# Inverse of a Linear Transformation

#### KataKoniK

Hi,

Is there a formula to do this? The textbook just says to "reverse" the action of T to get T^-1 (T inverse). Can someone explain to me in laymen terms, how to accomplish this? For example,

For T = [2x y]^T is T^-1 = [-2x y]^T?

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#### matt grime

Homework Helper
T may not be invertible (this T isn't) inverses only exiist for square matrices, adn not all of those. you may want to look up the adjugate matrix, or any basic linear algebra text book that'll talk about this when it discusses gaussian elimination.

#### Hurkyl

Staff Emeritus
Gold Member
I have a hunch that KataKoniK meant that T is the linear transformation that takes [x, y]^t to [2x, y]^t. Its inverse, if it exists, would take [2x, y]^t to [x, y]^t.

#### KataKoniK

To make my question more clear, in the following example, how come x + 5y becomes x - 5y? The question basically says to let T be the transformation induced by an invertible 2x2 matrix A. In each case, interpret T^-1 geometrically. For this question,

A = 1 5
0 1

http://img332.imageshack.us/img332/7757/ex3re.jpg [Broken]

Also, just to note that in my initial post, the matrix A for that one was

A = 2 0
0 1

Last edited by a moderator:

#### Hurkyl

Staff Emeritus
Gold Member
I suspect the answer will be clear once you interpret the transformation geometrically.

You could, of course, verify it by turning the crank: check that T T-1 is the identity.

#### KataKoniK

Hurkyl said:
I suspect the answer will be clear once you interpret the transformation geometrically.

You could, of course, verify it by turning the crank: check that T T-1 is the identity.
Dumb question, but how do we get the identity by doing T T-1? Isn't that undefined/does not exist? Unless I'm missing something here.

#### Hurkyl

Staff Emeritus
Gold Member
T and T^-1 are both functions R^2 --> R^2, right? I don't see any obstacle to composing them.

(Or equivalently, multiplying their matrix representations)

Technically, you should prove both T T^-1 and T^-1 T are both the identity.

Alright, thanks.

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