# Linear Algebra: Characterizations of invertible matrices

## Homework Statement

Let T be a linear transformation that maps Rn onto Rn.
Show that T-1 exists and maps Rn onto Rn.
Is T-1 also one to one?

## The Attempt at a Solution

I am not quite sure where to start. To prove that T is invertible I need to show that A is an invertible matrix. To show that A is an invertible matrix I can use The Invertible Matrix Theorem part i. "The linear transformation x → Ax maps Rn onto Rn
This theorem also says that if part i. is true then A is an invertible matrix.
Then, by theorem 9, T is invertible because A is invertible.
Is that enough to prove that T-1 exists? How do I show that it maps Rn onto Rn? Is it also 1-1?

I don't know that theorem 9 and part i. are! I'm not taking your class with you.

Theorem 9
Let T: R^n → R^n be a linear transformation and let A be the standard matrix for T. Then T is invertible if and only if A is an invertible matrix. In that case, the linear transformation S given by S(x) = A^-1 x is the unique function satisfying
(1) S(T(x)) = x for all x in R^n
(2) T(S(x)) = x for all x in R^n

part i. is from the Invertible Matrix Theorem
Let A be a square nxn matrix. Then the following statements are logically equivalent:
a. A is an invertible matrix
b. A is row equivalent to the nxn identity matrix
c. A has n pivot positions
d. The equation Ax = 0 has only the trivial solution
e. The columns of A form a linearly independent set
f. The linear transformation x→ Ax is one-to-one
g. The equation Ax= b has at least one solution for each b in R^n
h. The columns of A span R^n
i. The linear transformation x → Ax maps R^n onto R^n
j. There is an nxn matrix C such that CA = I
k. There is an nxn matrix D such that DA = I
l. A^T is an invertible matrix.

Okay, from what I understand of your proof, you are using part (i) to conclude that the matrix associated with T is invertible.

You are then using "Theorem 9" to conclude that because the matrix associated with T is invertible, T itself must be. So far, we have that T is invertible so that T-1 exists.

To show that it maps Rn onto Rn, note that because T is onto, given an arbitrary v1 in Rn, one can find v2 in Rn so that T(v1) = v2. Apply the inverse to both sides of this equation and see what you get. Note that v1 was arbitrary.

To show that it's 1-1, suppose that we have v1, v2 with T-1(v1) = T-1(v2). Again, just operate on both sides of the equation with T.