Show that if p(λ)=a0+a1 λ+a2 λ^2+λ^3=(λ-μ)^3, then [ 0 1 0 0 0 1

Show that if p(λ)=a0+a1 λ+a2 λ^2+λ^3=(λ-μ)^3, then [ 0 ...

Key Concepts

Companion Matrix

A companion matrix is associated with a polynomial and encapsulates its coefficients so that the characteristic polynomial of the matrix equals the given polynomial. It is commonly used to study the properties of the polynomial through matrix theory and is particularly valued for its role in control theory and numerical analysis.

Similarity Transformation

A similarity transformation involves expressing a matrix as V?¹AV, effectively changing the basis in which the linear transformation is represented. This technique is essential because many matrix properties, such as eigenvalues and the structure of Jordan blocks, are invariant under similarity, and it often simplifies the analysis of the matrix.

Jordan Block

A Jordan block is a square matrix corresponding to a single eigenvalue that may include nontrivial entries outside the main diagonal (typically ones immediately above the diagonal), indicating the presence of generalized eigenvectors. The arrangement of Jordan blocks in a matrix's Jordan normal form reveals the geometric multiplicity and algebraic multiplicity of its eigenvalues.

Generalized Eigenvectors

Generalized eigenvectors extend the concept of eigenvectors for cases where the matrix does not have a complete basis of eigenvectors, as in the presence of Jordan blocks. They form chains that are used to construct the Jordan canonical form and provide insight into the structure of the transformation beyond what ordinary eigenvectors offer.

Transcript

0:00 Hello there.

00:01 Okay, so for this exercise we need to consider the following to beta matrix, a defined as a, b, c, and d.

00:08 Okay, and we're going to assume that we have the this matrix have the following characteristic polynomial.

00:20 We need to show that the characteristic polynomial, if we evaluate in the characteristic polynomial, the matrix itself, this satisfy this is going to be equals to 0 okay and in order to prove this we need to will actually consider the the equation for the the actual expression for the for the characteristic polynomial of these matrix and we know that the characteristic polynomial for 2 by 2 matrices is defined as lambda square minus the trace of the matrix times lambda and plus the determinant of the matrix.

01:11 Now if we translate this into what we want to show, what we need to do here is just replace the lambda's by a's.

01:20 So we are going to have the following expression.

01:24 The characteristic polynomial evaluated in the matrix, it's equals to a square minus the trace of a times the matrix a and here plus the determinant of the matrix okay so these we need to show that is equal to zero okay so that's the question is this is equal to zero we need to show this so let's start to obtain what we needed for this exercise i'm going to write here the matrix a very quickly a b c and d so the first thing what's going to be a square so a square is defined as a time a so we have we have here the matrix multiplication ab c d times a matrix a b cd and this matrix multiplication result into a square plus b times c plus b that multiplies a plus a that multiplies a plus d here and this will be c a plus d and the last one is bc plus the square so this is going to be a square now what is the trace of this matrix that's very simple the trace of a is just the sum of the elements in the diagonal a plus d finally the determinant the determinant of this matrix is equal to ad minus b, c.

03:34 And we have all the components required to replace into the characteristic polynomial.

03:39 So if we consider these expressions, we have that the characteristic polynomial evaluated at a is equal to the following sum of matrices.

03:51 So here we have a square plus b c b a plus d c a plus d c a plus d here b times c and plus v square minus the trace the trace is given by a plus d a plus d a plus d the matrix, so we put just a, b, c and d, and plus the determinant, that is equal to ad minus b, c, and here we need to multiply by the identity matrix.

04:53 1 .10.

04:55 Okay, so as you can observe, we need to sum just component by component.

04:59 And what we obtain from this is that the characteristic polynomial evaluated at the matrix a is equals to the following matrix...

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