Question

Let \(T: V \rightarrow V\) be a linear transformation. Show that \(T^{2}=1_{V}\) if and only if \(T\) is invertible and \(T=T^{-1}\)

Short answer

If \(T^2 = 1_V\), then \(T\) is invertible and \(T = T^{-1}\), and vice versa.

Step-by-step solution

Understanding the Problem

We need to prove a bi-conditional statement: that \(T^{2}=1_{V}\) is equivalent to \(T\) being invertible and \(T = T^{-1}\). This means we'll show both implications separately.

First Implication: Showing \(T^2 = 1_V\) implies \(T\) is invertible and \(T = T^{-1}\)

Given \(T^2 = 1_V\), it means \(T(T(v)) = v\) for any \(v \in V\). Therefore, applying \(T\) again returns the original vector. This suggests \(T\) is its own inverse, hence \(T\) is invertible and \(T = T^{-1}\).

Second Implication: Showing \(T\) is invertible and \(T = T^{-1}\) implies \(T^2 = 1_V\)

Assume \(T\) is invertible and \(T = T^{-1}\). Then we have \(T^2 = T \circ T = T \circ T^{-1} = 1_V\), because composing \(T\) with its inverse gives the identity transformation. Hence, \(T^2 = 1_V\).

Conclusion of Both Implications

We have proven both directions of the equivalence: \(T^2 = 1_V\) means that \(T\) is invertible and \(T = T^{-1}\), and vice versa. This completes the proof of the bi-conditional statement.

Key concepts

Invertible Linear Map

Linear transformations are mappings between vector spaces that preserve vector addition and scalar multiplication. An "invertible linear map" is a special type of linear transformation that has an inverse. This means if you apply the transformation and then apply its inverse, you get back to where you started. A linear map \( T: V \rightarrow V \) is invertible if there exists another map \( T^{-1}: V \rightarrow V \) such that applying \( T \) followed by \( T^{-1} \) (or vice versa) gives the identity transformation on \( V \). Essentially, this process is like taking a step forward and then stepping back to your original position.

The conditions for a linear map to be invertible include:

• It must be bijective. This means it has a one-to-one correspondence with its output and input, so each output arises from exactly one input.
• The map must be both injective (no two different inputs give the same output) and surjective (every possible output is covered).

Understanding these conditions helps in identifying invertible transformations. They are crucial in linear algebra as they allow restoration of the original vector after transformation.

Identity Transformation

The "identity transformation" is one of the simplest, yet most important concepts in linear transformations. In mathematical terms, the identity transformation \( 1_V \) on a vector space \( V \) maps every vector to itself. This transformation acts like a "do-nothing" operation and is denoted by \( 1_V \) or simply \( I \). You can think of it as a mirror that reflects a vector unchanged.

Key features of identity transformation are:

• It satisfies \( 1_V(v) = v \) for every vector \( v \in V \).
• The identity transformation serves as the multiplicative identity for composition of transformations. This means composing any transformation with the identity transformation yields the original transformation: \( T \circ 1_V = T \) and \( 1_V \circ T = T \).
• It is always invertible, with the identity transformation itself being its inverse.

In the context of our exercise, why \( T^2 = 1_V \) indicates that \( T \) is its own inverse is tied to the notion that applying \( T \) twice is like doing nothing, bringing us back to the original vector.

Inverse Transformation

An "inverse transformation" refers to the counterpart of a linear transformation that reverses its effect. If \( T: V \rightarrow V \) is a linear transformation, then its inverse \( T^{-1}: V \rightarrow V \) satisfies \( T \circ T^{-1} = 1_V \) and \( T^{-1} \circ T = 1_V \). This means applying \( T \) and then \( T^{-1} \) should result in the identity transformation.

Calculating inverses is a vital skill in linear algebra:

• Not every linear transformation has an inverse. Only those that are invertible (i.e., bijective) do.
• An inverse "undoes" the effect of the original transformation. If \( T \) stretches or rotates a vector, \( T^{-1} \) reverses that stretching or rotating.
• The inverse is unique if it exists, meaning there is only one inverse transformation for a given map that satisfies the necessary properties.

In the proof from the exercise, showing that \( T = T^{-1} \) highlights how \( T \) reverses its own effects, combining to give you the identity transformation (just like doing nothing). This property is fundamental to recognizing invertible maps in various algebraic contexts.