For each of the following matrices, find all cigcnvalues and corresponding linearly independent

For each of the following matrices, find all cigcnvalues and...

Key Concepts

Linear Independence of Eigenvectors

The set of eigenvectors corresponding to the eigenvalues of a matrix must be linearly independent to form a basis for the vector space. This concept is particularly important in the context of matrix diagonalization, where a matrix possessing a full set of linearly independent eigenvectors can be represented in a diagonal form via a similarity transformation.

Algebraic and Geometric Multiplicity

Algebraic multiplicity refers to the number of times an eigenvalue appears as a root of the characteristic polynomial, while geometric multiplicity is the number of linearly independent eigenvectors associated with an eigenvalue. The comparison of these multiplicities is essential for determining whether a matrix is diagonalizable and understanding the complete structure of its eigenspaces.

Eigenvalues

Eigenvalues are scalars associated with a square matrix that provide insight into the matrix’s behavior, particularly in linear transformations. They are defined as the values ? for which there exists a nonzero vector v such that A*v = ?*v. Finding eigenvalues typically involves solving the equation det(A - ?I) = 0, which is known as the characteristic equation.

Eigenvectors

Eigenvectors are nonzero vectors that, when multiplied by the matrix, yield a scaled version of themselves corresponding to the eigenvalue. In other words, for a given eigenvalue ?, any vector v satisfying A*v = ?*v is called an eigenvector of A. These vectors are crucial in understanding the invariant directions of the transformation represented by the matrix.

Characteristic Polynomial

The characteristic polynomial of a matrix is obtained from the determinant of (A - ?I) and is a key tool in the process of finding eigenvalues. The roots of this polynomial equation provide the eigenvalues. The degree of the polynomial is equal to the size of the matrix and summarizes the algebraic structure of its eigenvalues.

Transcript

00:01 We're given matrices and we're asked to find all the eigenvalues in corresponding linearly independent eigenvectors for these matrices.

00:12 We're also asked when possible to find a non -singular matrix p that diagonalizes the matrix.

00:25 In part a, we're given the matrix a.

00:31 This is the 2x2 matrix 2 negative 3, 2 negative 5.

00:35 Now this is a 2x2 matrix so we can find a characteristic polynomial delta of t.

00:47 This is t squared minus the trace of t which is negative 3, so t squared plus 3 t plus the determinant of a which is negative 4, which we can factor as t minus, sorry t plus 4 times t minus 1.

01:07 And so the matrix a has eigenvalues lambda 1 equals negative 4 and lambda 2 equals 1.

01:18 I think nobody listening to this show is going to use that.

01:30 If there's a way to harass quickbooks payments.

01:39 Now to find the eigenvectors corresponding to these eigenvalues.

01:53 So they can continue robbing.

01:56 Which is in tax complicated is because they pay lobbyists so that they're soft.

02:00 So in the case of lambda equals negative 4, we're going to subtract negative 4 down the diagonal of a.

02:10 And so we get matrix m, which is a plus 4i, which gives us the 2x2 matrix 6 negative 3.

02:28 2 -1, and this corresponds to the homogenous system 6x minus 3 y equals 0, and 2x minus y equals 0, which just corresponds to the single equation 2x minus y equals 0.

02:52 So let's take, for example, x equals 1, then it follows at y equals 2, so such an eigenvalue could be v equals 1, 2.

03:09 And this is an eigenvector belonging to eigenvalue negative 4.

03:35 Now consider the other icon value, lambda 2 equals 1, we're going to subtract 1 down the diagonal of a to get the matrix m, so n is a minus 1i, or a minus i.

03:50 This is the 2x2 matrix 1, negative 3, 2, negative 6, which corresponds to the homogenous system, x minus 3y equals 0, and 2x minus 6 y equals 0, which corresponds to the single equation, x minus 3y equals 0.

04:12 To find an eigenvector, we have one degree of freedom.

04:19 Take y equal to 1, and then x is 3.

04:22 So we get the vector u, which is 3 ,1, and this is an eigenvector belonging to the eigenvalue, lambda 2, which is 1.

04:45 Okay, now in part b, that was unintentional, but in part b, we're given the matrix b, which is the 2x2 matrix 2 for negative 1, 6.

05:14 I should do that more.

05:16 This is also 2x2 matrix, so to find the characteristic polynomial delta of t, well, it's t squared minus the trace of b, which is 8 times t, plus the determinant of b, which is 16, and this is equal to t minus 4 squared.

05:39 Therefore, the zeros of our characteristic polynomial, which is just lambda equals 4, are our eigenvalues.

06:00 Now, for this eigenvalue, to find corresponding eigenvector, we'll subtract 4 down the diagonal of a, b.

06:09 So we get m, which is a minus 4i, which is the matrix negative 2, 4, negative 1, 2...

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