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 \(T, U \rightarrow U\) is a map satisfying the properties shown.

\(T : U \rightarrow\) (U a vector space) is linear if

\(T (v + w) = T (v) + T (w),\) and

\(T (cv) = cT (v),\) for vectors v, w in U and scalars c

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 \(B = {v_{1}, v_{2}, ..., v_{n}}\) be a in basis for U.

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

Then anu linear map T : B \rightarrow B can be represented as a (unique) square matrix T_{b} = [b_{ij}]{1 \leq i, j \leq n),\) defined by

\(T (v_{1}) = \sum_{j = 1}^{n} b_{ji} v_{j}

Step 4

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

\(T_{B} = [b_{ij}]\)

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 \((T_{B})=det ([b_{ij}]) = det([c_{ij}]) = Det(T_{C})\)

Let B, C are two bases of U.

We need to show det \((T_{B}) = det (T_{C}),\)

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 \(T_{B} = P^{-1} T_{C} P,\)

where P is the change of basis matrix.

Now

det \((T_{B}) = det(P^{-1} T_{C} P) = det (P)^{-1} det (P) = det (T_{C})\)

Step 6

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

\(T : U \rightarrow U\) invertible

\(\Leftrightarrow EE S (inverse\ of\ T)\ with\ TS = I\) (identity)

\(\Leftrightarrow\ det\ (TS) = det (I) = 1\)

\(\Leftrightarrow det (T) det (S) = 1\)

\(\Leftrightarrow det (T) \neq 0\)

Step 7

Now, \(S = T^{-1}\) and

det \((T) det (S) = 1\) (from the above)

det \((T^{-1}) = det (S) = \frac{1}{det(T)} = det (T)^{-1},\)

as required.

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 \(T, U \rightarrow U\) is a map satisfying the properties shown.

\(T : U \rightarrow\) (U a vector space) is linear if

\(T (v + w) = T (v) + T (w),\) and

\(T (cv) = cT (v),\) for vectors v, w in U and scalars c

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 \(B = {v_{1}, v_{2}, ..., v_{n}}\) be a in basis for U.

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

Then anu linear map T : B \rightarrow B can be represented as a (unique) square matrix T_{b} = [b_{ij}]{1 \leq i, j \leq n),\) defined by

\(T (v_{1}) = \sum_{j = 1}^{n} b_{ji} v_{j}

Step 4

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

\(T_{B} = [b_{ij}]\)

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 \((T_{B})=det ([b_{ij}]) = det([c_{ij}]) = Det(T_{C})\)

Let B, C are two bases of U.

We need to show det \((T_{B}) = det (T_{C}),\)

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 \(T_{B} = P^{-1} T_{C} P,\)

where P is the change of basis matrix.

Now

det \((T_{B}) = det(P^{-1} T_{C} P) = det (P)^{-1} det (P) = det (T_{C})\)

Step 6

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

\(T : U \rightarrow U\) invertible

\(\Leftrightarrow EE S (inverse\ of\ T)\ with\ TS = I\) (identity)

\(\Leftrightarrow\ det\ (TS) = det (I) = 1\)

\(\Leftrightarrow det (T) det (S) = 1\)

\(\Leftrightarrow det (T) \neq 0\)

Step 7

Now, \(S = T^{-1}\) and

det \((T) det (S) = 1\) (from the above)

det \((T^{-1}) = det (S) = \frac{1}{det(T)} = det (T)^{-1},\)

as required.