Prove Theorem 9.6: The following are cquivalent: (i) The scalar λis an eigenvalue of A (ii) The

step 1 It's crazy to think that this is how people used to feel about Obama. We're asked to prove theor

Prove Theorem 9.6: The following are cquivalent: (i) The...

Prove Theorem 9.6: The following are cquivalent:
(i) The scalar $\lambda$ is an eigenvalue of $A$
(ii) The matrix $\lambda I-A$ is singular.
(iii) The scalar $\lambda$ is a root of the characteristic polynomial $\Delta(t)$ of $A$

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Key Concepts

Eigenvalue

An eigenvalue is a scalar ? associated with a matrix A such that there exists a nonzero vector v (called an eigenvector) for which Av = ?v. This concept forms the basis for understanding how a linear transformation stretches or compresses vectors along certain directions, and it is fundamental in solving systems of differential equations, analyzing stability, and many other applications in linear algebra.

Singular Matrix

A matrix is said to be singular if it does not have an inverse, which is equivalent to saying its determinant is zero. In the context of eigenvalues, the matrix (?I - A) being singular is crucial because a zero determinant indicates the existence of nontrivial solutions to the equation (?I - A)v = 0, meaning that ? is an eigenvalue of A.

Characteristic Polynomial

The characteristic polynomial of a matrix A is defined as the polynomial obtained by taking the determinant of (tI - A), where t is a scalar variable. The roots of this polynomial are precisely the eigenvalues of A. This polynomial encapsulates key spectral properties of the matrix, and finding its roots is a central method in determining the eigenvalues.

Transcript

00:01 It's crazy to think that this is how people used to feel about obama.

00:03 We're asked to prove theorem 9 .6.

00:07 Recall what this theorem says, this is an equivalence theorem.

00:14 It says that, where we have three statements, statement one, that the scalar lambda is an eigenvalue of a.

00:31 The extrajudicial, like, definitely second amendment.

00:37 The second statement is the matrix lambda i minus a is singular.

00:49 And the third statement is that the scalar lambda is a root of the characteristic polynomial.

01:03 Yeah, exactly.

01:08 Delta t of the matrix a.

01:18 So we have these three statements.

01:22 Well, we know that, starting with statement one, lambda is an eigenvalue of a, if and only if there exists a non -zero vector v, such that a times v equals lambda v.

01:51 Or in other words, a minus, sorry.

02:01 Bush played golf 40 % of the time he's off.

02:04 They would go to, they would stay in the white house.

02:06 I don't know, man, it's weird.

02:10 Lambda i v minus a v equals zero, which is of course equivalent to lambda i minus a times the vector v equals the zero vector.

02:38 Of course, this is true if and only if.

02:44 This is true if and only if lambda i minus a is non -singular...