Finding matrix whose null space is spanned by two given vectors

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Let $ K = SP (-5,8,14,0),(-1,4,2,4) $

So based on the solutions given here and here it should be same as finding orthogonal vectors so $ (x,y,z,t) cdot (-5,8,14,0) = 0 \ (x,y,z,t) cdot (-1,4,2,4) = 0 $

and then I find those four vectors $ (frac103,0,1,0),(-frac83,0,0,1),(0,frac13,1,0),(0,-frac53,0,1) $

setting those vectors in a matrix isn't the solution though... Im not sure what am I missing?

linear-algebra orthogonality

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edited Jul 22 at 14:01

Rodrigo de Azevedo

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asked Jul 22 at 10:58

bm1125

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Let $ K = SP (-5,8,14,0),(-1,4,2,4) $

So based on the solutions given here and here it should be same as finding orthogonal vectors so $ (x,y,z,t) cdot (-5,8,14,0) = 0 \ (x,y,z,t) cdot (-1,4,2,4) = 0 $

and then I find those four vectors $ (frac103,0,1,0),(-frac83,0,0,1),(0,frac13,1,0),(0,-frac53,0,1) $

setting those vectors in a matrix isn't the solution though... Im not sure what am I missing?

linear-algebra orthogonality

share|cite|improve this question

edited Jul 22 at 14:01

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 10:58

bm1125

1216

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Let $ K = SP (-5,8,14,0),(-1,4,2,4) $

So based on the solutions given here and here it should be same as finding orthogonal vectors so $ (x,y,z,t) cdot (-5,8,14,0) = 0 \ (x,y,z,t) cdot (-1,4,2,4) = 0 $

and then I find those four vectors $ (frac103,0,1,0),(-frac83,0,0,1),(0,frac13,1,0),(0,-frac53,0,1) $

setting those vectors in a matrix isn't the solution though... Im not sure what am I missing?

linear-algebra orthogonality

share|cite|improve this question

edited Jul 22 at 14:01

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 10:58

bm1125

1216

Let $ K = SP (-5,8,14,0),(-1,4,2,4) $

So based on the solutions given here and here it should be same as finding orthogonal vectors so $ (x,y,z,t) cdot (-5,8,14,0) = 0 \ (x,y,z,t) cdot (-1,4,2,4) = 0 $

and then I find those four vectors $ (frac103,0,1,0),(-frac83,0,0,1),(0,frac13,1,0),(0,-frac53,0,1) $

setting those vectors in a matrix isn't the solution though... Im not sure what am I missing?

linear-algebra orthogonality

share|cite|improve this question

edited Jul 22 at 14:01

Rodrigo de Azevedo

12.6k41751

asked Jul 22 at 10:58

bm1125

1216

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 14:01

Rodrigo de Azevedo

12.6k41751

edited Jul 22 at 14:01

Rodrigo de Azevedo

12.6k41751

edited Jul 22 at 14:01

Rodrigo de Azevedo

12.6k41751

12.6k41751

asked Jul 22 at 10:58

bm1125

1216

asked Jul 22 at 10:58

bm1125

1216

asked Jul 22 at 10:58

bm1125

1216

1216

add a comment | 

add a comment | 

1 Answer
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A simple way to do this: use elementary column operations to transform the matrix $$beginbmatrix -5 & -1 \ 8 & 4\ 14 & 2\ 0&4 endbmatrix$$ into 'reverse' echelon form: $$beginbmatrix-frac514 & -frac114\frac47 & frac57\1&0\0&1endbmatrix.$$ Notice that this matrix is in the form $$beginbmatrix -F \ I endbmatrix,$$ and so if we multiply it by $$beginbmatrix I & F \ 0 & 0 endbmatrix,$$ (where the 0's in this case can be any amount of rows with zero entries) the result will be the zero matrix, which is what we want. So depending on how many rows we want in our resulting matrix we can select as many zero rows at the bottom as we want. It seems as if you were aiming for a $4 times 4$ matrix so let's add two rows of zero...so therefore $$beginbmatrix 1 & 0 & frac514 & frac114\0&1&-frac47 & -frac57\0&0&0&0\0&0&0&0 endbmatrix$$ would work. Notics that this matrix is in row reduced echelon form, and so you can left multiply it by any elementary matrix to get other matrices that would have the same null space. The theory for the above, you can find in the following article: Row reduction of a matrix and $A=CaB$ by Steven L Lee and Gilbert Strang.

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answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

A simple way to do this: use elementary column operations to transform the matrix $$beginbmatrix -5 & -1 \ 8 & 4\ 14 & 2\ 0&4 endbmatrix$$ into 'reverse' echelon form: $$beginbmatrix-frac514 & -frac114\frac47 & frac57\1&0\0&1endbmatrix.$$ Notice that this matrix is in the form $$beginbmatrix -F \ I endbmatrix,$$ and so if we multiply it by $$beginbmatrix I & F \ 0 & 0 endbmatrix,$$ (where the 0's in this case can be any amount of rows with zero entries) the result will be the zero matrix, which is what we want. So depending on how many rows we want in our resulting matrix we can select as many zero rows at the bottom as we want. It seems as if you were aiming for a $4 times 4$ matrix so let's add two rows of zero...so therefore $$beginbmatrix 1 & 0 & frac514 & frac114\0&1&-frac47 & -frac57\0&0&0&0\0&0&0&0 endbmatrix$$ would work. Notics that this matrix is in row reduced echelon form, and so you can left multiply it by any elementary matrix to get other matrices that would have the same null space. The theory for the above, you can find in the following article: Row reduction of a matrix and $A=CaB$ by Steven L Lee and Gilbert Strang.

share|cite|improve this answer

answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

add a comment | 

up vote
2
down vote



accepted

A simple way to do this: use elementary column operations to transform the matrix $$beginbmatrix -5 & -1 \ 8 & 4\ 14 & 2\ 0&4 endbmatrix$$ into 'reverse' echelon form: $$beginbmatrix-frac514 & -frac114\frac47 & frac57\1&0\0&1endbmatrix.$$ Notice that this matrix is in the form $$beginbmatrix -F \ I endbmatrix,$$ and so if we multiply it by $$beginbmatrix I & F \ 0 & 0 endbmatrix,$$ (where the 0's in this case can be any amount of rows with zero entries) the result will be the zero matrix, which is what we want. So depending on how many rows we want in our resulting matrix we can select as many zero rows at the bottom as we want. It seems as if you were aiming for a $4 times 4$ matrix so let's add two rows of zero...so therefore $$beginbmatrix 1 & 0 & frac514 & frac114\0&1&-frac47 & -frac57\0&0&0&0\0&0&0&0 endbmatrix$$ would work. Notics that this matrix is in row reduced echelon form, and so you can left multiply it by any elementary matrix to get other matrices that would have the same null space. The theory for the above, you can find in the following article: Row reduction of a matrix and $A=CaB$ by Steven L Lee and Gilbert Strang.

share|cite|improve this answer

answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

A simple way to do this: use elementary column operations to transform the matrix $$beginbmatrix -5 & -1 \ 8 & 4\ 14 & 2\ 0&4 endbmatrix$$ into 'reverse' echelon form: $$beginbmatrix-frac514 & -frac114\frac47 & frac57\1&0\0&1endbmatrix.$$ Notice that this matrix is in the form $$beginbmatrix -F \ I endbmatrix,$$ and so if we multiply it by $$beginbmatrix I & F \ 0 & 0 endbmatrix,$$ (where the 0's in this case can be any amount of rows with zero entries) the result will be the zero matrix, which is what we want. So depending on how many rows we want in our resulting matrix we can select as many zero rows at the bottom as we want. It seems as if you were aiming for a $4 times 4$ matrix so let's add two rows of zero...so therefore $$beginbmatrix 1 & 0 & frac514 & frac114\0&1&-frac47 & -frac57\0&0&0&0\0&0&0&0 endbmatrix$$ would work. Notics that this matrix is in row reduced echelon form, and so you can left multiply it by any elementary matrix to get other matrices that would have the same null space. The theory for the above, you can find in the following article: Row reduction of a matrix and $A=CaB$ by Steven L Lee and Gilbert Strang.

share|cite|improve this answer

answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

A simple way to do this: use elementary column operations to transform the matrix $$beginbmatrix -5 & -1 \ 8 & 4\ 14 & 2\ 0&4 endbmatrix$$ into 'reverse' echelon form: $$beginbmatrix-frac514 & -frac114\frac47 & frac57\1&0\0&1endbmatrix.$$ Notice that this matrix is in the form $$beginbmatrix -F \ I endbmatrix,$$ and so if we multiply it by $$beginbmatrix I & F \ 0 & 0 endbmatrix,$$ (where the 0's in this case can be any amount of rows with zero entries) the result will be the zero matrix, which is what we want. So depending on how many rows we want in our resulting matrix we can select as many zero rows at the bottom as we want. It seems as if you were aiming for a $4 times 4$ matrix so let's add two rows of zero...so therefore $$beginbmatrix 1 & 0 & frac514 & frac114\0&1&-frac47 & -frac57\0&0&0&0\0&0&0&0 endbmatrix$$ would work. Notics that this matrix is in row reduced echelon form, and so you can left multiply it by any elementary matrix to get other matrices that would have the same null space. The theory for the above, you can find in the following article: Row reduction of a matrix and $A=CaB$ by Steven L Lee and Gilbert Strang.

share|cite|improve this answer

answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

answered Jul 22 at 13:16

Christiaan Hattingh

3,782921

3,782921

add a comment | 

add a comment | 

 

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