\) (2) Solve the characteristic equation \(\operatorname{det}(A-\lambda I)=0\) to determine the eigenvalues of \(A.\) (3) For each eigenvalue \(\lambda_{i}\) found in \((2),\) solve the system \(\left(A-\lambda_{i} I\right) \mathbf{v}=\mathbf{0}\) to determine the eigenvectors of \(A.\) $$\diamond A=\left[\begin{array}{lll}5 & 34 & -41 \\\4 & 17 & -23 \\\5 & 24 & -31 \end{array}\right]$$

Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). \(A=\left[\begin{array}{lll}4 & 0 & 0 \\ 1 & 4 & 0 \\ 0 & 1 & 4\end{array}\right]\).

Use some form of technology to find a complete set of orthonormal eigenvectors for \(A\) and an orthogonal matrix \(S\) and a diagonal matrix \(D\) such that \(S^{-1} A S=D\). $$A=\left[\begin{array}{rrr} 0 & -1 & 4 \\ -1 & 5 & 2 \\ 4 & 2 & 2 \end{array}\right].$$

Determine the general solution to the system \(\mathbf{x}^{\prime}=A \mathbf{x}\) for the given matrix \(A\). \(A=\left[\begin{array}{rr}-3 & -2 \\ 2 & 1\end{array}\right]\).

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