Question

Given a $3 \times 3$ symmetric matrix, formulate two distinct ways of characterizing its middle eigenvalue $\lambda_2$.

    Given a $3 \times 3$ symmetric matrix, formulate two distinct ways of characterizing its middle eigenvalue $\lambda_2$.

Applied Linear Algebra (Undergraduate Texts in Mathematics)

Applied Linear Algebra (Undergraduate Texts in Mathematics)

Peter J. Olver,… 2nd Edition

Chapter 8, Problem 39

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Solved by Expert Nikhil Kumar Rajpurohit

03/17/2022

Step 1

A $3 \times 3$ symmetric matrix $A$ has real eigenvalues, and the eigenvectors corresponding to distinct eigenvalues are orthogonal. The eigenvalues can be ordered as $\lambda_1 \leq \lambda_2 \leq \lambda_3$. Show more…

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Given a $3 \times 3$ symmetric matrix, formulate two distinct ways of characterizing its middle eigenvalue $\lambda_2$.

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Key Concepts

Spectral Theorem

The spectral theorem asserts that every real symmetric matrix can be diagonalized by an orthogonal transformation. This guarantees that all eigenvalues are real and that the matrix has an orthonormal basis of eigenvectors. This property is foundational when discussing eigenvalue characterizations for symmetric matrices.

Courant-Fischer Theorem

The Courant-Fischer theorem provides a variational characterization of the eigenvalues of a symmetric matrix. It states that the k-th eigenvalue can be described both as a min-max (minimum over k-dimensional subspaces of the maximum Rayleigh quotient on that subspace) and as a max-min (maximum over (n-k+1)-dimensional subspaces of the minimum Rayleigh quotient on that subspace). For a 3×3 matrix, these formulations yield two distinct ways to characterize the middle eigenvalue.

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Transcript

00:01 In this question we have to given a matrix a.

00:08 We have to prove it.

00:09 Eigen vector corresponding to its eigen values are auto vulnerable to each other.

00:16 Are you willing to each other? so it's for its iron values.

00:20 We will go with characteristic equation a minus lambda is equal to zero.

00:25 A.

00:25 Is 3333 minus lambda i.

00:29 So lambda will be subtracted from its dying in the element and it's the women.

00:36 It should be zero.

00:37 So i can say three minus lambda scare minus 19 0.

00:41 If you call this i will get zero comma six, forgetting eigen vector corresponding to zero comma six.

00:48 Let lambda is equal to zero and its eigen vector is equal to x.

00:53 X comma y.

00:55 For its iron vector we will move with a minus lambda down into i into matrix x.

01:01 It should be geometrics, geo volumetrics.

01:07 So the value of a is 3333 -10.

01:12 So nothing will be subjected and it will be x into y.

01:17 It should be zero zero, x will be minus y.

01:22 So it's x will be x out.

01:25 Then it will be one and minus one question number one.

01:31 Okay and if i go for λ two is equal to six a minus lambda...

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