Difference between revisions of "Section 1.8 homework"

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(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 />
 
'''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'' 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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Solution:
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|-
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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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|}
 
<br />
 
<br />
 
'''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 />
 
'''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 />
 
<br />
 
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''Solution''
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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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:
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|-
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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 />
 
<br />
 
<br />
 
''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 />
 
''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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|}
 
<br />
 
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'''6.''' Let <math>V \neq \{0\}</math> be finite-dimensional and assume that  
 
'''6.''' Let <math>V \neq \{0\}</math> be finite-dimensional and assume that  
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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 />
 
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 />
 
<br />
 
<br />
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''Proof:'' I will use proof by contrapositive. The equivalent statement would then be &quot;`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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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Proof:
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|-
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|I will use proof by contrapositive. The equivalent statement would then be &quot;`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 />
 
<br />
 
<br />
 
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 />
 
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 />
 
<br />
 
<br />
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|}
  
 
'''13.''' Consider the map  
 
'''13.''' Consider the map  
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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 />
 
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 />
 
<br />
 
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''Proof:'' 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 />
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Proof:
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|-
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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 />
 
<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 />
 
Similarly if <math>z = \alpha + i \beta \in \mathbb{C}</math> and <math>a \in \mathbb{R}</math> then:<br />
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Therefore <math>\Psi</math> is <math>\mathbb{R}</math>-linear.<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.
 
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.
 +
|}

Latest revision as of 00:02, 16 November 2015

1. Let be linear maps between finite-dimensional vector spaces that satisfy . Is it true that ?

Solution:
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 such that but . Here is one example of functions: . . Then we have but so but .


4. Show that a linear map is one-to-one if and only if implies .

Proof:
First note that for any linear map because .


Proof: Suppose that is one-to-one. Then if we have by the note above so that we must have . Therefore implies . Now suppose that implies . If then by linearity of we have . But then by hypothesis that means which implies . Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} is one-to-one.


6. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V \neq \{0\}} be finite-dimensional and assume that

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_1,L_2,...,L_n: V \to V}

are linear operators. Show that if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_1 \circ L_2 \circ \cdots \circ L_n = 0} then at least one of the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_i} are not one-to-one.

Proof:
I will use proof by contrapositive. The equivalent statement would then be "`If all of the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_i} are one-to-one, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_1 \circ \cdots \circ L_n \ne 0} . 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_1,...,L_n} are all one-to-one. Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_1 \circ \cdots \circ L_n} is also a one-to-one function and so the only input that will give an output of 0 is the input Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} from problem 4. Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L_1 \circ \cdots \circ L_n \ne 0} and we are done.


If you don’t know the fact from set theory you can prove it as follows. Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f,g} are one-to-one functions. Consider the function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f \circ g} . Then to show this new function is one-to-one assume that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f \circ g(x) = f \circ g (y)} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(g(x)) = f(g(y))} . But since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is one-to-one that means the inputs to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} must be the same or in other words Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) = g(y)} . But then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g} is one-to-one so that means Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = y} and therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f \circ g} is one-to-one.

13. Consider the map Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi: \mathbb{C} \to \text{Mat}_{2 \times 2} (\mathbb{R})} defined by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi(\alpha + i \beta) = \begin{bmatrix} \alpha & -\beta \\ \beta & \alpha \end{bmatrix}} ( a) Show that this is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} -linear and one-to-one, but not onto. Find an example of a matrix in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{Mat}_{2 \times 2}(\mathbb{R})} that does not come from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{C}} .

Proof:
To show this is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}-} linear let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z_1 = \alpha_1+i \beta_1, z_2 = \alpha_2 + i \beta_2 \in \mathbb{C}} . Then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}
Similarly if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z = \alpha + i \beta \in \mathbb{C}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \in \mathbb{R}} then:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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)}
Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbb{R}} -linear.
Now to show Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} is not onto we notice that any matrix in the image of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} has top left and bottom right coordinate the same. So the simple matrix Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}} cannot possibly be in the image of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} . Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Psi} is not onto.

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