Suppose v is an eigenvector of a linear operator T belonging...

Key Concepts

Eigenvalue and Eigenvector

An eigenvector of a linear operator T is a nonzero vector v that satisfies the equation T(v) = ?v, where ? is a scalar called an eigenvalue. This condition is fundamental in understanding how linear transformations act on special vectors in a vector space, reducing complicated transformations to simple scaling operations.

Powers of Linear Operators

Taking powers of a linear operator involves applying the operator repeatedly. For example, if T is an operator and v is its eigenvector with eigenvalue ?, then T applied n times (denoted T^n) yields T^n(v) = ?^n v. This illustrates that the scalar multiple property of eigenvectors extends naturally to iterated applications of the operator.

Polynomial Functional Calculus

Polynomial functional calculus is the concept of applying a polynomial f(t) to a linear operator T, forming f(T) by substituting T for t in the polynomial. When applied to an eigenvector v of T with eigenvalue ?, the result is f(T)(v) = f(?)v, which shows that f(?) becomes an eigenvalue of f(T) corresponding to the eigenvector v. This principle is useful in simplifying and generalizing operator actions.

Transcript

00:03 We're told that v is an eigenvector of a linear operator t belonging to the eigenvalue lambda and we're asked to prove statements about v and lambda in part a we're asked to prove that for positive values of n v is an eigenvector of t to the n belonging to the eigenvalue lambda to the n.

00:47 Well, because v is an eigenvector of t belonging to eigenvalue lambda, we know that t times v is equal to lambda times v.

01:02 Therefore, t to the n applied to v, by definition, this is t to the n minus 1 of t of v, which is t to the n minus 1 applied to lambda v.

01:22 And we see that we can continue this process until we get all the way down to lambda to the n.

01:32 Well, not quite.

01:34 I'll add another step first.

01:37 So this is the same as t to the n minus 2 times t of lambda v.

01:43 And we can use, well, not even doing that.

01:50 I can use homogeneity to write this as lambda times t to the n minus 1 of v.

01:55 And this is lambda squared times t to the n minus 2 of v.

02:04 After homogeneity again, eventually we get lambda to the n times v.

02:10 Therefore, it follows that v is an eigenvector of t to the n belonging to eigenvalue lambda to the end.

02:32 And so we've proven part a and part b we have to show that f of landa is an item value of f of t or any polynomial f of little t.

02:46 So let f of little t be a polynomial.

02:54 Well, we have that you know that since the lambda is an eigenvalue of t, it follows that the characteristic polynomial of t as lambda has a root.

03:32 I'll take a different approach.

03:40 Because f of t is a polynomial, it can be written in the form f of t equals t to the n, well a sub n, t to the n plus a7 minus 1, t to the n minus 1, and so on, down to a1t plus a0.

04:03 And we have that f of t, by definition, this is a7t to the n plus a7 minus 1, t to the n minus 1, t to the n minus 1, and so on, all the way down to a1.

04:20 1 times t plus a 0 and then i'll multiply by i guess 1 this is the identity transformation i could also write id here now let's see if v is also an eigenvector of f of t well we have f of t applied to v well this is a a...

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