Determine an orthogonal matrix S such that S^T A S=diag(λ1, λ2, …, λn), where A denotes the give

Determine an orthogonal matrix S such that S^T A S=diag(λ1,...

Key Concepts

Matrix Diagonalization

Matrix diagonalization is the process of finding a matrix S such that S?¹AS is a diagonal matrix. For symmetric matrices, diagonalization is achieved using an orthogonal matrix composed of eigenvectors, simplifying many problems in linear algebra by reducing them to operations on diagonal matrices where each diagonal entry is an eigenvalue.

Eigenvalues and Eigenvectors

Eigenvalues are scalars associated with a matrix that indicate factors by which the eigenvectors are scaled during a linear transformation. Eigenvectors are non-zero vectors that change at most by a scalar factor when a linear transformation is applied. In the context of matrix diagonalization, the eigenvectors form an orthonormal basis used to construct the orthogonal matrix that diagonalizes the symmetric matrix.

Orthogonal Matrix

An orthogonal matrix is a square matrix whose columns and rows are orthonormal vectors. This property implies that the inverse of an orthogonal matrix is its transpose, which makes computations in transformations, rotations, and reflections simpler, especially in preserving lengths and angles in Euclidean space.

Spectral Theorem

The spectral theorem states that any real symmetric matrix can be diagonalized by an orthogonal matrix. This means that for a symmetric matrix, there exists an orthogonal matrix S such that S?AS is a diagonal matrix containing the eigenvalues of A. This theorem is fundamental in linear algebra for simplifying matrix computations and understanding the structure of linear transformations.

Transcript

00:01 We were given a matrix a, and we are asked to find a matrix s, such that matrix s is orthogonal, and such that s diagonalize is a.

00:18 The matrix a is 1 -1 -1 -1 -1 -1 -1 -1.

00:28 In order to find the matrix s, we define the eigenvalues of a.

00:34 Characteristic polynomial of a is the determinant of a minus lambda i, which is equal to the determinant of the determinant of matrix 1 minus lambda 1, negative 1, 1 minus lambda 1, and negative 1, 1, 1 minus lambda.

00:59 This is equal to expanding across the first row, 1 minus lambda times 1 minus lambda squared minus 1 squared or 1.

01:23 Minus determinant of 1 -1 negative 1, 1 minus lambda, or 1 times 1 minus lambda, minus 1 times negative 1, or plus 1, and minus the determinant of 1, 1 minus lambda, negative 1, 1, 1, which is 1 minus negative 1, minus 1, or plus lambda minus 1.

01:59 So we get 1 plus lambda, or instead of 1 plus lambda, there should be 1 minus lambda.

02:21 So 1 minus lambda times the difference of squares gives us 1 minus lambda minus 1, negative lambda times 1 minus lambda plus 1 or 2 minus lambda minus minus, minus negative lambda plus 2 or plus 2 minus lambda, and plus this should actually mean minus 2 minus lambda.

03:09 And this is also minus 2 minus lambda and factoring out 2 minus lambda we get 2 minus lambda times 1 minus lambda times negative negative lambda, which is lambda squared minus lambda minus two.

03:45 It's actually factored as 2 minus lambda, lambda minus 2 times lambda plus 1, which can also be written as the opposite of lambda plus 1 times 2 minus lambda.

04:16 So character's equation is when this polynomial that equals 0, and you know the polynomial has roots which are the eigenvalues and therefore we have eigenvalues, negative 1, multiplicity 1, and lambda 2 equals positive 2 of multiplicity 2.

04:44 Suppose that v1 or x1 is an eigenvector associated with eigenvalue lambda 1, and x1 satisfies the equation a minus lambda 1 i.

04:59 X1 equals zero vector.

05:03 Matrix a minus 9 to 1 i is the same as matrix a plus i is going to be 2, 1, negative 1, 1, 2, 1, negative 1, 2.

05:23 By gaussian elimination, matrix can be reduced to 1, 2, 1, 2, 1, 2, 1, 2, 1, negative 1, 2, and then by adding combinations of rows, we get 1 -2 -1 -0, 1 -2 -2 times 2, which is 1 -9 -4, or negative 3, and negative 1 -1 minus 2, or negative 3.

06:00 0, 1 plus 2 is 3, and 2 plus 1 is 3, which is further reduced to the equation, or matrix 1 -2 -1 -0, 1 -0, which is then reduced to the equation, or matrix 1 -2 -1 -0, which is then reduced to, which, to the matrix 1 -0 -1 minus 2 times 1 or negative 1 -0 -1 -1 -0 -1 -0 -0 -1 -0.

06:34 So we obtain the equations x11 -1 -1 minus x -13 is equal to 0, and x12 plus x -13 is equal to 0.

06:53 These equations imply that x11 is equal to x13, and that x12 is equal to negative x13.

07:04 If we take x13 through the parameter r, then the idecture x1 is equal to r, negative r, which is the same as r times the vector 1, negative 1.

07:20 If we take r to be 1, then our eigenvector x1 is the vector 1, negative 1, 1.

07:36 To obtain a unit vector, that's an eigenvector.

07:41 Find a norm of x1.

07:44 This is the square root of 1 squared, which is 1 plus negative 1, which is 1, 2, 1 square, which is 1, 4 root 3.

07:55 And so our unit vector u1 is 1 over root 3, negative 1 over root 3, 3, 1 over root 3.

08:13 Now suppose that x2 is an item vector associated with the item value lambda 2, then x2 satisfies the equation a minus lambda 2 i x2 is equal to the 0 vector.

08:34 The matrix a minus 9 to 2i is the same as the matrix a minus 2i.

08:40 So that we get matrix 1 minus 2, which is negative 1, 1, negative 1, 1, negative 1, 1, negative 1, 1, negative 1, negative 1, negative 1, negative 1, 0 ,000, 0 ,000.

09:02 Gaussian elimination, this matrix reduces to 1, negative 1 ,000, 0 ,000, so we obtain the equation x211 ,000, 0 ,000, so we obtain the equation, minus x22 plus x23 equal 0 and therefore you have the x21 is equal to x22 minus x23 take x22 to be the parameter r and x23 to be the parameter s then x2 is the vector r minus s r s which is equal to r times the vector 110 plus x times the vector negative 1 0 1 if we take r to be 0 and s to be 1 then we get the x2 is the vector negative 101 because this eigen space has multiplicity or igene values multiplicity of 2 the eigen space has dimensioned 2 so we want to find another eigenve vector of the space that's the linearly independent from x2.

10:36 Suppose this argumenter is x3, and we'll have the x3 satisfies the same equation as x2, and therefore has the same form as x2 does, but it has to be linearly independent.

10:52 And to find with linearly independent, this time take r21 and sb zero, then x3 is vector one and zero, and it's clear that x2, x2, that x2, and x3 are linearly independent and they form a basis for this eigenspaces.

11:22 However, x2, interproduct of x2 and x3, which is just the dot product, is negative 1 or negative 1, plus 0, plus 0, which is negative 1...

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