Suppose V is finite-dimensional and S, T ∈ℒ(V) . Prove that S T is invertible if and only if bot

Suppose V is finite-dimensional and S, T ∈ℒ(V) . Prove that...

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Key Concepts

Composition of Operators

The composition of two linear transformations has a direct relation with the invertibility of the individual maps. Specifically, if both operators are invertible, then their composition is invertible with the inverse given by the reverse composition of the individual inverses. Conversely, in a finite-dimensional setting, if the composition is invertible, then each operator must be invertible.

Finite-Dimensional Vector Spaces

These spaces possess the important property that any linear transformation is invertible if and only if it is both one-to-one and onto. This equivalence, which holds in finite-dimensional contexts, is key to linking the invertibility of individual operators and their product.

Invertible Linear Operators

A linear operator is considered invertible if there exists another operator that reverses its action, i.e., the composition in both orders yields the identity mapping. This bijective property, characterized by a trivial kernel and a full image, is central to understanding how transformations can be undone.

Transcript

00:01 All right, for this problem, we're told that t and s, they're linear transformations from the vector space v to v.

00:06 And we're also told that s is invertible.

00:09 And then we're asked to prove that t and s inverse times t times s, that those two matrices have the same eigenvalues with the same multiplicities.

00:19 So for starters, i just want to recall that eigenvalues of this matrix, or of any matrix, are the roots, multiplicities included, of the polynomial determinant of lambda i and then minus that matrix.

00:33 So somehow, if we could show that this determinant is equal to the determinant with just a t, then we'll have proved this statement right here.

00:44 Ok, and that's exactly what we'll do.

00:46 First, i just need to algebraically manipulate this and then use rules of the determinant.

00:52 So this is equal to the determinant of.

00:58 So the identity matrix, i can write this as s inverse times s and then minus s inverse times t times s.

01:08 And then i can factor out an s inverse.

01:16 I have lambda times s minus t times s.

01:20 And then i can factor out an s on the right.

01:26 Because remember that matrix multiplication, it is not commutative.

01:31 So i have a t and then times an s.

01:34 Ok, and then i can split this apart because the determinant is multiplicative...

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