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When is this linear transformation an isomorphism?

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data
    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.
    3. 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?
     
    Last edited: Apr 10, 2016
  2. jcsd
  3. Apr 10, 2016 #2

    pasmith

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    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].
     
  4. Apr 10, 2016 #3
    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?
     
    Last edited: Apr 10, 2016
  5. Apr 12, 2016 #4

    pasmith

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    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]
     
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