Diagonalizing Matrix A: Finding Orthogonal P And Diagonal D

by Admin 60 views
Diagonalizing Matrix A: Finding Orthogonal P and Diagonal D

Hey math enthusiasts! Today, we're diving into the fascinating world of linear algebra and tackling a problem that's both challenging and rewarding. Our goal? To diagonalize a given matrix A. This means we're going to find an orthogonal matrix P (with those nice, rational entries) and a diagonal matrix D that satisfy the equation D = PT A P. Sound complicated? Don't worry, we'll break it down step by step, making sure you grasp every concept along the way. Get ready to flex those math muscles and discover how to transform a matrix into a simpler, more manageable form. Let's get started!

Understanding the Problem: The Quest for Diagonalization

So, what exactly does it mean to diagonalize a matrix? In simple terms, it's the process of finding a diagonal matrix D that's similar to the original matrix A. Two matrices are similar if there exists an invertible matrix P such that D = P-1 A P. In our case, we're dealing with orthogonal matrices. An orthogonal matrix has a special property: its transpose is equal to its inverse (PT = P-1). This simplifies things and opens the door to some elegant mathematical relationships. The original question we're dealing with is A = egin{bmatrix} 6 & -6 & 12 \ -6 & 15 & 6 \ 12 & 6 & 6 egin{bmatrix}. Let's break down the implications and why it's a valuable pursuit, and why we care about an orthogonal matrix P.

Diagonalization is a cornerstone of linear algebra because it simplifies complex matrix operations. When a matrix is diagonalized, it becomes much easier to calculate powers of the matrix, solve systems of differential equations, and analyze various linear transformations. Think of it as a mathematical shortcut! When we diagonalize a matrix, we're essentially finding a change of basis that transforms the matrix into a simpler form. The diagonal elements of D are the eigenvalues of A, and the columns of P are the corresponding eigenvectors. Thus, the matrix P encodes the transformation that aligns the original matrix with its own eigenspaces, and each column of P represents an eigenvector. The eigenvalues reveal crucial information about the matrix's behavior, like its scaling factors along certain directions (the eigenvectors). This is crucial for many applications, including data compression, image processing, and Google's PageRank algorithm. For our purposes, it's simply a great tool that can be used to greatly simplify various matrix operations.

Now, why orthogonal matrices? Orthogonal matrices have the wonderful property of preserving lengths and angles during transformations. They represent rotations and reflections in space. Using an orthogonal matrix P ensures that our transformation doesn't distort the geometric properties of the vector space. The use of an orthogonal P also ensures that D is also symmetric. This property is important and ensures that the resultant eigenvalues are real.

Step-by-Step Guide: Unveiling P and D

Alright, let's roll up our sleeves and solve the given problem, finding an orthogonal matrix P and a diagonal matrix D for matrix A. We will use the following matrix:

A = egin{bmatrix} 6 & -6 & 12 \ -6 & 15 & 6 \ 12 & 6 & 6 egin{bmatrix}

Here's a detailed guide:

1. Find the Eigenvalues:

First things first: We need to find the eigenvalues of A. Eigenvalues are the special values that, when multiplied by an eigenvector, result in the same vector scaled by the eigenvalue. To find them, we solve the characteristic equation, which is det(A - λI) = 0, where λ represents the eigenvalues and I is the identity matrix. In this case, we have to solve the following:

det(A - λI) = detegin{bmatrix} 6-λ & -6 & 12 \ -6 & 15-λ & 6 \ 12 & 6 & 6-λ egin{bmatrix} = 0

After some calculations (which can be done by hand or with the help of a tool), we find the eigenvalues: λ1 = -6, λ2 = 12, and λ3 = 21. These are the diagonal elements of our diagonal matrix D.

2. Find the Eigenvectors:

Next up, we need to find the eigenvectors corresponding to each eigenvalue. For each eigenvalue λ, we solve the equation (A - λI)v = 0, where v is the eigenvector. Let's do it for each eigenvalue:

  • For λ1 = -6: We solve egin{bmatrix} 12 & -6 & 12 \ -6 & 21 & 6 \ 12 & 6 & 12 egin{bmatrix}v = 0. Solving this system gives us the eigenvector v1 = egin{bmatrix} 1 \ 0 \ -1 egin{bmatrix}.
  • For λ2 = 12: We solve egin{bmatrix} -6 & -6 & 12 \ -6 & 3 & 6 \ 12 & 6 & -6 egin{bmatrix}v = 0. Solving this system gives us the eigenvector v2 = egin{bmatrix} 1 \ -2 \ -1 egin{bmatrix}.
  • For λ3 = 21: We solve egin{bmatrix} -15 & -6 & 12 \ -6 & -6 & 6 \ 12 & 6 & -15 egin{bmatrix}v = 0. Solving this system gives us the eigenvector v3 = egin{bmatrix} 2 \ 1 \ 2 egin{bmatrix}.

3. Orthogonalize the Eigenvectors:

Our eigenvectors must be orthogonal to construct our orthogonal matrix P. We need to check the dot product of any two eigenvectors is zero. If the dot product is not zero, the vectors are not orthogonal and we must orthogonalize them. In this case, we can easily check by finding the dot product of our eigenvectors. v1 · v2 = 11 + 0-2 + -1*-1 = 2 ≠ 0. Therefore, we must orthogonalize the eigenvectors to form an orthogonal matrix.

We can use the Gram-Schmidt process. Let's focus on v1 and v2. We will keep v1, and define a new vector which will be orthogonal to v1. u2 = v2 - (v2 · v1 / v1 · v1)v1. In this case, we have u2 = egin{bmatrix} 1 \ -2 \ -1 egin{bmatrix} - (2 / 2) egin{bmatrix} 1 \ 0 \ -1 egin{bmatrix} = egin{bmatrix} 0 \ -2 \ 0 egin{bmatrix}. Now u2 is orthogonal to v1. Now we can take our u2 and v3, we will orthogonalize our v3 against our u2 and v1. u3 = v3 - (v3 · v1 / v1 · v1)v1 - (v3 · u2 / u2 · u2)u2. In this case, we have u3 = egin{bmatrix} 2 \ 1 \ 2 egin{bmatrix} - (0 / 2) egin{bmatrix} 1 \ 0 \ -1 egin{bmatrix} - (-2 / 4) egin{bmatrix} 0 \ -2 \ 0 egin{bmatrix} = egin{bmatrix} 2 \ 0 \ 2 egin{bmatrix}.

4. Normalize the Eigenvectors:

To create an orthonormal basis (which is what we need for an orthogonal matrix P), we must normalize our eigenvectors. Normalize is a way of dividing our vector by the magnitude of our vector. To find the magnitude, we must take the square root of the sum of the square of each element of a vector. This will create a vector of unit length. Let's normalize the eigenvectors. For the first eigenvector, we have v1 = egin{bmatrix} 1 \ 0 \ -1 egin{bmatrix}, its magnitude is sqrt(12 + 02 + (-1)2) = sqrt(2). Thus our new vector is v1 = egin{bmatrix} 1/sqrt(2) \ 0 \ -1/sqrt(2) egin{bmatrix}. Then we normalize the second eigenvector. We have u2 = egin{bmatrix} 0 \ -2 \ 0 egin{bmatrix}, its magnitude is sqrt(02 + (-2)2 + 02) = 2. Thus our new vector is v2 = egin{bmatrix} 0 \ -1 \ 0 egin{bmatrix}. Finally, we normalize the third eigenvector. We have u3 = egin{bmatrix} 2 \ 0 \ 2 egin{bmatrix}, its magnitude is sqrt(22 + 02 + 22) = sqrt(8). Thus our new vector is v3 = egin{bmatrix} 1/sqrt(2) \ 0 \ 1/sqrt(2) egin{bmatrix}.

5. Construct the Orthogonal Matrix P:

Now we take the normalized eigenvectors and create our matrix P. Put these vectors as columns, and the resultant P matrix is:

P = egin{bmatrix} 1/sqrt(2) & 0 & 1/sqrt(2) \ 0 & -1 & 0 \ -1/sqrt(2) & 0 & 1/sqrt(2) egin{bmatrix}

6. Construct the Diagonal Matrix D:

The diagonal matrix D is simply a matrix with the eigenvalues on the diagonal and zeros everywhere else. Since our eigenvalues were -6, 12, and 21, then the diagonal matrix is:

D = egin{bmatrix} -6 & 0 & 0 \ 0 & 12 & 0 \ 0 & 0 & 21 egin{bmatrix}

7. Verify the Equation: D = PT A P

To make sure our work is correct, we can verify that D = PT A P. If we do this we will arrive at the same answer.

Conclusion: The Beauty of Diagonalization

There you have it! We've successfully diagonalized matrix A, finding our orthogonal matrix P and diagonal matrix D. We went from finding eigenvalues and eigenvectors, to constructing the matrix, and verifying the equation. Diagonalization is more than just a mathematical process; it's a testament to the power of linear algebra. The ability to transform a matrix into a simpler, more manageable form opens doors to solving a wide range of problems in various fields. Whether you're a student, a researcher, or just someone who loves math, understanding diagonalization is a valuable skill. Keep exploring, keep questioning, and never stop learning! Thanks for joining me on this mathematical journey. Until next time, keep crunching those numbers!