How to prove that the cross product of two vectors is a linear transformation?
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linear-algebra linear-transformations cross-product
asked Feb 15 '16 at 2:57
Derick
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I'm stuck, could I please have a hint?
linear-algebra linear-transformations cross-product
asked Feb 15 '16 at 2:57
Derick
625
1
It's a bilinear transformation on the pair of vectors; it's a linear transformation on the tensor product of two vectors. $qquad$
– Michael Hardy
Feb 15 '16 at 3:00
@DerickWhat is a bilinear transformation? What are tensor products?
– Derick
Feb 15 '16 at 3:03
Given a specific vector, $v$, the transformation $L(x) = vtimes x$ will indeed be linear. If you can come up with a matrix such that if you multiply a vector by, the result is the cross product of the vector, then that acts as sufficient proof that it is a linear transformation.
– JMoravitz
Feb 15 '16 at 3:03
@Derick : A bilinear transformation is a function of two vector variables that is linear in each variable separately. That means if you hold one of them constant and let the other one vary, then it's a linear function of that other one. The difference between an ordered pair of vectors and a tensor product of vectors is that if you multiply one of the vectors by a nonzero scalar and the other by the reciprocal of that scalar, then you get a different ordered pair but the same tensor product. $qquad$
– Michael Hardy
Feb 15 '16 at 3:06
@Derick To the question about orthogonality, no... why would it? You are able to take the cross product of any two vectors, whether they are orthogonal or not. For the second, it is more than just that.
– JMoravitz
Feb 15 '16 at 3:15
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm stuck, could I please have a hint?
linear-algebra linear-transformations cross-product
asked Feb 15 '16 at 2:57
Derick
625
I'm stuck, could I please have a hint?
linear-algebra linear-transformations cross-product
asked Feb 15 '16 at 2:57
Derick
625
asked Feb 15 '16 at 2:57
Derick
625
asked Feb 15 '16 at 2:57
Derick
625
asked Feb 15 '16 at 2:57
Derick
625
625
1
It's a bilinear transformation on the pair of vectors; it's a linear transformation on the tensor product of two vectors. $qquad$
– Michael Hardy
Feb 15 '16 at 3:00
@DerickWhat is a bilinear transformation? What are tensor products?
– Derick
Feb 15 '16 at 3:03
Given a specific vector, $v$, the transformation $L(x) = vtimes x$ will indeed be linear. If you can come up with a matrix such that if you multiply a vector by, the result is the cross product of the vector, then that acts as sufficient proof that it is a linear transformation.
– JMoravitz
Feb 15 '16 at 3:03
@Derick : A bilinear transformation is a function of two vector variables that is linear in each variable separately. That means if you hold one of them constant and let the other one vary, then it's a linear function of that other one. The difference between an ordered pair of vectors and a tensor product of vectors is that if you multiply one of the vectors by a nonzero scalar and the other by the reciprocal of that scalar, then you get a different ordered pair but the same tensor product. $qquad$
– Michael Hardy
Feb 15 '16 at 3:06
@Derick To the question about orthogonality, no... why would it? You are able to take the cross product of any two vectors, whether they are orthogonal or not. For the second, it is more than just that.
– JMoravitz
Feb 15 '16 at 3:15
add a comment |Â
1
It's a bilinear transformation on the pair of vectors; it's a linear transformation on the tensor product of two vectors. $qquad$
– Michael Hardy
Feb 15 '16 at 3:00
@DerickWhat is a bilinear transformation? What are tensor products?
– Derick
Feb 15 '16 at 3:03
Given a specific vector, $v$, the transformation $L(x) = vtimes x$ will indeed be linear. If you can come up with a matrix such that if you multiply a vector by, the result is the cross product of the vector, then that acts as sufficient proof that it is a linear transformation.
– JMoravitz
Feb 15 '16 at 3:03
@Derick : A bilinear transformation is a function of two vector variables that is linear in each variable separately. That means if you hold one of them constant and let the other one vary, then it's a linear function of that other one. The difference between an ordered pair of vectors and a tensor product of vectors is that if you multiply one of the vectors by a nonzero scalar and the other by the reciprocal of that scalar, then you get a different ordered pair but the same tensor product. $qquad$
– Michael Hardy
Feb 15 '16 at 3:06
@Derick To the question about orthogonality, no... why would it? You are able to take the cross product of any two vectors, whether they are orthogonal or not. For the second, it is more than just that.
– JMoravitz
Feb 15 '16 at 3:15
1
1
It's a bilinear transformation on the pair of vectors; it's a linear transformation on the tensor product of two vectors. $qquad$
– Michael Hardy
Feb 15 '16 at 3:00
It's a bilinear transformation on the pair of vectors; it's a linear transformation on the tensor product of two vectors. $qquad$
– Michael Hardy
Feb 15 '16 at 3:00
@DerickWhat is a bilinear transformation? What are tensor products?
– Derick
Feb 15 '16 at 3:03
@DerickWhat is a bilinear transformation? What are tensor products?
– Derick
Feb 15 '16 at 3:03
Given a specific vector, $v$, the transformation $L(x) = vtimes x$ will indeed be linear. If you can come up with a matrix such that if you multiply a vector by, the result is the cross product of the vector, then that acts as sufficient proof that it is a linear transformation.
– JMoravitz
Feb 15 '16 at 3:03
Given a specific vector, $v$, the transformation $L(x) = vtimes x$ will indeed be linear. If you can come up with a matrix such that if you multiply a vector by, the result is the cross product of the vector, then that acts as sufficient proof that it is a linear transformation.
– JMoravitz
Feb 15 '16 at 3:03
@Derick : A bilinear transformation is a function of two vector variables that is linear in each variable separately. That means if you hold one of them constant and let the other one vary, then it's a linear function of that other one. The difference between an ordered pair of vectors and a tensor product of vectors is that if you multiply one of the vectors by a nonzero scalar and the other by the reciprocal of that scalar, then you get a different ordered pair but the same tensor product. $qquad$
– Michael Hardy
Feb 15 '16 at 3:06
@Derick : A bilinear transformation is a function of two vector variables that is linear in each variable separately. That means if you hold one of them constant and let the other one vary, then it's a linear function of that other one. The difference between an ordered pair of vectors and a tensor product of vectors is that if you multiply one of the vectors by a nonzero scalar and the other by the reciprocal of that scalar, then you get a different ordered pair but the same tensor product. $qquad$
– Michael Hardy
Feb 15 '16 at 3:06
@Derick To the question about orthogonality, no... why would it? You are able to take the cross product of any two vectors, whether they are orthogonal or not. For the second, it is more than just that.
– JMoravitz
Feb 15 '16 at 3:15
@Derick To the question about orthogonality, no... why would it? You are able to take the cross product of any two vectors, whether they are orthogonal or not. For the second, it is more than just that.
– JMoravitz
Feb 15 '16 at 3:15
add a comment |Â
3 Answers
3
active
oldest
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up vote
5
down vote
(tacitly assuming that we are working with the traditional cross product of two vectors in $mathbbR^3$)
Assuming that one of the vectors is held constant. Let $v$ be the fixed vector. We wish to prove that $L(x) = vtimes x$ is a linear transformation of $x$ from $mathbbR^3$ to $mathbbR^3$.
Let $v = beginbmatrixv_1\v_2\v_3endbmatrix$. Let $x = beginbmatrixx_1\x_2\x_3endbmatrix$.
The cross product $vtimes x = detleft(beginbmatrixhati&hatj&hatk\
v_1&v_2&v_3\x_1&x_2&x_3endbmatrixright) = beginbmatrixv_2x_3-v_3x_2\-v_1x_3+v_3x_1\v_1x_2-v_2x_1endbmatrix = x_1beginbmatrix0\v_3\-v_2endbmatrix+x_2beginbmatrix-v_3\0\v_1endbmatrix+x_3beginbmatrixstar\star\starendbmatrix$
Letting you fill in the $star$'s above yourself.
How can we construct a matrix then such that $Ax = vtimes x$?
If we successfully construct such a matrix, then that shows that the transformation is indeed a linear transformation.
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
add a comment |Â
up vote
0
down vote
For a function $L: mathbbR^3 to mathbbR^3$ to be a linear transformation, you have to verify two properties:
- $L(a overrightarrowv)=aL(overrightarrowv)$ for any scalar $a$ and any vector $overrightarrowv$
- $L(overrightarrowv+overrightarroww)=L(overrightarrowv)+L(overrightarroww)$ for any two vectors $overrightarrowv$ and $overrightarroww$
In this case, fix some vector $overrightarrowu$, and consider the linear transformation defined by $L(overrightarrowv)= overrightarrowu times overrightarrowv$. To show that the cross-product is linear, you need to show that properties (1) and (2) above hold; in other words, you need to verify that:
- $overrightarrowu times (aoverrightarrowv)=a ( overrightarrowu times overrightarrowv)$
- $overrightarrowu times (overrightarrowv + overrightarroww)=(overrightarrowutimes overrightarrowv)+(overrightarrowutimes overrightarroww)$
Can you take it from there?
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
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up vote
0
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To prove that $vecatimes(vecb + vecc) = vecatimesvecb + vecatimesvecc$, we have to make these vectors in Caertesian component using an orthonormal natural base $hatx_j ~$. I shall use Einstein Summation Convention, that is $veca = a_jhatx_j$, $vecb = b_jhatx_j$, and $vecc = c_jhatx_j$, without summation symbol $sum$. Now, we get
$vecatimes(vecb + vecc)$
$= a_jhatx_jtimeshatx_k(b_k + c_k)$
$= a_j(b_k + c_k)hatx_jtimeshatx_k$
$= (a_jb_k + a_jc_k)hatx_jtimeshatx_k$
$= a_jb_khatx_jtimeshatx_k + a_jc_khatx_jtimeshatx_k$
$= (a_jhatx_j)times(b_khatx_k) + (a_jhatx_j)times(c_khatx_k)$
$= vecatimesvecb + vecatimesvecc$.
answered Aug 14 at 4:36
Rotor Rotator
1
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
(tacitly assuming that we are working with the traditional cross product of two vectors in $mathbbR^3$)
Assuming that one of the vectors is held constant. Let $v$ be the fixed vector. We wish to prove that $L(x) = vtimes x$ is a linear transformation of $x$ from $mathbbR^3$ to $mathbbR^3$.
Let $v = beginbmatrixv_1\v_2\v_3endbmatrix$. Let $x = beginbmatrixx_1\x_2\x_3endbmatrix$.
The cross product $vtimes x = detleft(beginbmatrixhati&hatj&hatk\
v_1&v_2&v_3\x_1&x_2&x_3endbmatrixright) = beginbmatrixv_2x_3-v_3x_2\-v_1x_3+v_3x_1\v_1x_2-v_2x_1endbmatrix = x_1beginbmatrix0\v_3\-v_2endbmatrix+x_2beginbmatrix-v_3\0\v_1endbmatrix+x_3beginbmatrixstar\star\starendbmatrix$
Letting you fill in the $star$'s above yourself.
How can we construct a matrix then such that $Ax = vtimes x$?
If we successfully construct such a matrix, then that shows that the transformation is indeed a linear transformation.
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
add a comment |Â
up vote
5
down vote
(tacitly assuming that we are working with the traditional cross product of two vectors in $mathbbR^3$)
Assuming that one of the vectors is held constant. Let $v$ be the fixed vector. We wish to prove that $L(x) = vtimes x$ is a linear transformation of $x$ from $mathbbR^3$ to $mathbbR^3$.
Let $v = beginbmatrixv_1\v_2\v_3endbmatrix$. Let $x = beginbmatrixx_1\x_2\x_3endbmatrix$.
The cross product $vtimes x = detleft(beginbmatrixhati&hatj&hatk\
v_1&v_2&v_3\x_1&x_2&x_3endbmatrixright) = beginbmatrixv_2x_3-v_3x_2\-v_1x_3+v_3x_1\v_1x_2-v_2x_1endbmatrix = x_1beginbmatrix0\v_3\-v_2endbmatrix+x_2beginbmatrix-v_3\0\v_1endbmatrix+x_3beginbmatrixstar\star\starendbmatrix$
Letting you fill in the $star$'s above yourself.
How can we construct a matrix then such that $Ax = vtimes x$?
If we successfully construct such a matrix, then that shows that the transformation is indeed a linear transformation.
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
add a comment |Â
up vote
5
down vote
up vote
5
down vote
(tacitly assuming that we are working with the traditional cross product of two vectors in $mathbbR^3$)
Assuming that one of the vectors is held constant. Let $v$ be the fixed vector. We wish to prove that $L(x) = vtimes x$ is a linear transformation of $x$ from $mathbbR^3$ to $mathbbR^3$.
Let $v = beginbmatrixv_1\v_2\v_3endbmatrix$. Let $x = beginbmatrixx_1\x_2\x_3endbmatrix$.
The cross product $vtimes x = detleft(beginbmatrixhati&hatj&hatk\
v_1&v_2&v_3\x_1&x_2&x_3endbmatrixright) = beginbmatrixv_2x_3-v_3x_2\-v_1x_3+v_3x_1\v_1x_2-v_2x_1endbmatrix = x_1beginbmatrix0\v_3\-v_2endbmatrix+x_2beginbmatrix-v_3\0\v_1endbmatrix+x_3beginbmatrixstar\star\starendbmatrix$
Letting you fill in the $star$'s above yourself.
How can we construct a matrix then such that $Ax = vtimes x$?
If we successfully construct such a matrix, then that shows that the transformation is indeed a linear transformation.
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
(tacitly assuming that we are working with the traditional cross product of two vectors in $mathbbR^3$)
Assuming that one of the vectors is held constant. Let $v$ be the fixed vector. We wish to prove that $L(x) = vtimes x$ is a linear transformation of $x$ from $mathbbR^3$ to $mathbbR^3$.
Let $v = beginbmatrixv_1\v_2\v_3endbmatrix$. Let $x = beginbmatrixx_1\x_2\x_3endbmatrix$.
The cross product $vtimes x = detleft(beginbmatrixhati&hatj&hatk\
v_1&v_2&v_3\x_1&x_2&x_3endbmatrixright) = beginbmatrixv_2x_3-v_3x_2\-v_1x_3+v_3x_1\v_1x_2-v_2x_1endbmatrix = x_1beginbmatrix0\v_3\-v_2endbmatrix+x_2beginbmatrix-v_3\0\v_1endbmatrix+x_3beginbmatrixstar\star\starendbmatrix$
Letting you fill in the $star$'s above yourself.
How can we construct a matrix then such that $Ax = vtimes x$?
If we successfully construct such a matrix, then that shows that the transformation is indeed a linear transformation.
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
answered Feb 15 '16 at 3:12
JMoravitz
44.4k33481
44.4k33481
add a comment |Â
add a comment |Â
up vote
0
down vote
For a function $L: mathbbR^3 to mathbbR^3$ to be a linear transformation, you have to verify two properties:
- $L(a overrightarrowv)=aL(overrightarrowv)$ for any scalar $a$ and any vector $overrightarrowv$
- $L(overrightarrowv+overrightarroww)=L(overrightarrowv)+L(overrightarroww)$ for any two vectors $overrightarrowv$ and $overrightarroww$
In this case, fix some vector $overrightarrowu$, and consider the linear transformation defined by $L(overrightarrowv)= overrightarrowu times overrightarrowv$. To show that the cross-product is linear, you need to show that properties (1) and (2) above hold; in other words, you need to verify that:
- $overrightarrowu times (aoverrightarrowv)=a ( overrightarrowu times overrightarrowv)$
- $overrightarrowu times (overrightarrowv + overrightarroww)=(overrightarrowutimes overrightarrowv)+(overrightarrowutimes overrightarroww)$
Can you take it from there?
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
add a comment |Â
up vote
0
down vote
For a function $L: mathbbR^3 to mathbbR^3$ to be a linear transformation, you have to verify two properties:
- $L(a overrightarrowv)=aL(overrightarrowv)$ for any scalar $a$ and any vector $overrightarrowv$
- $L(overrightarrowv+overrightarroww)=L(overrightarrowv)+L(overrightarroww)$ for any two vectors $overrightarrowv$ and $overrightarroww$
In this case, fix some vector $overrightarrowu$, and consider the linear transformation defined by $L(overrightarrowv)= overrightarrowu times overrightarrowv$. To show that the cross-product is linear, you need to show that properties (1) and (2) above hold; in other words, you need to verify that:
- $overrightarrowu times (aoverrightarrowv)=a ( overrightarrowu times overrightarrowv)$
- $overrightarrowu times (overrightarrowv + overrightarroww)=(overrightarrowutimes overrightarrowv)+(overrightarrowutimes overrightarroww)$
Can you take it from there?
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For a function $L: mathbbR^3 to mathbbR^3$ to be a linear transformation, you have to verify two properties:
- $L(a overrightarrowv)=aL(overrightarrowv)$ for any scalar $a$ and any vector $overrightarrowv$
- $L(overrightarrowv+overrightarroww)=L(overrightarrowv)+L(overrightarroww)$ for any two vectors $overrightarrowv$ and $overrightarroww$
In this case, fix some vector $overrightarrowu$, and consider the linear transformation defined by $L(overrightarrowv)= overrightarrowu times overrightarrowv$. To show that the cross-product is linear, you need to show that properties (1) and (2) above hold; in other words, you need to verify that:
- $overrightarrowu times (aoverrightarrowv)=a ( overrightarrowu times overrightarrowv)$
- $overrightarrowu times (overrightarrowv + overrightarroww)=(overrightarrowutimes overrightarrowv)+(overrightarrowutimes overrightarroww)$
Can you take it from there?
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
For a function $L: mathbbR^3 to mathbbR^3$ to be a linear transformation, you have to verify two properties:
- $L(a overrightarrowv)=aL(overrightarrowv)$ for any scalar $a$ and any vector $overrightarrowv$
- $L(overrightarrowv+overrightarroww)=L(overrightarrowv)+L(overrightarroww)$ for any two vectors $overrightarrowv$ and $overrightarroww$
In this case, fix some vector $overrightarrowu$, and consider the linear transformation defined by $L(overrightarrowv)= overrightarrowu times overrightarrowv$. To show that the cross-product is linear, you need to show that properties (1) and (2) above hold; in other words, you need to verify that:
- $overrightarrowu times (aoverrightarrowv)=a ( overrightarrowu times overrightarrowv)$
- $overrightarrowu times (overrightarrowv + overrightarroww)=(overrightarrowutimes overrightarrowv)+(overrightarrowutimes overrightarroww)$
Can you take it from there?
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
answered Feb 15 '16 at 3:24
mweiss
17.2k23268
17.2k23268
add a comment |Â
add a comment |Â
up vote
0
down vote
To prove that $vecatimes(vecb + vecc) = vecatimesvecb + vecatimesvecc$, we have to make these vectors in Caertesian component using an orthonormal natural base $hatx_j ~$. I shall use Einstein Summation Convention, that is $veca = a_jhatx_j$, $vecb = b_jhatx_j$, and $vecc = c_jhatx_j$, without summation symbol $sum$. Now, we get
$vecatimes(vecb + vecc)$
$= a_jhatx_jtimeshatx_k(b_k + c_k)$
$= a_j(b_k + c_k)hatx_jtimeshatx_k$
$= (a_jb_k + a_jc_k)hatx_jtimeshatx_k$
$= a_jb_khatx_jtimeshatx_k + a_jc_khatx_jtimeshatx_k$
$= (a_jhatx_j)times(b_khatx_k) + (a_jhatx_j)times(c_khatx_k)$
$= vecatimesvecb + vecatimesvecc$.
answered Aug 14 at 4:36
Rotor Rotator
1
add a comment |Â
up vote
0
down vote
To prove that $vecatimes(vecb + vecc) = vecatimesvecb + vecatimesvecc$, we have to make these vectors in Caertesian component using an orthonormal natural base $hatx_j ~$. I shall use Einstein Summation Convention, that is $veca = a_jhatx_j$, $vecb = b_jhatx_j$, and $vecc = c_jhatx_j$, without summation symbol $sum$. Now, we get
$vecatimes(vecb + vecc)$
$= a_jhatx_jtimeshatx_k(b_k + c_k)$
$= a_j(b_k + c_k)hatx_jtimeshatx_k$
$= (a_jb_k + a_jc_k)hatx_jtimeshatx_k$
$= a_jb_khatx_jtimeshatx_k + a_jc_khatx_jtimeshatx_k$
$= (a_jhatx_j)times(b_khatx_k) + (a_jhatx_j)times(c_khatx_k)$
$= vecatimesvecb + vecatimesvecc$.
answered Aug 14 at 4:36
Rotor Rotator
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To prove that $vecatimes(vecb + vecc) = vecatimesvecb + vecatimesvecc$, we have to make these vectors in Caertesian component using an orthonormal natural base $hatx_j ~$. I shall use Einstein Summation Convention, that is $veca = a_jhatx_j$, $vecb = b_jhatx_j$, and $vecc = c_jhatx_j$, without summation symbol $sum$. Now, we get
$vecatimes(vecb + vecc)$
$= a_jhatx_jtimeshatx_k(b_k + c_k)$
$= a_j(b_k + c_k)hatx_jtimeshatx_k$
$= (a_jb_k + a_jc_k)hatx_jtimeshatx_k$
$= a_jb_khatx_jtimeshatx_k + a_jc_khatx_jtimeshatx_k$
$= (a_jhatx_j)times(b_khatx_k) + (a_jhatx_j)times(c_khatx_k)$
$= vecatimesvecb + vecatimesvecc$.
answered Aug 14 at 4:36
Rotor Rotator
1
To prove that $vecatimes(vecb + vecc) = vecatimesvecb + vecatimesvecc$, we have to make these vectors in Caertesian component using an orthonormal natural base $hatx_j ~$. I shall use Einstein Summation Convention, that is $veca = a_jhatx_j$, $vecb = b_jhatx_j$, and $vecc = c_jhatx_j$, without summation symbol $sum$. Now, we get
$vecatimes(vecb + vecc)$
$= a_jhatx_jtimeshatx_k(b_k + c_k)$
$= a_j(b_k + c_k)hatx_jtimeshatx_k$
$= (a_jb_k + a_jc_k)hatx_jtimeshatx_k$
$= a_jb_khatx_jtimeshatx_k + a_jc_khatx_jtimeshatx_k$
$= (a_jhatx_j)times(b_khatx_k) + (a_jhatx_j)times(c_khatx_k)$
$= vecatimesvecb + vecatimesvecc$.
answered Aug 14 at 4:36
Rotor Rotator
1
answered Aug 14 at 4:36
Rotor Rotator
1
answered Aug 14 at 4:36
Rotor Rotator
1
answered Aug 14 at 4:36
Rotor Rotator
1
1
add a comment |Â
add a comment |Â
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1
It's a bilinear transformation on the pair of vectors; it's a linear transformation on the tensor product of two vectors. $qquad$
– Michael Hardy
Feb 15 '16 at 3:00
@DerickWhat is a bilinear transformation? What are tensor products?
– Derick
Feb 15 '16 at 3:03
Given a specific vector, $v$, the transformation $L(x) = vtimes x$ will indeed be linear. If you can come up with a matrix such that if you multiply a vector by, the result is the cross product of the vector, then that acts as sufficient proof that it is a linear transformation.
– JMoravitz
Feb 15 '16 at 3:03
@Derick : A bilinear transformation is a function of two vector variables that is linear in each variable separately. That means if you hold one of them constant and let the other one vary, then it's a linear function of that other one. The difference between an ordered pair of vectors and a tensor product of vectors is that if you multiply one of the vectors by a nonzero scalar and the other by the reciprocal of that scalar, then you get a different ordered pair but the same tensor product. $qquad$
– Michael Hardy
Feb 15 '16 at 3:06
@Derick To the question about orthogonality, no... why would it? You are able to take the cross product of any two vectors, whether they are orthogonal or not. For the second, it is more than just that.
– JMoravitz
Feb 15 '16 at 3:15