Finding a linear map…
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I am given that the set $U = { (x_1, x_2, x_3, x_4, x_5 in mathbb{R}^5 : x_1 - 2x_2 = 0 : text{and} : x_4 - 3x_5 = 0 }$.
I have already shown that $U$ is a subspace of $mathbb{R}^5$. Now I want to find a linear map whose null space is equal to $U$. My idea so far...
Define the linear map $T in mathcal{L}(mathbb{R}^5)$ to be
$$T(x_1, x_2, x_3, x_4, x_5) = (x_1, 0, 0, x_4, 0).$$
Note that for $u in U$ we have that $u = (2x_2, x_2, x_3, 3x_5, x_5)$. Hence, $Tu = 0$ and we conclude that null$:T = U$.
Is this correct?
linear-algebra linear-transformations
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
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add a comment |
$begingroup$
I am given that the set $U = { (x_1, x_2, x_3, x_4, x_5 in mathbb{R}^5 : x_1 - 2x_2 = 0 : text{and} : x_4 - 3x_5 = 0 }$.
I have already shown that $U$ is a subspace of $mathbb{R}^5$. Now I want to find a linear map whose null space is equal to $U$. My idea so far...
Define the linear map $T in mathcal{L}(mathbb{R}^5)$ to be
$$T(x_1, x_2, x_3, x_4, x_5) = (x_1, 0, 0, x_4, 0).$$
Note that for $u in U$ we have that $u = (2x_2, x_2, x_3, 3x_5, x_5)$. Hence, $Tu = 0$ and we conclude that null$:T = U$.
Is this correct?
linear-algebra linear-transformations
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
$endgroup$
$begingroup$
Hint: The null space of a matrix is the orthogonal complement of its row space.
$endgroup$
– amd
Jan 1 at 0:32
add a comment |
$begingroup$
I am given that the set $U = { (x_1, x_2, x_3, x_4, x_5 in mathbb{R}^5 : x_1 - 2x_2 = 0 : text{and} : x_4 - 3x_5 = 0 }$.
I have already shown that $U$ is a subspace of $mathbb{R}^5$. Now I want to find a linear map whose null space is equal to $U$. My idea so far...
Define the linear map $T in mathcal{L}(mathbb{R}^5)$ to be
$$T(x_1, x_2, x_3, x_4, x_5) = (x_1, 0, 0, x_4, 0).$$
Note that for $u in U$ we have that $u = (2x_2, x_2, x_3, 3x_5, x_5)$. Hence, $Tu = 0$ and we conclude that null$:T = U$.
Is this correct?
linear-algebra linear-transformations
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
$endgroup$
I am given that the set $U = { (x_1, x_2, x_3, x_4, x_5 in mathbb{R}^5 : x_1 - 2x_2 = 0 : text{and} : x_4 - 3x_5 = 0 }$.
I have already shown that $U$ is a subspace of $mathbb{R}^5$. Now I want to find a linear map whose null space is equal to $U$. My idea so far...
Define the linear map $T in mathcal{L}(mathbb{R}^5)$ to be
$$T(x_1, x_2, x_3, x_4, x_5) = (x_1, 0, 0, x_4, 0).$$
Note that for $u in U$ we have that $u = (2x_2, x_2, x_3, 3x_5, x_5)$. Hence, $Tu = 0$ and we conclude that null$:T = U$.
Is this correct?
linear-algebra linear-transformations
linear-algebra linear-transformations
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
asked Dec 31 '18 at 18:41
Taylor McMillanTaylor McMillan
695
695
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Hint: The null space of a matrix is the orthogonal complement of its row space.
$endgroup$
– amd
Jan 1 at 0:32
add a comment |
$begingroup$
Hint: The null space of a matrix is the orthogonal complement of its row space.
$endgroup$
– amd
Jan 1 at 0:32
$begingroup$
Hint: The null space of a matrix is the orthogonal complement of its row space.
$endgroup$
– amd
Jan 1 at 0:32
$begingroup$
Hint: The null space of a matrix is the orthogonal complement of its row space.
$endgroup$
– amd
Jan 1 at 0:32
add a comment |
1 Answer
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No, it is not correct. From $u=(2x_2,x_2,x_3,3x_5,x_5)$, what you can deduce is that $T(u)=(2x_2,0,0,3x_5,0)$, not that $T(u)=(0,0,0,0,0)$.
Consider$$begin{array}{rccc}Tcolon&mathbb{R}^5&longrightarrow&mathbb{R}^2\&(x_1,x_2,x_3,x_4,x_5)&mapsto&(x_1-2x_2,x_4-3x_5)end{array}$$instead.
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
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Oh, I see now. Thank you very much!
$endgroup$
– Taylor McMillan
Dec 31 '18 at 19:03
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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$begingroup$
No, it is not correct. From $u=(2x_2,x_2,x_3,3x_5,x_5)$, what you can deduce is that $T(u)=(2x_2,0,0,3x_5,0)$, not that $T(u)=(0,0,0,0,0)$.
Consider$$begin{array}{rccc}Tcolon&mathbb{R}^5&longrightarrow&mathbb{R}^2\&(x_1,x_2,x_3,x_4,x_5)&mapsto&(x_1-2x_2,x_4-3x_5)end{array}$$instead.
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
$endgroup$
$begingroup$
Oh, I see now. Thank you very much!
$endgroup$
– Taylor McMillan
Dec 31 '18 at 19:03
add a comment |
$begingroup$
No, it is not correct. From $u=(2x_2,x_2,x_3,3x_5,x_5)$, what you can deduce is that $T(u)=(2x_2,0,0,3x_5,0)$, not that $T(u)=(0,0,0,0,0)$.
Consider$$begin{array}{rccc}Tcolon&mathbb{R}^5&longrightarrow&mathbb{R}^2\&(x_1,x_2,x_3,x_4,x_5)&mapsto&(x_1-2x_2,x_4-3x_5)end{array}$$instead.
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
$endgroup$
$begingroup$
Oh, I see now. Thank you very much!
$endgroup$
– Taylor McMillan
Dec 31 '18 at 19:03
add a comment |
$begingroup$
No, it is not correct. From $u=(2x_2,x_2,x_3,3x_5,x_5)$, what you can deduce is that $T(u)=(2x_2,0,0,3x_5,0)$, not that $T(u)=(0,0,0,0,0)$.
Consider$$begin{array}{rccc}Tcolon&mathbb{R}^5&longrightarrow&mathbb{R}^2\&(x_1,x_2,x_3,x_4,x_5)&mapsto&(x_1-2x_2,x_4-3x_5)end{array}$$instead.
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
$endgroup$
No, it is not correct. From $u=(2x_2,x_2,x_3,3x_5,x_5)$, what you can deduce is that $T(u)=(2x_2,0,0,3x_5,0)$, not that $T(u)=(0,0,0,0,0)$.
Consider$$begin{array}{rccc}Tcolon&mathbb{R}^5&longrightarrow&mathbb{R}^2\&(x_1,x_2,x_3,x_4,x_5)&mapsto&(x_1-2x_2,x_4-3x_5)end{array}$$instead.
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
answered Dec 31 '18 at 18:46
José Carlos SantosJosé Carlos Santos
164k22131235
164k22131235
$begingroup$
Oh, I see now. Thank you very much!
$endgroup$
– Taylor McMillan
Dec 31 '18 at 19:03
add a comment |
$begingroup$
Oh, I see now. Thank you very much!
$endgroup$
– Taylor McMillan
Dec 31 '18 at 19:03
$begingroup$
Oh, I see now. Thank you very much!
$endgroup$
– Taylor McMillan
Dec 31 '18 at 19:03
$begingroup$
Oh, I see now. Thank you very much!
$endgroup$
– Taylor McMillan
Dec 31 '18 at 19:03
add a comment |
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$begingroup$
Hint: The null space of a matrix is the orthogonal complement of its row space.
$endgroup$
– amd
Jan 1 at 0:32