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 />
 
<br />
 
<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 />
 +
|}
 
<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 />
 
<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 />
+
!Proof:
 +
|-
 +
|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 />
 
<br />
 
'''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;"
 +
!Proof:
 +
|-
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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 />
 
<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.
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|}

Latest revision as of 23:02, 15 November 2015

1. 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 L, K: V \to V} be linear maps between finite-dimensional vector spaces that satisfy 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 \circ K = 0} . Is it true 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 K \circ L = 0} ?

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 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,B} such 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 AB = 0} but 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 BA \ne 0} . Here is one example of functions: 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 = \mathbb{R}^2} . 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(x,y) = (x,0), K(x,y) = (0,x+y)} . Then we have 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 \circ K (x,y) = L(0,x+y) = (0,0)} but 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 K \circ L(x,y) = K(x,0) = (0,x+0)} so 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 \circ K = 0} but 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 K \circ L \ne 0} .


4. Show that a linear 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 L: V \to W} is one-to-one if and only 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(x)= 0} implies 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 = 0} .

Proof:
First note that for any linear 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 L(0) = 0} because 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(0) = L(0 \cdot 0 ) = 0 \cdot L(0) = 0} .


Proof: 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 (\Rightarrow)} 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} is one-to-one. Then 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(x) = 0} we have 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(x) = L(0)} by the note above so that we must have 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 = 0} . 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(x) = 0} implies 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 = 0} . 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 (\Leftarrow)} Now 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(x) = 0} implies 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 = 0} . 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(x) = L(y)} then by linearity 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 L} we have 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(x-y) = L(x) - L(y) = 0} . But then by hypothesis 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 = 0} which implies 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} . 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 is one-to-one so that means and therefore is one-to-one.

13. Consider the map defined by ( a) Show that this is -linear and one-to-one, but not onto. Find an example of a matrix in that does not come from .

Proof:
To show this is linear let . Then:


Similarly if and then:

Therefore is -linear.
Now to show is not onto we notice that any matrix in the image of has top left and bottom right coordinate the same. So the simple matrix cannot possibly be in the image of . Therefore is not onto.

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