Section 1.8 - Revision history

MathAdmin: Created page with "'''1.''' Let $L, K: V \to V$ be linear maps between finite-dimensional vector spaces that satisfy $L \circ K = 0$ . Is it true that $K \circ L = 0$
2015-11-16T07:16:31Z

Created page with "'''1.''' Let <math>L, K: V \to V</math> be linear maps between finite-dimensional vector spaces that satisfy <math>L \circ K = 0</math>. Is it true that <math>K \circ L = 0</m..."

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'''1.''' Let <math>L, K: V \to V</math> be linear maps between finite-dimensional vector spaces that satisfy <math>L \circ K = 0</math>. Is it true that <math>K \circ L = 0</math>?<br />
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!Solution:
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|No. in general composition of functions is not commutative. By the theorem that any linear map can be expressed as a matrix, finding a counterexample comes down to finding two matrices <math>A,B</math> such that <math>AB = 0</math> but <math>BA \ne 0</math>. Here is one example of functions: <math>V = \mathbb{R}^2</math>. <math>L(x,y) = (x,0), K(x,y) = (0,x+y)</math>. Then we have <math>L \circ K (x,y) = L(0,x+y) = (0,0)</math> but <math>K \circ L(x,y) = K(x,0) = (0,x+0)</math> so <math>L \circ K = 0</math> but <math>K \circ L \ne 0</math>.<br />
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'''4.''' Show that a linear map <math>L: V \to W</math> is one-to-one if and only if <math>L(x)= 0</math> implies <math>x = 0</math>.<br />
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!Proof:
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|First note that for any linear map <math>L(0) = 0</math> because <math>L(0) = L(0 \cdot 0 ) = 0 \cdot L(0) = 0</math>.<br />
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''Proof:'' <math>(\Rightarrow)</math>Suppose that <math>L</math> is one-to-one. Then if <math>L(x) = 0</math> we have <math>L(x) = L(0)</math> by the note above so that we must have <math>x = 0</math>. Therefore <math>L(x) = 0</math> implies <math>x = 0</math>. <math>(\Leftarrow)</math> Now suppose that <math>L(x) = 0</math> implies <math>x = 0</math>. If <math>L(x) = L(y)</math> then by linearity of <math>L</math> we have <math>L(x-y) = L(x) - L(y) = 0</math>. But then by hypothesis that means <math>x - y = 0</math> which implies <math>x = y</math>. Therefore <math>L</math> is one-to-one.<br />
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'''6.''' Let <math>V \neq \{0\}</math> be finite-dimensional and assume that

<math>L_1,L_2,...,L_n: V \to V</math>

are linear operators. Show that if <math>L_1 \circ L_2 \circ \cdots \circ L_n = 0</math> then at least one of the <math>L_i</math> are not one-to-one.<br />
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Proof:
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|I will use proof by contrapositive. The equivalent statement would then be "`If all of the <math>L_i</math> are one-to-one, then <math>L_1 \circ \cdots \circ L_n \ne 0</math>. Then this becomes very easy if you know the fact from set theory that the composition of one-to-one functions is a one-to-one function. This gives the following. Suppose that <math>L_1,...,L_n</math> are all one-to-one. Then <math>L_1 \circ \cdots \circ L_n</math> is also a one-to-one function and so the only input that will give an output of 0 is the input <math>0</math> from problem 4. Therefore <math>L_1 \circ \cdots \circ L_n \ne 0</math> and we are done.<br />
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If you don’t know the fact from set theory you can prove it as follows. Suppose <math>f,g</math> are one-to-one functions. Consider the function <math>f \circ g</math>. Then to show this new function is one-to-one assume that <math>f \circ g(x) = f \circ g (y)</math>. Then <math>f(g(x)) = f(g(y))</math>. But since <math>f</math> is one-to-one that means the inputs to <math>f</math> must be the same or in other words <math>g(x) = g(y)</math>. But then <math>g</math> is one-to-one so that means <math>x = y</math> and therefore <math>f \circ g</math> is one-to-one.<br />
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'''13.''' Consider the map
<math>\Psi: \mathbb{C} \to \text{Mat}_{2 \times 2} (\mathbb{R})</math>
defined by
<math>
\Psi(\alpha + i \beta) = \begin{bmatrix} \alpha & -\beta \\ \beta & \alpha \end{bmatrix}</math> (
a) Show that this is <math>\mathbb{R}</math>-linear and one-to-one, but not onto. Find an example of a matrix in <math>\text{Mat}_{2 \times 2}(\mathbb{R})</math> that does not come from <math>\mathbb{C}</math>.<br />
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Proof:
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|To show this is <math>\mathbb{R}-</math>linear let <math>z_1 = \alpha_1+i \beta_1, z_2 = \alpha_2 + i \beta_2 \in \mathbb{C}</math>. Then:<br />
<math>\Psi(z_1+z_2) = \Psi(\alpha_1+i\beta_1 + \alpha_2+i\beta_2) = \Psi(\alpha_1 + \alpha_2 + i(\beta_1 + \beta_2)) = \begin{bmatrix} \alpha_1+\alpha_2 & -\beta_1-\beta_2 \\ \beta_1+\beta_2 & \alpha_1+\alpha_2 \end{bmatrix} = \begin{bmatrix} \alpha_1 & -\beta_1 \\ \beta_1 & \alpha_1 \end{bmatrix} +\begin{bmatrix} \alpha_2 & -\beta_2 \\ \beta_2 & \alpha_2 \end{bmatrix} = \Psi(z_1) + \Psi(z_2)</math><br />
Similarly if <math>z = \alpha + i \beta \in \mathbb{C}</math> and <math>a \in \mathbb{R}</math> then:<br />
<math>\Psi(a z) = \Psi (a \alpha + i a\beta) = \begin{bmatrix} a\alpha & -a\beta \\ a\beta & a\alpha \end{bmatrix} = a \begin{bmatrix} \alpha & -\beta \\ \beta & \alpha \end{bmatrix} = a \Psi(z)</math><br />
Therefore <math>\Psi</math> is <math>\mathbb{R}</math>-linear.<br />
Now to show <math>\Psi</math> is not onto we notice that any matrix in the image of <math>\Psi</math> has top left and bottom right coordinate the same. So the simple matrix <math>\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}</math> cannot possibly be in the image of <math>\Psi</math>. Therefore <math>\Psi</math> is not onto.
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