Difference between revisions of "Section 1.11 Homework"

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(Created page with "'''7.''' Show that two 2-dimensional subspaces of a 3-dimensional subspace must have nontrivial intersection.<br /> <br /> ''Proof:'' (by contradiction) Suppose <math>M,N</mat...")
 
 
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'''7.''' Show that two 2-dimensional subspaces of a 3-dimensional subspace must have nontrivial intersection.<br />
 
'''7.''' Show that two 2-dimensional subspaces of a 3-dimensional subspace must have nontrivial intersection.<br />
 
<br />
 
<br />
−
''Proof:'' (by contradiction) Suppose <math>M,N</math> are both 2-dimensional subspaces of a 3-dimension vector space <math>V</math> and assume that <math>M,N</math> have trivial intersection. Then <math>M+N</math> is also a subspace of <math>V</math>, and since <math>M,N</math> have a trivial intersection <math>M+N = M \oplus N</math>. But then:<br />
+
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Proof:
 +
|-
 +
|(by contradiction) Suppose <math>M,N</math> are both 2-dimensional subspaces of a 3-dimension vector space <math>V</math> and assume that <math>M,N</math> have trivial intersection. Then <math>M+N</math> is also a subspace of <math>V</math>, and since <math>M,N</math> have a trivial intersection <math>M+N = M \oplus N</math>. But then:<br />
 
<math>\dim (M+ N) = \dim M + \dim N = 2 + 2</math>. However subspaces must have a smaller dimension than the whole vector space and <math>4 > 3</math>. This is a contradiction and so <math>M,N</math> must have trivial intersection.<br />
 
<math>\dim (M+ N) = \dim M + \dim N = 2 + 2</math>. However subspaces must have a smaller dimension than the whole vector space and <math>4 > 3</math>. This is a contradiction and so <math>M,N</math> must have trivial intersection.<br />
 +
|}
 
<br />
 
<br />
−
8. Let <math>M_1,M_2 \subset V</math> be subspaces of a finite dimensional vector space <math>V</math>. Show that <math>\dim (M_1 \cap M_2) + \dim (M_1 \cup M_2) = \dim M_1 + \dim M_2</math>.<br />
+
 
−
<br />
 
−
Proof: Define the linear map <math>L: M_1 \times M_2 \to V</math> by <math>L(x_1,x_2) = x_1 - x_2</math>. Then by dimension formula <math>\dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(K)</math> First note that in general <math>\dim (V \times W) = \dim V + \dim W</math>. This fact I won’t prove here but is why <math>\dim \mathbb{R}^2 = 1+1 = 2</math>. Now <math>\ker(L) = \{(x_1,x_2): L(x_1,x_2) = 0\}</math>. That is, <math>(x_1,x_2) \in \ker(L)</math> iff <math>x_1 - x_2 = 0 \Rightarrow x_1 = x_2</math>. But since <math>x_1 \in M_1</math> and <math>x_2 \in M_2</math> and they are actually the same vector, <math>x_1 = x_2</math>, then we must have <math>x_1 = x_2 \in M_1 \cap M_2</math>. That says that the elements of the kernel are ordered pairs where the first and second component are equal and must be in <math>M_1 \cap M_2</math>. Then we can write <math>\ker(L) = \{ (x,x) : x \in M_1 \cap M_2\}</math>. I claim that this is isomorphic to <math>M_1 \cap M_2</math>. To prove this consider the function <math>\phi: M_1 \cap M_2 \to \ker(L)</math> as <math>\phi(x) = (x,x)</math>. This map <math>\phi</math> is an isomorphism which you can check. Since we have an isomorphism, the dimensions must equal and so <math>\dim(M_1 \cap M_2) = \dim(\ker(L))</math>. Finally let us examine <math>\text{im}(L) = \{x_1 - x_2: x_1 \in M_1, x_2 \in M_2\}</math>. I claim that <math>\text{im}(L) = M_1 + M_2</math>. Note, this is equal and not just isomorphic. To see this, we note that if <math>x_2 \in M_2</math> then <math>-x_2 \in M_2</math> by subspace property. So then any <math>x_1 + x_2 \in M_1 + M_2</math> is also equal to <math>x_1 - (-x_2) \in \text{im}(L)</math>. So these sets do indeed contain the exact same elements. That means <math>\dim (M_1 + M_2) = \dim \text{im}(L)</math>. Putting this all together gives:<br />
 
−
<math>\dim M_1 + \dim M_2 = \dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(L) = \dim (M_1 \cap M_2) + \dim(M_1 + M_2)</math>.<br />
 
−
<br />
 
 
'''8.''' Let <math>M_1,M_2 \subset V</math> be subspaces of a finite dimensional vector space <math>V</math>. Show that <math>\dim (M_1 \cap M_2) + \dim (M_1 \cup M_2) = \dim M_1 + \dim M_2</math>.<br />
 
'''8.''' Let <math>M_1,M_2 \subset V</math> be subspaces of a finite dimensional vector space <math>V</math>. Show that <math>\dim (M_1 \cap M_2) + \dim (M_1 \cup M_2) = \dim M_1 + \dim M_2</math>.<br />
 
<br />
 
<br />
−
''Proof:'' Define the linear map <math>L: M_1 \times M_2 \to V</math> by <math>L(x_1,x_2) = x_1 - x_2</math>. Then by dimension formula <math>\dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(K)</math> First note that in general <math>\dim (V \times W) = \dim V + \dim W</math>. This fact I won’t prove here but is why <math>\dim \mathbb{R}^2 = 1+1 = 2</math>. Now <math>\ker(L) = \{(x_1,x_2): L(x_1,x_2) = 0\}</math>. That is, <math>(x_1,x_2) \in \ker(L)</math> iff <math>x_1 - x_2 = 0 \Rightarrow x_1 = x_2</math>. But since <math>x_1 \in M_1</math> and <math>x_2 \in M_2</math> and they are actually the same vector, <math>x_1 = x_2</math>, then we must have <math>x_1 = x_2 \in M_1 \cap M_2</math>. That says that the elements of the kernel are ordered pairs where the first and second component are equal and must be in <math>M_1 \cap M_2</math>. Then we can write <math>\ker(L) = \{ (x,x) : x \in M_1 \cap M_2\}</math>. I claim that this is isomorphic to <math>M_1 \cap M_2</math>. To prove this consider the function <math>\phi: M_1 \cap M_2 \to \ker(L)</math> as <math>\phi(x) = (x,x)</math>. This map <math>\phi</math> is an isomorphism which you can check. Since we have an isomorphism, the dimensions must equal and so <math>\dim(M_1 \cap M_2) = \dim(\ker(L))</math>. Finally let us examine <math>\text{im}(L) = \{x_1 - x_2: x_1 \in M_1, x_2 \in M_2\}</math>. I claim that <math>\text{im}(L) = M_1 + M_2</math>. Note, this is equal and not just isomorphic. To see this, we note that if <math>x_2 \in M_2</math> then <math>-x_2 \in M_2</math> by subspace property. So then any <math>x_1 + x_2 \in M_1 + M_2</math> is also equal to <math>x_1 - (-x_2) \in \text{im}(L)</math>. So these sets do indeed contain the exact same elements. That means <math>\dim (M_1 + M_2) = \dim \text{im}(L)</math>. Putting this all together gives:<br />
+
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Proof:
 +
|-
 +
|Define the linear map <math>L: M_1 \times M_2 \to V</math> by <math>L(x_1,x_2) = x_1 - x_2</math>. Then by dimension formula <math>\dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(K)</math> First note that in general <math>\dim (V \times W) = \dim V + \dim W</math>. This fact I won’t prove here but is why <math>\dim \mathbb{R}^2 = 1+1 = 2</math>. Now <math>\ker(L) = \{(x_1,x_2): L(x_1,x_2) = 0\}</math>. That is, <math>(x_1,x_2) \in \ker(L)</math> iff <math>x_1 - x_2 = 0 \Rightarrow x_1 = x_2</math>. But since <math>x_1 \in M_1</math> and <math>x_2 \in M_2</math> and they are actually the same vector, <math>x_1 = x_2</math>, then we must have <math>x_1 = x_2 \in M_1 \cap M_2</math>. That says that the elements of the kernel are ordered pairs where the first and second component are equal and must be in <math>M_1 \cap M_2</math>. Then we can write <math>\ker(L) = \{ (x,x) : x \in M_1 \cap M_2\}</math>. I claim that this is isomorphic to <math>M_1 \cap M_2</math>. To prove this consider the function <math>\phi: M_1 \cap M_2 \to \ker(L)</math> as <math>\phi(x) = (x,x)</math>. This map <math>\phi</math> is an isomorphism which you can check. Since we have an isomorphism, the dimensions must equal and so <math>\dim(M_1 \cap M_2) = \dim(\ker(L))</math>. Finally let us examine <math>\text{im}(L) = \{x_1 - x_2: x_1 \in M_1, x_2 \in M_2\}</math>. I claim that <math>\text{im}(L) = M_1 + M_2</math>. Note, this is equal and not just isomorphic. To see this, we note that if <math>x_2 \in M_2</math> then <math>-x_2 \in M_2</math> by subspace property. So then any <math>x_1 + x_2 \in M_1 + M_2</math> is also equal to <math>x_1 - (-x_2) \in \text{im}(L)</math>. So these sets do indeed contain the exact same elements. That means <math>\dim (M_1 + M_2) = \dim \text{im}(L)</math>. Putting this all together gives:<br />
 
<math>\dim M_1 + \dim M_2 = \dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(L) = \dim (M_1 \cap M_2) + \dim(M_1 + M_2)</math>.<br />
 
<math>\dim M_1 + \dim M_2 = \dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(L) = \dim (M_1 \cap M_2) + \dim(M_1 + M_2)</math>.<br />
 +
|}
 
<br />
 
<br />
  
 
'''16.''' Show that the matrix<br />
 
'''16.''' Show that the matrix<br />
−
<math>\begin{bmatrix} 0 & 1 \\ 0 & 1\end{bmatrix</math>
+
<math>\begin{bmatrix} 0 & 1 \\ 0 & 1\end{bmatrix}</math>
 
as a linear map satisfies <math>\ker(L) = \text{im}(L)</math>.<br />
 
as a linear map satisfies <math>\ker(L) = \text{im}(L)</math>.<br />
 
<br />
 
<br />
−
''Proof:'' The matrix is already in eschelon form and has one pivot in the second column. That means that a basis for the column space which is the same as the image would be the second column. In other words, <math>\text{im}(L) = \text{Span} \left (\begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )</math>. Now for the kernel space. Writing out the equation <math>Lx = 0</math> reads <math>0x_1 + 1x_2 = 0</math> or in other words <math>x_2 = 0</math>. Then an arbitrary element of the kernel <math>\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix}</math>. So again <math>\ker(L) = \text{Span} \left (\begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )</math>. In other words, <math>\ker(L) = \text{im}(L)</math>.<br />
+
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Proof:
 +
|-
 +
|The matrix is already in eschelon form and has one pivot in the second column. That means that a basis for the column space which is the same as the image would be the second column. In other words, <math>\text{im}(L) = \text{Span} \left (\begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )</math>. Now for the kernel space. Writing out the equation <math>Lx = 0</math> reads <math>0x_1 + 1x_2 = 0</math> or in other words <math>x_2 = 0</math>. Then an arbitrary element of the kernel <math>\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix}</math>. So again <math>\ker(L) = \text{Span} \left (\begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )</math>. In other words, <math>\ker(L) = \text{im}(L)</math>.<br />
 +
|}
 
<br />
 
<br />
  
−
17. Show that<br />
+
'''17.''' Show that<br />
 
<math>\begin{bmatrix} 0 & 0 \\ \alpha & 1\end{bmatrix}</math>
 
<math>\begin{bmatrix} 0 & 0 \\ \alpha & 1\end{bmatrix}</math>
 
defines a projection for all <math>\alpha \in \mathbb{F}</math>. Compute the kernel and image.<br />
 
defines a projection for all <math>\alpha \in \mathbb{F}</math>. Compute the kernel and image.<br />
 
<br />
 
<br />
−
First I will deal with the case <math>\alpha = 0</math>. In this case the matrix is <math>\begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}</math> and we see by the procedure in the last problem that: <math>\text{im} (L) = \text{Span} \left (\begin{bmatrix} 0 \\ 1 \end{bmatrix} \right )</math> and <math>\ker(L) = \text{Span} \left ( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )</math>.<br />
+
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Proof:
 +
|-
 +
|First I will deal with the case <math>\alpha = 0</math>. In this case the matrix is <math>\begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}</math> and we see by the procedure in the last problem that: <math>\text{im} (L) = \text{Span} \left (\begin{bmatrix} 0 \\ 1 \end{bmatrix} \right )</math> and <math>\ker(L) = \text{Span} \left ( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )</math>.<br />
 
<br />
 
<br />
 
Now for the case <math>\alpha \ne 0</math>. Then we still have only one pivot and either column can form a basis for the image. Using the second column makes it look nicer, and is the same as the previous case. <math>\text{im} (L) = \text{Span} \left (\begin{bmatrix} 0 \\ 1 \end{bmatrix} \right )</math>. The difference is when we write out the equation <math>Lx = 0</math> to find the kernel, we get <math>\alpha x_1 + x_2 = 0</math>. With <math>x_2</math> as our free variable this means <math>x_1 = -\frac{1}{\alpha} x_2 </math> so that a basis for the kernel is <math>\ker(L) = \text{Span} \left ( \begin{bmatrix} -\frac{1}{\alpha} \\ 1 \end{bmatrix} \right )</math>.
 
Now for the case <math>\alpha \ne 0</math>. Then we still have only one pivot and either column can form a basis for the image. Using the second column makes it look nicer, and is the same as the previous case. <math>\text{im} (L) = \text{Span} \left (\begin{bmatrix} 0 \\ 1 \end{bmatrix} \right )</math>. The difference is when we write out the equation <math>Lx = 0</math> to find the kernel, we get <math>\alpha x_1 + x_2 = 0</math>. With <math>x_2</math> as our free variable this means <math>x_1 = -\frac{1}{\alpha} x_2 </math> so that a basis for the kernel is <math>\ker(L) = \text{Span} \left ( \begin{bmatrix} -\frac{1}{\alpha} \\ 1 \end{bmatrix} \right )</math>.
 +
|}

Latest revision as of 23:05, 15 November 2015

7. Show that two 2-dimensional subspaces of a 3-dimensional subspace must have nontrivial intersection.

Proof:
(by contradiction) Suppose are both 2-dimensional subspaces of a 3-dimension vector space 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} 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 M,N} have trivial intersection. 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 M+N} is also a subspace 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 V} , and 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 M,N} have a trivial intersection 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 M+N = M \oplus N} . 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 \dim (M+ N) = \dim M + \dim N = 2 + 2} . However subspaces must have a smaller dimension than the whole vector space 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 4 > 3} . This is a contradiction and 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 M,N} must have trivial intersection.


8. 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 M_1,M_2 \subset V} be subspaces of a finite dimensional vector space 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} . Show 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 \dim (M_1 \cap M_2) + \dim (M_1 \cup M_2) = \dim M_1 + \dim M_2} .

Proof:
Define the 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: M_1 \times M_2 \to V} 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 L(x_1,x_2) = x_1 - x_2} . Then by dimension formula 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 \dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(K)} First note that in general 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 \dim (V \times W) = \dim V + \dim W} . This fact I won’t prove here but is why 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 \dim \mathbb{R}^2 = 1+1 = 2} . Now 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 \ker(L) = \{(x_1,x_2): L(x_1,x_2) = 0\}} . That 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 (x_1,x_2) \in \ker(L)} iff 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_1 - x_2 = 0 \Rightarrow x_1 = x_2} . 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 x_1 \in M_1} 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 x_2 \in M_2} and they are actually the same vector, 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_1 = x_2} , then 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_1 = x_2 \in M_1 \cap M_2} . That says that the elements of the kernel are ordered pairs where the first and second component are equal and must be 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 M_1 \cap M_2} . Then we can write 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 \ker(L) = \{ (x,x) : x \in M_1 \cap M_2\}} . I claim that this is isomorphic 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 M_1 \cap M_2} . To prove this 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 \phi: M_1 \cap M_2 \to \ker(L)} as 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 \phi(x) = (x,x)} . This 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 \phi} is an isomorphism which you can check. Since we have an isomorphism, the dimensions must equal and 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 \dim(M_1 \cap M_2) = \dim(\ker(L))} . Finally let us examine 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{im}(L) = \{x_1 - x_2: x_1 \in M_1, x_2 \in M_2\}} . I claim 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 \text{im}(L) = M_1 + M_2} . Note, this is equal and not just isomorphic. To see this, we note 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 x_2 \in M_2} 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 -x_2 \in M_2} by subspace property. So then any 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_1 + x_2 \in M_1 + M_2} is also equal 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 x_1 - (-x_2) \in \text{im}(L)} . So these sets do indeed contain the exact same elements. 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 \dim (M_1 + M_2) = \dim \text{im}(L)} . Putting this all together gives:

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 \dim M_1 + \dim M_2 = \dim(M_1 \times M_2) = \dim \ker(L) + \dim \text{im}(L) = \dim (M_1 \cap M_2) + \dim(M_1 + M_2)} .


16. Show that the 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} 0 & 1 \\ 0 & 1\end{bmatrix}} as a linear map satisfies 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 \ker(L) = \text{im}(L)} .

Proof:
The matrix is already in eschelon form and has one pivot in the second column. That means that a basis for the column space which is the same as the image would be the second column. 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 \text{im}(L) = \text{Span} \left (\begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )} . Now for the kernel space. Writing out the equation 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 Lx = 0} reads 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 0x_1 + 1x_2 = 0} 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 x_2 = 0} . Then an arbitrary element of the kernel 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} x_1 \\ x_2 \end{bmatrix} = x_2 \begin{bmatrix} 1 \\ 0 \end{bmatrix}} . So again 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 \ker(L) = \text{Span} \left (\begin{bmatrix} 1 \\ 0 \end{bmatrix} \right )} . 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 \ker(L) = \text{im}(L)} .


17. Show 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 \begin{bmatrix} 0 & 0 \\ \alpha & 1\end{bmatrix}} defines a projection for all . Compute the kernel and image.

Proof:
First I will deal with the case . In this case the matrix is and we see by the procedure in the last problem that: and .


Now for the case . Then we still have only one pivot and either column can form a basis for the image. Using the second column makes it look nicer, and is the same as the previous case. . The difference is when we write out the equation to find the kernel, we get . With as our free variable this means so that a basis for the kernel is .

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