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How to determine if a set of five $2times2$ matrices is independent
Clash Royale CLAN TAG#URR8PPP
up vote
10
down vote
favorite
$$S=biggleft[beginmatrix1&2\2&1endmatrixright], left[beginmatrix2&1\-1&2endmatrixright], left[beginmatrix0&1\1&2endmatrixright],left[beginmatrix1&0\1&1endmatrixright],
left[beginmatrix1&4\0&3endmatrixright]bigg$$
How can I determine if a set of five $2times2$ matrices are independent?
linear-algebra matrices
edited Jul 26 at 9:05
TheSimpliFire
9,60461851
asked Jul 25 at 12:58
Mi_11
544
add a comment |Â
up vote
10
down vote
favorite
$$S=biggleft[beginmatrix1&2\2&1endmatrixright], left[beginmatrix2&1\-1&2endmatrixright], left[beginmatrix0&1\1&2endmatrixright],left[beginmatrix1&0\1&1endmatrixright],
left[beginmatrix1&4\0&3endmatrixright]bigg$$
How can I determine if a set of five $2times2$ matrices are independent?
linear-algebra matrices
edited Jul 26 at 9:05
TheSimpliFire
9,60461851
asked Jul 25 at 12:58
Mi_11
544
27
They can't be, View them as length-4 vectors and conclude.
– Parcly Taxel
Jul 25 at 12:59
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
$$S=biggleft[beginmatrix1&2\2&1endmatrixright], left[beginmatrix2&1\-1&2endmatrixright], left[beginmatrix0&1\1&2endmatrixright],left[beginmatrix1&0\1&1endmatrixright],
left[beginmatrix1&4\0&3endmatrixright]bigg$$
How can I determine if a set of five $2times2$ matrices are independent?
linear-algebra matrices
edited Jul 26 at 9:05
TheSimpliFire
9,60461851
asked Jul 25 at 12:58
Mi_11
544
$$S=biggleft[beginmatrix1&2\2&1endmatrixright], left[beginmatrix2&1\-1&2endmatrixright], left[beginmatrix0&1\1&2endmatrixright],left[beginmatrix1&0\1&1endmatrixright],
left[beginmatrix1&4\0&3endmatrixright]bigg$$
How can I determine if a set of five $2times2$ matrices are independent?
linear-algebra matrices
edited Jul 26 at 9:05
TheSimpliFire
9,60461851
asked Jul 25 at 12:58
Mi_11
544
edited Jul 26 at 9:05
TheSimpliFire
9,60461851
edited Jul 26 at 9:05
TheSimpliFire
9,60461851
edited Jul 26 at 9:05
TheSimpliFire
9,60461851
9,60461851
asked Jul 25 at 12:58
Mi_11
544
asked Jul 25 at 12:58
Mi_11
544
asked Jul 25 at 12:58
Mi_11
544
544
27
They can't be, View them as length-4 vectors and conclude.
– Parcly Taxel
Jul 25 at 12:59
add a comment |Â
27
They can't be, View them as length-4 vectors and conclude.
– Parcly Taxel
Jul 25 at 12:59
27
27
They can't be, View them as length-4 vectors and conclude.
– Parcly Taxel
Jul 25 at 12:59
They can't be, View them as length-4 vectors and conclude.
– Parcly Taxel
Jul 25 at 12:59
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
41
down vote
Since the space of all $2times2$ matrices is $4$-dimensional, every set of $5$ such matrices is linearly dependent.
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
3
@Mi_11 consider accepting it as valid
– Ander Biguri
Jul 25 at 15:43
add a comment |Â
up vote
6
down vote
As has been pointed out, four matrices form a basis for the $2times2$ matrices (the easiest would be
$$
left[beginmatrix1&0\0&0endmatrixright], left[beginmatrix0&1\0&0endmatrixright], left[beginmatrix0&0\1&0endmatrixright], left[beginmatrix0&0\0&1endmatrixright]
$$) so five matrices cannot be linearly dependent.
In your case the dependence is
$$
left[beginmatrix1&2\2&1endmatrixright] + left[beginmatrix2&1\-1&2endmatrixright] + left[beginmatrix0&1\1&2endmatrixright] - 2left[beginmatrix1&0\1&1endmatrixright] -
left[beginmatrix1&4\0&3endmatrixright] =
left[beginmatrix0&0\0&0endmatrixright].
$$
answered Jul 25 at 15:11
Charles
23.3k448111
add a comment |Â
up vote
2
down vote
As the others have said, this set of $5$ must be linearly dependent because the dimension of the space of all $2times 2$ matrices is $4$.
More generally, how do you show that a set of vectors is linearly dependent or independent? Create a linear combination of the vectors, set it equal to $0$, and try to solve it.
$$
a_1X_1 + a_2X_2 + dotsb+ a_nX_n = 0
$$
If the only possible solution is $a_1 = a_2 = dotsb = a_n = 0$ then the set is independent. If a different solution exists then the set is dependent.
edited Jul 25 at 21:23
Foobaz John
18k41245
answered Jul 25 at 13:56
nurdyguy
1214
add a comment |Â
up vote
0
down vote
Stretch out the matrices to complete the rows of the following matrix
$$newcommandadjoperatornameadj
M(v)=beginbmatrix
v_1&1&2&2&1\
v_2&2&1&-1&2\
v_3&0&1&1&2\
v_4&1&0&1&1\
v_5&1&4&0&3
endbmatrixtag1
$$
Note that the top row of the adjugate of $M(v)$
$$
beginbmatrix
-14&-14&-14&28&14
endbmatrixtag2
$$
is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $adj M(v)$. By Laplace's Formula,
$$
det M(v)=ucdot vtag3
$$
Setting $v$ to be any of the fixed columns of $M(v)$ gives $det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.
We can rewrite the dot product of $-frac114u$ with the fixed columns of $M(v)$ as
$$
1 overbracebeginbmatrix
1&2\2&1
endbmatrix^textrow $1$
+
1 overbracebeginbmatrix
2&1\-1&2
endbmatrix^textrow $2$
+
1 overbracebeginbmatrix
0&1\1&2
endbmatrix^textrow $3$
-2 overbracebeginbmatrix
1&0\1&1
endbmatrix^textrow $4$
-
1 overbracebeginbmatrix
1&4\0&3
endbmatrix^textrow $5$
=
beginbmatrix
0&0\0&0
endbmatrixtag4
$$
answered Jul 26 at 20:50
robjohn♦
258k26297612
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
41
down vote
Since the space of all $2times2$ matrices is $4$-dimensional, every set of $5$ such matrices is linearly dependent.
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
3
@Mi_11 consider accepting it as valid
– Ander Biguri
Jul 25 at 15:43
add a comment |Â
up vote
41
down vote
Since the space of all $2times2$ matrices is $4$-dimensional, every set of $5$ such matrices is linearly dependent.
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
3
@Mi_11 consider accepting it as valid
– Ander Biguri
Jul 25 at 15:43
add a comment |Â
up vote
41
down vote
up vote
41
down vote
Since the space of all $2times2$ matrices is $4$-dimensional, every set of $5$ such matrices is linearly dependent.
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
Since the space of all $2times2$ matrices is $4$-dimensional, every set of $5$ such matrices is linearly dependent.
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
answered Jul 25 at 13:00
José Carlos Santos
113k1697176
113k1697176
3
@Mi_11 consider accepting it as valid
– Ander Biguri
Jul 25 at 15:43
add a comment |Â
3
@Mi_11 consider accepting it as valid
– Ander Biguri
Jul 25 at 15:43
3
3
@Mi_11 consider accepting it as valid
– Ander Biguri
Jul 25 at 15:43
@Mi_11 consider accepting it as valid
– Ander Biguri
Jul 25 at 15:43
add a comment |Â
up vote
6
down vote
As has been pointed out, four matrices form a basis for the $2times2$ matrices (the easiest would be
$$
left[beginmatrix1&0\0&0endmatrixright], left[beginmatrix0&1\0&0endmatrixright], left[beginmatrix0&0\1&0endmatrixright], left[beginmatrix0&0\0&1endmatrixright]
$$) so five matrices cannot be linearly dependent.
In your case the dependence is
$$
left[beginmatrix1&2\2&1endmatrixright] + left[beginmatrix2&1\-1&2endmatrixright] + left[beginmatrix0&1\1&2endmatrixright] - 2left[beginmatrix1&0\1&1endmatrixright] -
left[beginmatrix1&4\0&3endmatrixright] =
left[beginmatrix0&0\0&0endmatrixright].
$$
answered Jul 25 at 15:11
Charles
23.3k448111
add a comment |Â
up vote
6
down vote
As has been pointed out, four matrices form a basis for the $2times2$ matrices (the easiest would be
$$
left[beginmatrix1&0\0&0endmatrixright], left[beginmatrix0&1\0&0endmatrixright], left[beginmatrix0&0\1&0endmatrixright], left[beginmatrix0&0\0&1endmatrixright]
$$) so five matrices cannot be linearly dependent.
In your case the dependence is
$$
left[beginmatrix1&2\2&1endmatrixright] + left[beginmatrix2&1\-1&2endmatrixright] + left[beginmatrix0&1\1&2endmatrixright] - 2left[beginmatrix1&0\1&1endmatrixright] -
left[beginmatrix1&4\0&3endmatrixright] =
left[beginmatrix0&0\0&0endmatrixright].
$$
answered Jul 25 at 15:11
Charles
23.3k448111
add a comment |Â
up vote
6
down vote
up vote
6
down vote
As has been pointed out, four matrices form a basis for the $2times2$ matrices (the easiest would be
$$
left[beginmatrix1&0\0&0endmatrixright], left[beginmatrix0&1\0&0endmatrixright], left[beginmatrix0&0\1&0endmatrixright], left[beginmatrix0&0\0&1endmatrixright]
$$) so five matrices cannot be linearly dependent.
In your case the dependence is
$$
left[beginmatrix1&2\2&1endmatrixright] + left[beginmatrix2&1\-1&2endmatrixright] + left[beginmatrix0&1\1&2endmatrixright] - 2left[beginmatrix1&0\1&1endmatrixright] -
left[beginmatrix1&4\0&3endmatrixright] =
left[beginmatrix0&0\0&0endmatrixright].
$$
answered Jul 25 at 15:11
Charles
23.3k448111
As has been pointed out, four matrices form a basis for the $2times2$ matrices (the easiest would be
$$
left[beginmatrix1&0\0&0endmatrixright], left[beginmatrix0&1\0&0endmatrixright], left[beginmatrix0&0\1&0endmatrixright], left[beginmatrix0&0\0&1endmatrixright]
$$) so five matrices cannot be linearly dependent.
In your case the dependence is
$$
left[beginmatrix1&2\2&1endmatrixright] + left[beginmatrix2&1\-1&2endmatrixright] + left[beginmatrix0&1\1&2endmatrixright] - 2left[beginmatrix1&0\1&1endmatrixright] -
left[beginmatrix1&4\0&3endmatrixright] =
left[beginmatrix0&0\0&0endmatrixright].
$$
answered Jul 25 at 15:11
Charles
23.3k448111
answered Jul 25 at 15:11
Charles
23.3k448111
answered Jul 25 at 15:11
Charles
23.3k448111
answered Jul 25 at 15:11
Charles
23.3k448111
23.3k448111
add a comment |Â
add a comment |Â
up vote
2
down vote
As the others have said, this set of $5$ must be linearly dependent because the dimension of the space of all $2times 2$ matrices is $4$.
More generally, how do you show that a set of vectors is linearly dependent or independent? Create a linear combination of the vectors, set it equal to $0$, and try to solve it.
$$
a_1X_1 + a_2X_2 + dotsb+ a_nX_n = 0
$$
If the only possible solution is $a_1 = a_2 = dotsb = a_n = 0$ then the set is independent. If a different solution exists then the set is dependent.
edited Jul 25 at 21:23
Foobaz John
18k41245
answered Jul 25 at 13:56
nurdyguy
1214
add a comment |Â
up vote
2
down vote
As the others have said, this set of $5$ must be linearly dependent because the dimension of the space of all $2times 2$ matrices is $4$.
More generally, how do you show that a set of vectors is linearly dependent or independent? Create a linear combination of the vectors, set it equal to $0$, and try to solve it.
$$
a_1X_1 + a_2X_2 + dotsb+ a_nX_n = 0
$$
If the only possible solution is $a_1 = a_2 = dotsb = a_n = 0$ then the set is independent. If a different solution exists then the set is dependent.
edited Jul 25 at 21:23
Foobaz John
18k41245
answered Jul 25 at 13:56
nurdyguy
1214
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As the others have said, this set of $5$ must be linearly dependent because the dimension of the space of all $2times 2$ matrices is $4$.
More generally, how do you show that a set of vectors is linearly dependent or independent? Create a linear combination of the vectors, set it equal to $0$, and try to solve it.
$$
a_1X_1 + a_2X_2 + dotsb+ a_nX_n = 0
$$
If the only possible solution is $a_1 = a_2 = dotsb = a_n = 0$ then the set is independent. If a different solution exists then the set is dependent.
edited Jul 25 at 21:23
Foobaz John
18k41245
answered Jul 25 at 13:56
nurdyguy
1214
As the others have said, this set of $5$ must be linearly dependent because the dimension of the space of all $2times 2$ matrices is $4$.
More generally, how do you show that a set of vectors is linearly dependent or independent? Create a linear combination of the vectors, set it equal to $0$, and try to solve it.
$$
a_1X_1 + a_2X_2 + dotsb+ a_nX_n = 0
$$
If the only possible solution is $a_1 = a_2 = dotsb = a_n = 0$ then the set is independent. If a different solution exists then the set is dependent.
edited Jul 25 at 21:23
Foobaz John
18k41245
answered Jul 25 at 13:56
nurdyguy
1214
edited Jul 25 at 21:23
Foobaz John
18k41245
edited Jul 25 at 21:23
Foobaz John
18k41245
edited Jul 25 at 21:23
Foobaz John
18k41245
18k41245
answered Jul 25 at 13:56
nurdyguy
1214
answered Jul 25 at 13:56
nurdyguy
1214
answered Jul 25 at 13:56
nurdyguy
1214
1214
add a comment |Â
add a comment |Â
up vote
0
down vote
Stretch out the matrices to complete the rows of the following matrix
$$newcommandadjoperatornameadj
M(v)=beginbmatrix
v_1&1&2&2&1\
v_2&2&1&-1&2\
v_3&0&1&1&2\
v_4&1&0&1&1\
v_5&1&4&0&3
endbmatrixtag1
$$
Note that the top row of the adjugate of $M(v)$
$$
beginbmatrix
-14&-14&-14&28&14
endbmatrixtag2
$$
is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $adj M(v)$. By Laplace's Formula,
$$
det M(v)=ucdot vtag3
$$
Setting $v$ to be any of the fixed columns of $M(v)$ gives $det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.
We can rewrite the dot product of $-frac114u$ with the fixed columns of $M(v)$ as
$$
1 overbracebeginbmatrix
1&2\2&1
endbmatrix^textrow $1$
+
1 overbracebeginbmatrix
2&1\-1&2
endbmatrix^textrow $2$
+
1 overbracebeginbmatrix
0&1\1&2
endbmatrix^textrow $3$
-2 overbracebeginbmatrix
1&0\1&1
endbmatrix^textrow $4$
-
1 overbracebeginbmatrix
1&4\0&3
endbmatrix^textrow $5$
=
beginbmatrix
0&0\0&0
endbmatrixtag4
$$
answered Jul 26 at 20:50
robjohn♦
258k26297612
add a comment |Â
up vote
0
down vote
Stretch out the matrices to complete the rows of the following matrix
$$newcommandadjoperatornameadj
M(v)=beginbmatrix
v_1&1&2&2&1\
v_2&2&1&-1&2\
v_3&0&1&1&2\
v_4&1&0&1&1\
v_5&1&4&0&3
endbmatrixtag1
$$
Note that the top row of the adjugate of $M(v)$
$$
beginbmatrix
-14&-14&-14&28&14
endbmatrixtag2
$$
is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $adj M(v)$. By Laplace's Formula,
$$
det M(v)=ucdot vtag3
$$
Setting $v$ to be any of the fixed columns of $M(v)$ gives $det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.
We can rewrite the dot product of $-frac114u$ with the fixed columns of $M(v)$ as
$$
1 overbracebeginbmatrix
1&2\2&1
endbmatrix^textrow $1$
+
1 overbracebeginbmatrix
2&1\-1&2
endbmatrix^textrow $2$
+
1 overbracebeginbmatrix
0&1\1&2
endbmatrix^textrow $3$
-2 overbracebeginbmatrix
1&0\1&1
endbmatrix^textrow $4$
-
1 overbracebeginbmatrix
1&4\0&3
endbmatrix^textrow $5$
=
beginbmatrix
0&0\0&0
endbmatrixtag4
$$
answered Jul 26 at 20:50
robjohn♦
258k26297612
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Stretch out the matrices to complete the rows of the following matrix
$$newcommandadjoperatornameadj
M(v)=beginbmatrix
v_1&1&2&2&1\
v_2&2&1&-1&2\
v_3&0&1&1&2\
v_4&1&0&1&1\
v_5&1&4&0&3
endbmatrixtag1
$$
Note that the top row of the adjugate of $M(v)$
$$
beginbmatrix
-14&-14&-14&28&14
endbmatrixtag2
$$
is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $adj M(v)$. By Laplace's Formula,
$$
det M(v)=ucdot vtag3
$$
Setting $v$ to be any of the fixed columns of $M(v)$ gives $det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.
We can rewrite the dot product of $-frac114u$ with the fixed columns of $M(v)$ as
$$
1 overbracebeginbmatrix
1&2\2&1
endbmatrix^textrow $1$
+
1 overbracebeginbmatrix
2&1\-1&2
endbmatrix^textrow $2$
+
1 overbracebeginbmatrix
0&1\1&2
endbmatrix^textrow $3$
-2 overbracebeginbmatrix
1&0\1&1
endbmatrix^textrow $4$
-
1 overbracebeginbmatrix
1&4\0&3
endbmatrix^textrow $5$
=
beginbmatrix
0&0\0&0
endbmatrixtag4
$$
answered Jul 26 at 20:50
robjohn♦
258k26297612
Stretch out the matrices to complete the rows of the following matrix
$$newcommandadjoperatornameadj
M(v)=beginbmatrix
v_1&1&2&2&1\
v_2&2&1&-1&2\
v_3&0&1&1&2\
v_4&1&0&1&1\
v_5&1&4&0&3
endbmatrixtag1
$$
Note that the top row of the adjugate of $M(v)$
$$
beginbmatrix
-14&-14&-14&28&14
endbmatrixtag2
$$
is independent of $v$ because it consists of cofactors of the elements of the left column of $M(v)$. Let $u$ be the top row of $adj M(v)$. By Laplace's Formula,
$$
det M(v)=ucdot vtag3
$$
Setting $v$ to be any of the fixed columns of $M(v)$ gives $det M(v)=0$ because of duplicate columns. Thus, $u$ is perpendicular to all the fixed columns of $M(v)$.
We can rewrite the dot product of $-frac114u$ with the fixed columns of $M(v)$ as
$$
1 overbracebeginbmatrix
1&2\2&1
endbmatrix^textrow $1$
+
1 overbracebeginbmatrix
2&1\-1&2
endbmatrix^textrow $2$
+
1 overbracebeginbmatrix
0&1\1&2
endbmatrix^textrow $3$
-2 overbracebeginbmatrix
1&0\1&1
endbmatrix^textrow $4$
-
1 overbracebeginbmatrix
1&4\0&3
endbmatrix^textrow $5$
=
beginbmatrix
0&0\0&0
endbmatrixtag4
$$
answered Jul 26 at 20:50
robjohn♦
258k26297612
edited Jul 26 at 20:59
answered Jul 26 at 20:50
robjohn♦
258k26297612
answered Jul 26 at 20:50
robjohn♦
258k26297612
answered Jul 26 at 20:50
robjohn♦
258k26297612
258k26297612
add a comment |Â
add a comment |Â
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27
They can't be, View them as length-4 vectors and conclude.
– Parcly Taxel
Jul 25 at 12:59