Ramsey

2021-02-11

Let $T:U\to U$ be a linear transformation and let beta be a basis of U Define the determinant det(T) of T as

det$(T)=det([T{]}_{\beta}).$

Show ta det (T) is well-defined, i. e. that it does not depend on the choice of the basis beta

Prove that T is invertible if and only if det$(T)\ne 0.$ If T is invertible, show that det $({T}^{-1})=\frac{1}{det(T)}$

det

Show ta det (T) is well-defined, i. e. that it does not depend on the choice of the basis beta

Prove that T is invertible if and only if det

wornoutwomanC

Skilled2021-02-12Added 81 answers

Step 1

To establish some basic facts about the determinant of a linear transformation T from U to U (U a vector space).

Step 2

A linear transformation

Step 3

Let B be a basis of U. Then T is represented as a square matrix of size n (n= dimension of U)

Let

(The vectors in B are linearly independent and they span U).

Then anu linear map T : B

Step 4

With the above notation, we define det(T) = determinant of the matrix

In order that det (T) is well-defined , we need to show that this definition does not depend on the choice of the basis B. In other words, if C is some other basis for U, we need to prove that.

Det

Let B, C are two bases of U.

We need to show det

which in turn proves that (T) is well - defined.

(det (T) is independent of the basis)

Step 5

Here , we have used the standard properties of determinants of matrices .

Proof: Let B,C are two bases of U.

Then

where P is the change of basis matrix.

Now

det

Step 6

Proof of T is invertible iff det (T) is not zero

Step 7

Now,

det

det

as required.

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