When is this linear transformation an isomorphism?

In summary, the matrix form of the linear transformation L from ℝ2 to ℝ2 is given by L(x1, x2)T=(1, 2 ; 3, α)(x1, x2)T=Ax. To determine at what values of α L is an isomorphism, we can check if the determinant of the matrix A is non-zero. When α=0, the inverse of A is \begin{pmatrix} 0 & \frac13 \\ \frac12 & - \frac16\end{pmatrix}, and the kernel and injectivity of L are both trivial. Therefore, L is an isomorphism for all values of α except for α=6 and α=0
  • #1
lep11
380
7

Homework Statement


Let L: ℝ2→ℝ2 such that L(x1, x2)T=(1, 2 ; 3, α)(x1, x2)T=Ax
Determine at what values of α is L an isomorphism. Obviously L is given in matrix form.

The Attempt at a Solution


First of all a quick check, dim (ℝ2)=dim(ℝ2)=2 Ok.

An isomorphism means linear transformation which is bijective. ##Det (A)=α-6## so for A to be invertible and therefore L to be bijective must hold ##α≠6.## On the other hand it suffices to determine when L is injective because L is bijective iff it's injective. If ##α## was 0, ker(L) would include infinitely many vectors and L would neither be injective nor bijective.
So ##α≠6.## and ##α≠0.##
What's the best approach to this particular problem?
 
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  • #2
A map from [itex]\mathbb{R}^n[/itex] to itself is invertible if and only if its determinant is non-zero. You have shown that the determinant of your map is [itex]\alpha - 6[/itex].

When [itex]\alpha = 0[/itex] the inverse of [itex]\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}[/itex] is [itex]\begin{pmatrix} \frac12 & -\frac16 \\ 0 & \frac13 \end{pmatrix}[/itex].
 
  • #3
pasmith said:
A map from [itex]\mathbb{R}^n[/itex] to itself is invertible if and only if its determinant is non-zero. You have shown that the determinant of your map is [itex]\alpha - 6[/itex].
We haven't actually been told that in our class (yet?) so we haven't proven that either. Is it that simple? So is the final solution that L is an isomorphism when α∈ℝ\{6}?

pasmith said:
When [itex]\alpha = 0[/itex] the inverse of [itex]\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix}[/itex] is [itex]\begin{pmatrix} \frac12 & -\frac16 \\ 0 & \frac13 \end{pmatrix}[/itex].
What about kernel of L and injectivity if α=0?
 
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  • #4
lep11 said:
We haven't actually been told that in our class (yet?) so we haven't proven that either. Is it that simple? So is the final solution that L is an isomorphism when α∈ℝ\{6}?What about kernel of L and injectivity if α=0?

The kernel is trivial: if [tex]\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = 0[/tex] then the bottom row gives [itex]3x = 0[/itex], so that [itex]x = 0[/itex]. The top row then gives [itex]x + 2y = 2y = 0[/itex] so [itex]y = 0[/itex] also.

Similarly it is injective: if [tex]\begin{pmatrix} 1 & 2 \\ 3 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a \\ b\end{pmatrix}[/tex] then [itex]x = \frac13 b[/itex] and [itex]y = \frac12a - \frac16b[/itex].

(The inverse given in my previous post appears to be incorrect; as appears from the above it should be [itex]\begin{pmatrix} 0 & \frac13 \\ \frac12 & - \frac16\end{pmatrix}[/itex]
 

1. What is a linear transformation?

A linear transformation is a mathematical function that maps a vector space to another vector space, while preserving the linear structure and operations of the original space.

2. What is an isomorphism?

An isomorphism is a bijective linear transformation between two vector spaces, which means it is both one-to-one and onto. This means that every element in the original vector space has a unique corresponding element in the transformed vector space, and vice versa.

3. How can you determine if a linear transformation is an isomorphism?

A linear transformation is an isomorphism if and only if it is both injective (one-to-one) and surjective (onto). This means that it must preserve both the linear structure and the dimension of the vector space.

4. What are the conditions for a linear transformation to be an isomorphism?

In addition to being both injective and surjective, a linear transformation must also be a homomorphism, meaning it preserves the operations of addition and scalar multiplication. It must also be a bijection, meaning it has a unique inverse transformation.

5. Why is it important to determine if a linear transformation is an isomorphism?

An isomorphism is a special type of linear transformation that preserves the structure and properties of the original vector space. This can be useful in various mathematical and scientific applications, such as solving systems of linear equations and understanding the relationship between different vector spaces.

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