Mixing
Rank-Deficiency in an Augmented Matrix comprised of Two Full Rank Matrices
Clash Royale CLAN TAG#URR8PPP
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I have a $boldsymbolM_6times 6$ matrix structured as follows:
$$boldsymbolA = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3endbmatrix$$
$$boldsymbolB = beginbmatrixb_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
$$boldsymbolM = beginbmatrixboldsymbolA\boldsymbolBendbmatrix = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3\b_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
I populated $boldsymbolA$, $boldsymbolB$ with several different combinations of values for $a_i, b_i$, e.g., $a_i$ = $[1, 2, 3]$, $b_i$ = $[4, 5, 6]$, to find the rank of $boldsymbolM$ using MATLAB.
It turns out that $Rank(boldsymbolA)=3$ and $Rank(boldsymbolB)=3$, but, $boldsymbolM$ is rank-deficient with $Rank(boldsymbolM)=5$. I was not able to find the dependent row in $boldsymbolM$. All rows look like linearly independent.
I tried to find the rank of M([r1, r2, r3, r4, r5],:) but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Could someone kindly help me how I can find the row in $boldsymbolM$ that is linearly dependent and makes $boldsymbolM$ rank deficient?
linear-algebra vector-spaces matlab matrix-rank
asked Aug 1 at 2:17
AFP
19811
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up vote
2
down vote
favorite
I have a $boldsymbolM_6times 6$ matrix structured as follows:
$$boldsymbolA = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3endbmatrix$$
$$boldsymbolB = beginbmatrixb_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
$$boldsymbolM = beginbmatrixboldsymbolA\boldsymbolBendbmatrix = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3\b_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
I populated $boldsymbolA$, $boldsymbolB$ with several different combinations of values for $a_i, b_i$, e.g., $a_i$ = $[1, 2, 3]$, $b_i$ = $[4, 5, 6]$, to find the rank of $boldsymbolM$ using MATLAB.
It turns out that $Rank(boldsymbolA)=3$ and $Rank(boldsymbolB)=3$, but, $boldsymbolM$ is rank-deficient with $Rank(boldsymbolM)=5$. I was not able to find the dependent row in $boldsymbolM$. All rows look like linearly independent.
I tried to find the rank of M([r1, r2, r3, r4, r5],:) but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Could someone kindly help me how I can find the row in $boldsymbolM$ that is linearly dependent and makes $boldsymbolM$ rank deficient?
linear-algebra vector-spaces matlab matrix-rank
asked Aug 1 at 2:17
AFP
19811
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a $boldsymbolM_6times 6$ matrix structured as follows:
$$boldsymbolA = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3endbmatrix$$
$$boldsymbolB = beginbmatrixb_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
$$boldsymbolM = beginbmatrixboldsymbolA\boldsymbolBendbmatrix = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3\b_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
I populated $boldsymbolA$, $boldsymbolB$ with several different combinations of values for $a_i, b_i$, e.g., $a_i$ = $[1, 2, 3]$, $b_i$ = $[4, 5, 6]$, to find the rank of $boldsymbolM$ using MATLAB.
It turns out that $Rank(boldsymbolA)=3$ and $Rank(boldsymbolB)=3$, but, $boldsymbolM$ is rank-deficient with $Rank(boldsymbolM)=5$. I was not able to find the dependent row in $boldsymbolM$. All rows look like linearly independent.
I tried to find the rank of M([r1, r2, r3, r4, r5],:) but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Could someone kindly help me how I can find the row in $boldsymbolM$ that is linearly dependent and makes $boldsymbolM$ rank deficient?
linear-algebra vector-spaces matlab matrix-rank
asked Aug 1 at 2:17
AFP
19811
I have a $boldsymbolM_6times 6$ matrix structured as follows:
$$boldsymbolA = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3endbmatrix$$
$$boldsymbolB = beginbmatrixb_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
$$boldsymbolM = beginbmatrixboldsymbolA\boldsymbolBendbmatrix = beginbmatrixa_1 & a_2 & a_3 & 0 & 0 & 0\0 & a_1 & 0 & a_2 & a_3 & 0\0 & 0 & a_1 & 0 & a_2 & a_3\b_1 & b_2 & b_3 & 0 & 0 & 0\0 & b_1 & 0 & b_2 & b_3 & 0\0 & 0 & b_1 & 0 & b_2 & b_3endbmatrix$$
I populated $boldsymbolA$, $boldsymbolB$ with several different combinations of values for $a_i, b_i$, e.g., $a_i$ = $[1, 2, 3]$, $b_i$ = $[4, 5, 6]$, to find the rank of $boldsymbolM$ using MATLAB.
It turns out that $Rank(boldsymbolA)=3$ and $Rank(boldsymbolB)=3$, but, $boldsymbolM$ is rank-deficient with $Rank(boldsymbolM)=5$. I was not able to find the dependent row in $boldsymbolM$. All rows look like linearly independent.
I tried to find the rank of M([r1, r2, r3, r4, r5],:) but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Could someone kindly help me how I can find the row in $boldsymbolM$ that is linearly dependent and makes $boldsymbolM$ rank deficient?
linear-algebra vector-spaces matlab matrix-rank
asked Aug 1 at 2:17
AFP
19811
edited Aug 1 at 2:23
asked Aug 1 at 2:17
AFP
19811
asked Aug 1 at 2:17
AFP
19811
asked Aug 1 at 2:17
AFP
19811
19811
add a comment |Â
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2 Answers
2
active
oldest
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up vote
3
down vote
accepted
I tried to find the rank of
M([r1, r2, r3, r4, r5],:)but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Not quite. It means that there are linear combination of the six rows (or the six columns) that sum to zero. Therefore, any one of the rows (or columns) can be expressed as a linear combination of the other 5 rows (or 5 columns).
Could someone kindly help me how I can find the row in M that is linearly dependent and makes M rank deficient?
As stated above, any one of the columns can be expressed as a nontrivial linear combination of the other 5 columns. ("Nontrivial" i.e. the coefficients of the linear combination are not all zero.) I will show how to find such a linear combination.
The reduced row echelon form $M$ is
$$ textrmrref(M) = left[ begin arraycccccc 1&0&0&0&0&-dfrac left( a_2
b_3-a_3b_2 right) ^2 left( b_2a_1-
b_1a_2 right) ^2\ 0&1&0&0&0&dfrac
left( b_3a_1-b_1a_3 right) left( a_2
b_3-a_3b_2 right) left( b_2a_1-
b_1a_2 right) ^2\ 0&0&1&0&0&-
dfrac a_2b_3-a_3b_2b_2a_1-
b_1a_2\ 0&0&0&1&0&-dfrac left( b_3
a_1-b_1a_3 right) ^2 left( b_2a_1
-b_1a_2 right) ^2\ 0&0&0&0&1&
dfrac b_3a_1-b_1a_3b_2a_1-
b_1a_2\ 0&0&0&0&0&0end array right] $$
(I computed this using the Maple software.) The fact that the last row is all zero implies that the $M$ is rank-deficient.
Moreover, the $textrmrref(M)$ implies that
$$ vecx = left[ begin arrayc dfrac left( a_2b_3-a_3
b_2 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3
a_1-b_1a_3 right) left( a_2b_3-a_3
b_2 right) s left( b_2a_1-b_1a_2
right) ^2\ dfrac left( a_2b_3
-a_3b_2 right) sb_2a_1-b_1a_2
\ dfrac left( b_3a_1-b_1
a_3 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3a_1
-b_1a_3 right) sb_2a_1-b_1a_2
\ send array right] $$
spans the null space of $M$: in other words, for any real number $s$, $$ M vecx = vec0. tag* $$
Equation (*) is equivalent with the following equation
$$ sum_n=1^6 x_nvecM_n = 0, tag$dagger$ $$
where "$vecM_n$" denotes the $n$th column of the matrix $M$. Since $x_n$ is a function of $s$, simply substitute one nonzero value of $s$. Then, Equation ($dagger$) indicates a nontrivial linear combination of the column vectors of $M$ that sums to zero.
answered Aug 1 at 3:06
Kyle
1,079617
2
I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows.
– Theo Bendit
Aug 1 at 3:08
1
I didn't think to try that - the RREF of $A^top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1 at 3:10
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up vote
3
down vote
I was hoping Kyle might mention it, but
$$b_1r_1 + b_2 r_2 + b_3 r_3 = a_1r_4 + a_2 r_5 + a_3 r_6,$$
which gives us our linear dependency.
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
Thank you. Simple and to the point.
– AFP
Aug 1 at 6:57
I wish I could mark two answers to my question.
– AFP
Aug 1 at 18:24
1
Keep Kyle's. He shows more or less how I got to this conclusion. :-)
– Theo Bendit
Aug 1 at 23:57
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
I tried to find the rank of
M([r1, r2, r3, r4, r5],:)but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Not quite. It means that there are linear combination of the six rows (or the six columns) that sum to zero. Therefore, any one of the rows (or columns) can be expressed as a linear combination of the other 5 rows (or 5 columns).
Could someone kindly help me how I can find the row in M that is linearly dependent and makes M rank deficient?
As stated above, any one of the columns can be expressed as a nontrivial linear combination of the other 5 columns. ("Nontrivial" i.e. the coefficients of the linear combination are not all zero.) I will show how to find such a linear combination.
The reduced row echelon form $M$ is
$$ textrmrref(M) = left[ begin arraycccccc 1&0&0&0&0&-dfrac left( a_2
b_3-a_3b_2 right) ^2 left( b_2a_1-
b_1a_2 right) ^2\ 0&1&0&0&0&dfrac
left( b_3a_1-b_1a_3 right) left( a_2
b_3-a_3b_2 right) left( b_2a_1-
b_1a_2 right) ^2\ 0&0&1&0&0&-
dfrac a_2b_3-a_3b_2b_2a_1-
b_1a_2\ 0&0&0&1&0&-dfrac left( b_3
a_1-b_1a_3 right) ^2 left( b_2a_1
-b_1a_2 right) ^2\ 0&0&0&0&1&
dfrac b_3a_1-b_1a_3b_2a_1-
b_1a_2\ 0&0&0&0&0&0end array right] $$
(I computed this using the Maple software.) The fact that the last row is all zero implies that the $M$ is rank-deficient.
Moreover, the $textrmrref(M)$ implies that
$$ vecx = left[ begin arrayc dfrac left( a_2b_3-a_3
b_2 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3
a_1-b_1a_3 right) left( a_2b_3-a_3
b_2 right) s left( b_2a_1-b_1a_2
right) ^2\ dfrac left( a_2b_3
-a_3b_2 right) sb_2a_1-b_1a_2
\ dfrac left( b_3a_1-b_1
a_3 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3a_1
-b_1a_3 right) sb_2a_1-b_1a_2
\ send array right] $$
spans the null space of $M$: in other words, for any real number $s$, $$ M vecx = vec0. tag* $$
Equation (*) is equivalent with the following equation
$$ sum_n=1^6 x_nvecM_n = 0, tag$dagger$ $$
where "$vecM_n$" denotes the $n$th column of the matrix $M$. Since $x_n$ is a function of $s$, simply substitute one nonzero value of $s$. Then, Equation ($dagger$) indicates a nontrivial linear combination of the column vectors of $M$ that sums to zero.
answered Aug 1 at 3:06
Kyle
1,079617
2
I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows.
– Theo Bendit
Aug 1 at 3:08
1
I didn't think to try that - the RREF of $A^top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1 at 3:10
add a comment |Â
up vote
3
down vote
accepted
I tried to find the rank of
M([r1, r2, r3, r4, r5],:)but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Not quite. It means that there are linear combination of the six rows (or the six columns) that sum to zero. Therefore, any one of the rows (or columns) can be expressed as a linear combination of the other 5 rows (or 5 columns).
Could someone kindly help me how I can find the row in M that is linearly dependent and makes M rank deficient?
As stated above, any one of the columns can be expressed as a nontrivial linear combination of the other 5 columns. ("Nontrivial" i.e. the coefficients of the linear combination are not all zero.) I will show how to find such a linear combination.
The reduced row echelon form $M$ is
$$ textrmrref(M) = left[ begin arraycccccc 1&0&0&0&0&-dfrac left( a_2
b_3-a_3b_2 right) ^2 left( b_2a_1-
b_1a_2 right) ^2\ 0&1&0&0&0&dfrac
left( b_3a_1-b_1a_3 right) left( a_2
b_3-a_3b_2 right) left( b_2a_1-
b_1a_2 right) ^2\ 0&0&1&0&0&-
dfrac a_2b_3-a_3b_2b_2a_1-
b_1a_2\ 0&0&0&1&0&-dfrac left( b_3
a_1-b_1a_3 right) ^2 left( b_2a_1
-b_1a_2 right) ^2\ 0&0&0&0&1&
dfrac b_3a_1-b_1a_3b_2a_1-
b_1a_2\ 0&0&0&0&0&0end array right] $$
(I computed this using the Maple software.) The fact that the last row is all zero implies that the $M$ is rank-deficient.
Moreover, the $textrmrref(M)$ implies that
$$ vecx = left[ begin arrayc dfrac left( a_2b_3-a_3
b_2 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3
a_1-b_1a_3 right) left( a_2b_3-a_3
b_2 right) s left( b_2a_1-b_1a_2
right) ^2\ dfrac left( a_2b_3
-a_3b_2 right) sb_2a_1-b_1a_2
\ dfrac left( b_3a_1-b_1
a_3 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3a_1
-b_1a_3 right) sb_2a_1-b_1a_2
\ send array right] $$
spans the null space of $M$: in other words, for any real number $s$, $$ M vecx = vec0. tag* $$
Equation (*) is equivalent with the following equation
$$ sum_n=1^6 x_nvecM_n = 0, tag$dagger$ $$
where "$vecM_n$" denotes the $n$th column of the matrix $M$. Since $x_n$ is a function of $s$, simply substitute one nonzero value of $s$. Then, Equation ($dagger$) indicates a nontrivial linear combination of the column vectors of $M$ that sums to zero.
answered Aug 1 at 3:06
Kyle
1,079617
2
I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows.
– Theo Bendit
Aug 1 at 3:08
1
I didn't think to try that - the RREF of $A^top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1 at 3:10
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
I tried to find the rank of
M([r1, r2, r3, r4, r5],:)but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Not quite. It means that there are linear combination of the six rows (or the six columns) that sum to zero. Therefore, any one of the rows (or columns) can be expressed as a linear combination of the other 5 rows (or 5 columns).
Could someone kindly help me how I can find the row in M that is linearly dependent and makes M rank deficient?
As stated above, any one of the columns can be expressed as a nontrivial linear combination of the other 5 columns. ("Nontrivial" i.e. the coefficients of the linear combination are not all zero.) I will show how to find such a linear combination.
The reduced row echelon form $M$ is
$$ textrmrref(M) = left[ begin arraycccccc 1&0&0&0&0&-dfrac left( a_2
b_3-a_3b_2 right) ^2 left( b_2a_1-
b_1a_2 right) ^2\ 0&1&0&0&0&dfrac
left( b_3a_1-b_1a_3 right) left( a_2
b_3-a_3b_2 right) left( b_2a_1-
b_1a_2 right) ^2\ 0&0&1&0&0&-
dfrac a_2b_3-a_3b_2b_2a_1-
b_1a_2\ 0&0&0&1&0&-dfrac left( b_3
a_1-b_1a_3 right) ^2 left( b_2a_1
-b_1a_2 right) ^2\ 0&0&0&0&1&
dfrac b_3a_1-b_1a_3b_2a_1-
b_1a_2\ 0&0&0&0&0&0end array right] $$
(I computed this using the Maple software.) The fact that the last row is all zero implies that the $M$ is rank-deficient.
Moreover, the $textrmrref(M)$ implies that
$$ vecx = left[ begin arrayc dfrac left( a_2b_3-a_3
b_2 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3
a_1-b_1a_3 right) left( a_2b_3-a_3
b_2 right) s left( b_2a_1-b_1a_2
right) ^2\ dfrac left( a_2b_3
-a_3b_2 right) sb_2a_1-b_1a_2
\ dfrac left( b_3a_1-b_1
a_3 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3a_1
-b_1a_3 right) sb_2a_1-b_1a_2
\ send array right] $$
spans the null space of $M$: in other words, for any real number $s$, $$ M vecx = vec0. tag* $$
Equation (*) is equivalent with the following equation
$$ sum_n=1^6 x_nvecM_n = 0, tag$dagger$ $$
where "$vecM_n$" denotes the $n$th column of the matrix $M$. Since $x_n$ is a function of $s$, simply substitute one nonzero value of $s$. Then, Equation ($dagger$) indicates a nontrivial linear combination of the column vectors of $M$ that sums to zero.
answered Aug 1 at 3:06
Kyle
1,079617
I tried to find the rank of
M([r1, r2, r3, r4, r5],:)but no matter what combination of 5 rows I pick, the rank is 5 which implies that any 5 rows combinations are linearly independent. Does this mean the 6th row is a combination of other 5 rows?
Not quite. It means that there are linear combination of the six rows (or the six columns) that sum to zero. Therefore, any one of the rows (or columns) can be expressed as a linear combination of the other 5 rows (or 5 columns).
Could someone kindly help me how I can find the row in M that is linearly dependent and makes M rank deficient?
As stated above, any one of the columns can be expressed as a nontrivial linear combination of the other 5 columns. ("Nontrivial" i.e. the coefficients of the linear combination are not all zero.) I will show how to find such a linear combination.
The reduced row echelon form $M$ is
$$ textrmrref(M) = left[ begin arraycccccc 1&0&0&0&0&-dfrac left( a_2
b_3-a_3b_2 right) ^2 left( b_2a_1-
b_1a_2 right) ^2\ 0&1&0&0&0&dfrac
left( b_3a_1-b_1a_3 right) left( a_2
b_3-a_3b_2 right) left( b_2a_1-
b_1a_2 right) ^2\ 0&0&1&0&0&-
dfrac a_2b_3-a_3b_2b_2a_1-
b_1a_2\ 0&0&0&1&0&-dfrac left( b_3
a_1-b_1a_3 right) ^2 left( b_2a_1
-b_1a_2 right) ^2\ 0&0&0&0&1&
dfrac b_3a_1-b_1a_3b_2a_1-
b_1a_2\ 0&0&0&0&0&0end array right] $$
(I computed this using the Maple software.) The fact that the last row is all zero implies that the $M$ is rank-deficient.
Moreover, the $textrmrref(M)$ implies that
$$ vecx = left[ begin arrayc dfrac left( a_2b_3-a_3
b_2 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3
a_1-b_1a_3 right) left( a_2b_3-a_3
b_2 right) s left( b_2a_1-b_1a_2
right) ^2\ dfrac left( a_2b_3
-a_3b_2 right) sb_2a_1-b_1a_2
\ dfrac left( b_3a_1-b_1
a_3 right) ^2s left( b_2a_1-b_1a_2
right) ^2\ -dfrac left( b_3a_1
-b_1a_3 right) sb_2a_1-b_1a_2
\ send array right] $$
spans the null space of $M$: in other words, for any real number $s$, $$ M vecx = vec0. tag* $$
Equation (*) is equivalent with the following equation
$$ sum_n=1^6 x_nvecM_n = 0, tag$dagger$ $$
where "$vecM_n$" denotes the $n$th column of the matrix $M$. Since $x_n$ is a function of $s$, simply substitute one nonzero value of $s$. Then, Equation ($dagger$) indicates a nontrivial linear combination of the column vectors of $M$ that sums to zero.
answered Aug 1 at 3:06
Kyle
1,079617
answered Aug 1 at 3:06
Kyle
1,079617
answered Aug 1 at 3:06
Kyle
1,079617
answered Aug 1 at 3:06
Kyle
1,079617
1,079617
2
I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows.
– Theo Bendit
Aug 1 at 3:08
1
I didn't think to try that - the RREF of $A^top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1 at 3:10
add a comment |Â
2
I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows.
– Theo Bendit
Aug 1 at 3:08
1
I didn't think to try that - the RREF of $A^top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1 at 3:10
2
2
I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows.
– Theo Bendit
Aug 1 at 3:08
I was just in the middle of doing the same thing! Try looking at the RREF of the transpose, for a much nicer linear combination of rows.
– Theo Bendit
Aug 1 at 3:08
1
1
I didn't think to try that - the RREF of $A^top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1 at 3:10
I didn't think to try that - the RREF of $A^top$ does look much nicer. Thank you @TheoBendit!
– Kyle
Aug 1 at 3:10
add a comment |Â
up vote
3
down vote
I was hoping Kyle might mention it, but
$$b_1r_1 + b_2 r_2 + b_3 r_3 = a_1r_4 + a_2 r_5 + a_3 r_6,$$
which gives us our linear dependency.
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
Thank you. Simple and to the point.
– AFP
Aug 1 at 6:57
I wish I could mark two answers to my question.
– AFP
Aug 1 at 18:24
1
Keep Kyle's. He shows more or less how I got to this conclusion. :-)
– Theo Bendit
Aug 1 at 23:57
add a comment |Â
up vote
3
down vote
I was hoping Kyle might mention it, but
$$b_1r_1 + b_2 r_2 + b_3 r_3 = a_1r_4 + a_2 r_5 + a_3 r_6,$$
which gives us our linear dependency.
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
Thank you. Simple and to the point.
– AFP
Aug 1 at 6:57
I wish I could mark two answers to my question.
– AFP
Aug 1 at 18:24
1
Keep Kyle's. He shows more or less how I got to this conclusion. :-)
– Theo Bendit
Aug 1 at 23:57
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I was hoping Kyle might mention it, but
$$b_1r_1 + b_2 r_2 + b_3 r_3 = a_1r_4 + a_2 r_5 + a_3 r_6,$$
which gives us our linear dependency.
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
I was hoping Kyle might mention it, but
$$b_1r_1 + b_2 r_2 + b_3 r_3 = a_1r_4 + a_2 r_5 + a_3 r_6,$$
which gives us our linear dependency.
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
answered Aug 1 at 3:42
Theo Bendit
11.7k1841
11.7k1841
Thank you. Simple and to the point.
– AFP
Aug 1 at 6:57
I wish I could mark two answers to my question.
– AFP
Aug 1 at 18:24
1
Keep Kyle's. He shows more or less how I got to this conclusion. :-)
– Theo Bendit
Aug 1 at 23:57
add a comment |Â
Thank you. Simple and to the point.
– AFP
Aug 1 at 6:57
I wish I could mark two answers to my question.
– AFP
Aug 1 at 18:24
1
Keep Kyle's. He shows more or less how I got to this conclusion. :-)
– Theo Bendit
Aug 1 at 23:57
Thank you. Simple and to the point.
– AFP
Aug 1 at 6:57
Thank you. Simple and to the point.
– AFP
Aug 1 at 6:57
I wish I could mark two answers to my question.
– AFP
Aug 1 at 18:24
I wish I could mark two answers to my question.
– AFP
Aug 1 at 18:24
1
1
Keep Kyle's. He shows more or less how I got to this conclusion. :-)
– Theo Bendit
Aug 1 at 23:57
Keep Kyle's. He shows more or less how I got to this conclusion. :-)
– Theo Bendit
Aug 1 at 23:57
add a comment |Â
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