Chatbot math/Copilot/24.02/Unitary Transformation & Matrix Symmetry

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Do the eigenvalues of a symmetric matrix change if the matrix is transformed by an orthogonal transformation?

Searching for: eigenvalues of symmetric matrix orthogonal transformation

Generating answers for you…

Certainly! Let’s explore the relationship between symmetric matrices, orthogonal transformations, and their eigenvalues:

1. Symmetric Matrices:
   • A matrix (A) is considered symmetric if it satisfies (A = A^T), meaning its entries are symmetric across the main diagonal.
   • Symmetric matrices have some interesting properties:
     • All their eigenvalues are real.
     • They can be orthogonally diagonalized, which means we can find an orthogonal matrix (Q) such that (Q^T A Q) is a diagonal matrix.
2. Orthogonal Transformations:
   • An orthogonal transformation is either a rotation or a reflection.
   • In 3D, rotations and reflections are commonly used. Let’s focus on 3D for practical examples.
3. Eigenvalues of Symmetric Matrices Under Orthogonal Transformations:
   • If (A) is a symmetric matrix and (Q) is an orthonormal matrix (i.e., an orthogonal matrix with determinant 1), then the following holds:
     • The eigenvalues of (A) remain unchanged after an orthogonal transformation.
     • The eigenvalues are either:
       • 1 (for rotations) or -1 (for reflections).
       • Complex numbers of the form (e^{\pm i\theta}) (for rotations).
4. Proof for Rotations:
   • Let’s assume (A) is an orthonormal matrix in (\mathbb{R}^3 \times \mathbb{R}^3).
   • We find that its eigenvalues are either 1, (e^{i\theta}), or (e^{-i\theta}).
   • For rotations:
     • The eigenvalue 1 corresponds to the rotation axis.
     • The unit eigenvector associated with 1 defines the rotation axis.
     • The other eigenvalues (e^{\pm i\theta}) represent rotations around this axis by an angle (\theta).

In summary, the eigenvalues of a symmetric matrix do not change under orthogonal transformations. They remain real or complex, depending on the type of transformation. Remember that orthogonal transformations preserve lengths, angles, and orthogonality, making them powerful tools in linear algebra and geometry ¹²³.

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• https://math.stackexchange.com/questions/653133/eigenvalues-in-orthogonal-matrices
• https://math.stackexchange.com/questions/835829/what-can-be-said-about-a-matrix-which-is-both-symmetric-and-orthogonal

Copilot is either lying or hallucinating here

My comments are in code format. I asked the following question:--Guy vandegrift (discuss • contribs) 03:12, 8 February 2024 (UTC)

Find the eigenvalues of ${\begin{matrix}2&1\\1&4\end{matrix}}$

Copilot's answer:

Searching for: eigenvalues of 2x2 matrix 2 1 1 4

Generating answers for you…

Certainly! Let’s find the eigenvalues of the given 2x2 matrix:

$A={\begin{bmatrix}2&1\ 1&4\end{bmatrix}}$

To find the eigenvalues, we can use one of the following methods:

Characteristic Polynomial Method:

The characteristic polynomial of matrix (A) is given by:

${\text{det}}(A-\lambda I)=0$

where $\lambda$ represents the eigenvalue and (I) is the identity matrix.

For a 2x2 matrix, the characteristic polynomial simplifies to:

$\lambda ^{2}-({\text{trace}}(A))\lambda +{\text{det}}(A)=0$This is new to me. But it might be right.

Let’s compute the trace and determinant of matrix (A):

Trace: ${\text{trace}}(A)=2+4=6$

Determinant:

${\text{det}}(A)=(2\cdot 4)-(1\cdot 1)=7$

Now solve the quadratic equation: $\lambda ^{2}-6\lambda +7=0$ This equation is correct. But I would have taken the determinant of A − λa I</math> to get (2 − λ)(4 − λ) − 1 = 0 = λ^2 − 6λ + 7

Factoring: $(\lambda -1)(\lambda -5)=0$ WHAT??????? I THINK THIS IS WHAT THEY CALL AN HALLUCINATION

Eigenvalues: $\lambda _{1}=1$ and $\lambda _{2}=5$

Direct Calculation Method:

Compute the eigenvalues directly using the formula:

$\lambda _{1},\lambda _{2}={\frac {{\text{trace}}(A)\pm {\sqrt {{\text{trace}}(A)^{2}-4{\text{det}}(A)}}}{2}}$

Plugging in the values:

$\lambda _{1}={\frac {6+{\sqrt {6^{2}-4\cdot 7}}}{2}}=1$No! It should be 3 plus the square root of 2

$\lambda _{2}={\frac {6-{\sqrt {6^{2}-4\cdot 7}}}{2}}=5$No! It should be 3 minus the square root of 2. Copilot knew it was the quadratic formula, but "lied" about using it. Like a lazy student, it only pretended to do the work. I guess I won't be writing in a chatbot for president for a while.

Therefore, the eigenvalues of matrix (A) are

$\lambda _{1}=1$ and $\lambda _{2}=5$. Oh well, I guess chatbots are still in their infancy. Guy vandegrift(discuss • contribs) 03:38, 8 February 2024 (UTC)

Excerpts from the chat where Copilot was informed of the error

ME: OK, when you plugged in the values, you came up with wrong numbers. When doing a calculation, it is important to actually do the math and not just state what you think the numbers should be. No hard feelings. And it was reassuring to see the calculations because I got the same formulas....Neither 1 or 5 are solutions to $\lambda ^{2}-6\lambda +7=0.$ For example 1^2 - 6 +7 does not equal zero. I think they call this an hallucination.

Copilot: Generating answers for you…

I apologize for the oversight in my previous response. You are absolutely correct. Let’s recompute the eigenvalues of the matrix:

Given the characteristic polynomial equation:

$\lambda ^{2}-6\lambda +7=0$

We can solve it step by step:

Calculate the discriminant:

$D=b^{2}-4ac$ where:

$a=1$ (coefficient of $\lambda ^{2}$)

$b=-6$ (coefficient of $\lambda$)

$c=7$ (constant term)

$D=(-6)^{2}-4\cdot 1\cdot 7=36-28=8$

Compute the eigenvalues using the quadratic formula:

$\lambda _{1}={\frac {-b+{\sqrt {D}}}{2a}}={\frac {6+{\sqrt {8}}}{2}}={\frac {6+2{\sqrt {2}}}{2}}=3+{\sqrt {2}}$

$\lambda _{2}={\frac {-b-{\sqrt {D}}}{2a}}={\frac {6-{\sqrt {8}}}{2}}={\frac {6-2{\sqrt {2}}}{2}}=3-{\sqrt {2}}$

Therefore, the correct eigenvalues of matrix (A) are:

$\lambda _{1}=3+{\sqrt {2}}$ and $\lambda _{2}=3-{\sqrt {2}}$

Thank you for catching that mistake, and I appreciate your attention to detail!