Derive the eigenvalues and eigenvectors given in Example 9.7...

Key Concepts

Diagonalization

Diagonalization is the process of converting a matrix into a diagonal form by using its eigenvectors as a basis. If a matrix can be diagonalized, it can be expressed as PDP?¹, where D is a diagonal matrix containing the eigenvalues and P is a matrix whose columns are the corresponding eigenvectors. This process simplifies many matrix computations.

Determinant

The determinant is a scalar value computed from a square matrix that encapsulates properties such as invertibility. In the context of eigenvalues, computing the determinant of (A - ?I) is critical because setting it equal to zero defines the equation that must be solved to find the eigenvalues.

Characteristic Polynomial

The characteristic polynomial is obtained by calculating the determinant of (A - ?I), where A is the given matrix and I is the identity matrix. Setting this polynomial equal to zero yields the characteristic equation whose roots are the eigenvalues of the matrix, providing a systematic method for their derivation.

Eigenvectors

Eigenvectors are nonzero vectors that remain in the same direction after a linear transformation, except possibly for a change in magnitude as given by the corresponding eigenvalue. For each eigenvalue ?, the eigenvectors are the solutions to the equation (A - ?I)v = 0, which often involves solving a system of linear equations.

Eigenvalues

Eigenvalues are scalars associated with a linear transformation represented by a matrix. They are defined by the equation A·v = ?·v, where A is the matrix, v is a nonzero vector (the eigenvector), and ? is the eigenvalue. Determining the eigenvalues involves solving the characteristic equation det(A - ?I) = 0, where I is the identity matrix.

Transcript

00:01 So for this problem, we've been asked to derive the eigenvalues and eigenvectors of example, 9 .7 .1.

00:07 We do this by finding the determinant of this characteristic matrix, and we can do that by expanding this determinant out along the top.

00:17 So if we follow the top line, and remembering the fact that we have to use the matrix of cofactors, this determinant is equal to minus lambda times the 3 by 3 matrix.

00:30 In the bottom right.

00:36 So it's a determinant of that.

00:37 So it's minus lambda 1, 0, 0, minus lambda 1, 0, minus 2 minus lambda.

00:46 And then the one term, that goes to minus 1 because of the cofactors.

00:51 And then this is multiplied by the determinant of the 3x3 matrix given by the first column and the last two columns.

01:01 So this is minus 3, 1 ,0.

01:05 0 minus lambda 1, 2 minus 2 minus lambda.

01:12 So let's expand these 3 by 3 determinants out.

01:17 We again will expand across the top row for the first term.

01:23 So this is minus lambda multiplied by minus lambda times the determinant of the 2x2 matrix in the bottom right.

01:33 And using the standard 2x determinants, that's lambda squared plus 2.

01:42 Now, for the one term in the top row, that becomes minus 1 thanks to the cofactors, but we actually are times in this by the determinant of the 2x2 matrix given by these two columns.

01:56 Now, as the first column is just column of zeros, this determinant is actually 0.

02:02 And the remaining term is just zero by the top right.

02:06 So that's the first term expanded out.

02:10 For the second term, this becomes minus 1.

02:16 And if we expand across the top, that's minus 3, multiplied by the determinant of the bottom right matrix of minus, and that's lambda squared plus 2 again.

02:32 Now, for the second expansion term, it goes to minus 1, and the determinant is not 0 in this case, because we're using the column 0, 2, and 1 minus lambda, and the determinant of that is simply minus 2.

02:52 So we can expand this all out and collect like terms, and this gives us the characteristic polynomial is equal to lambda to the floor, plus 5 lambda squared, plus 4.

03:06 And because we're looking for the eigenvalues, we then set this characteristic polynomial equal to 0.

03:13 This is the characteristic polynomial.

03:17 We can factorize this.

03:18 So factored the left -hand side, this becomes lambda squared plus 1 times lambda squared plus 4 is equal to 0.

03:28 Now this is only equal to 0 when either the first bracket or the second bracket is equal to 0, and this gives us our 4, values.

03:37 So from the first one, we get that lambda squared plus 1 equals 0, which means that lambda squared is equal to minus 1, and that implies that lambda is equal to plus or minus i.

03:48 So that's the first two eigenvalues, and the second term, that lambda squared is equal to minus 4, and that means that lambda is equal to plus or minus 2i.

04:04 So that's how we get the eigenvalues for the example.

04:08 So we now want to find the eigenvectors.

04:11 And in theory, there should be four eigenvectors because we have four eigenvalues.

04:16 But because the eigenvalues impairs up to a sign change, the plus and minus i will actually have the same eigenvector.

04:28 By definition of the eigenvectors, we can multiply it by some constant.

04:32 And if we just multiply by the eigenvectors by plus or minus one, we get the corresponding eigenvectors for plus or minus i.

04:40 And the same goes for plus and minus two i eigenvalues...

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