Mixing
Decide with proof whether linearly independent over F
Consider the following vectors
w1=(1,1,1,1)
w2 = (1,0,1,0)
w3 = (1,1,2,2)
w4 = (-1,1,4,1)
Decide with proof whether the 4 vectors are linearly independent.
I am unsure of how to do this
analysis vector-spaces field-theory vectors
asked Nov 21 '18 at 18:43
user607735user607735
153
add a comment |
Consider the following vectors
w1=(1,1,1,1)
w2 = (1,0,1,0)
w3 = (1,1,2,2)
w4 = (-1,1,4,1)
Decide with proof whether the 4 vectors are linearly independent.
I am unsure of how to do this
analysis vector-spaces field-theory vectors
asked Nov 21 '18 at 18:43
user607735user607735
153
1
Are you familiar with the definition of linear independence?
– Anurag A
Nov 21 '18 at 18:45
To test for linear independence of $n$ vectors in $Bbb R^n$, you can simply form a matrix using those vectors and find the determinant of it. If the determinant is zero then they are dependent. If the determinant is nonzero then they are independent. This is an application of a more general result about independence of vectors and the rank of a matrix formed from those vectors, often seen via row reduction.
– JMoravitz
Nov 21 '18 at 19:17
add a comment |
Consider the following vectors
w1=(1,1,1,1)
w2 = (1,0,1,0)
w3 = (1,1,2,2)
w4 = (-1,1,4,1)
Decide with proof whether the 4 vectors are linearly independent.
I am unsure of how to do this
analysis vector-spaces field-theory vectors
asked Nov 21 '18 at 18:43
user607735user607735
153
Consider the following vectors
w1=(1,1,1,1)
w2 = (1,0,1,0)
w3 = (1,1,2,2)
w4 = (-1,1,4,1)
Decide with proof whether the 4 vectors are linearly independent.
I am unsure of how to do this
analysis vector-spaces field-theory vectors
analysis vector-spaces field-theory vectors
asked Nov 21 '18 at 18:43
user607735user607735
153
asked Nov 21 '18 at 18:43
user607735user607735
153
asked Nov 21 '18 at 18:43
user607735user607735
153
asked Nov 21 '18 at 18:43
user607735user607735
153
asked Nov 21 '18 at 18:43
user607735user607735
153
153
1
Are you familiar with the definition of linear independence?
– Anurag A
Nov 21 '18 at 18:45
To test for linear independence of $n$ vectors in $Bbb R^n$, you can simply form a matrix using those vectors and find the determinant of it. If the determinant is zero then they are dependent. If the determinant is nonzero then they are independent. This is an application of a more general result about independence of vectors and the rank of a matrix formed from those vectors, often seen via row reduction.
– JMoravitz
Nov 21 '18 at 19:17
add a comment |
1
Are you familiar with the definition of linear independence?
– Anurag A
Nov 21 '18 at 18:45
To test for linear independence of $n$ vectors in $Bbb R^n$, you can simply form a matrix using those vectors and find the determinant of it. If the determinant is zero then they are dependent. If the determinant is nonzero then they are independent. This is an application of a more general result about independence of vectors and the rank of a matrix formed from those vectors, often seen via row reduction.
– JMoravitz
Nov 21 '18 at 19:17
1
1
Are you familiar with the definition of linear independence?
– Anurag A
Nov 21 '18 at 18:45
Are you familiar with the definition of linear independence?
– Anurag A
Nov 21 '18 at 18:45
To test for linear independence of $n$ vectors in $Bbb R^n$, you can simply form a matrix using those vectors and find the determinant of it. If the determinant is zero then they are dependent. If the determinant is nonzero then they are independent. This is an application of a more general result about independence of vectors and the rank of a matrix formed from those vectors, often seen via row reduction.
– JMoravitz
Nov 21 '18 at 19:17
To test for linear independence of $n$ vectors in $Bbb R^n$, you can simply form a matrix using those vectors and find the determinant of it. If the determinant is zero then they are dependent. If the determinant is nonzero then they are independent. This is an application of a more general result about independence of vectors and the rank of a matrix formed from those vectors, often seen via row reduction.
– JMoravitz
Nov 21 '18 at 19:17
add a comment |
1 Answer
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To decide if the four vectors $(1,1,1,1)^T, (1,0,1,0)^T,(1,1,2,2)^T,(-1,1,4,1)^T$, you need to show that $$c_1pmatrix{1\1\1\1} + c_2pmatrix{1\0\1\0} + c_3pmatrix{1\1\2\2} + c_4pmatrix{-1\1\4\1} = pmatrix{0\0\0\0}$$ only has the trivial solution $c_1 = c_2 = c_3 = c_4 = 0$. The left-side can be rewritten to yield $$pmatrix{c_1 + c_2 + c_3 - c_4\ c_1 + c_3 + c_4\c_1 + c_2 + 2c_3 + 4c_4\ c_1 + 2c_3 + c_4} = pmatrix{0\0\0\0},$$ which we can write as a linear system $$pmatrix{1 & 1 & 1 & -1 & 0\ 1 & 0 & 1 & 1 & 0\1 & 1 & 2 & 4 & 0\1 & 0 & 2 & 1 & 0},$$ from which you can use elementary row operations to find the reduced-row echelon form to determine the solution.
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
add a comment |
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To decide if the four vectors $(1,1,1,1)^T, (1,0,1,0)^T,(1,1,2,2)^T,(-1,1,4,1)^T$, you need to show that $$c_1pmatrix{1\1\1\1} + c_2pmatrix{1\0\1\0} + c_3pmatrix{1\1\2\2} + c_4pmatrix{-1\1\4\1} = pmatrix{0\0\0\0}$$ only has the trivial solution $c_1 = c_2 = c_3 = c_4 = 0$. The left-side can be rewritten to yield $$pmatrix{c_1 + c_2 + c_3 - c_4\ c_1 + c_3 + c_4\c_1 + c_2 + 2c_3 + 4c_4\ c_1 + 2c_3 + c_4} = pmatrix{0\0\0\0},$$ which we can write as a linear system $$pmatrix{1 & 1 & 1 & -1 & 0\ 1 & 0 & 1 & 1 & 0\1 & 1 & 2 & 4 & 0\1 & 0 & 2 & 1 & 0},$$ from which you can use elementary row operations to find the reduced-row echelon form to determine the solution.
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
add a comment |
To decide if the four vectors $(1,1,1,1)^T, (1,0,1,0)^T,(1,1,2,2)^T,(-1,1,4,1)^T$, you need to show that $$c_1pmatrix{1\1\1\1} + c_2pmatrix{1\0\1\0} + c_3pmatrix{1\1\2\2} + c_4pmatrix{-1\1\4\1} = pmatrix{0\0\0\0}$$ only has the trivial solution $c_1 = c_2 = c_3 = c_4 = 0$. The left-side can be rewritten to yield $$pmatrix{c_1 + c_2 + c_3 - c_4\ c_1 + c_3 + c_4\c_1 + c_2 + 2c_3 + 4c_4\ c_1 + 2c_3 + c_4} = pmatrix{0\0\0\0},$$ which we can write as a linear system $$pmatrix{1 & 1 & 1 & -1 & 0\ 1 & 0 & 1 & 1 & 0\1 & 1 & 2 & 4 & 0\1 & 0 & 2 & 1 & 0},$$ from which you can use elementary row operations to find the reduced-row echelon form to determine the solution.
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
add a comment |
To decide if the four vectors $(1,1,1,1)^T, (1,0,1,0)^T,(1,1,2,2)^T,(-1,1,4,1)^T$, you need to show that $$c_1pmatrix{1\1\1\1} + c_2pmatrix{1\0\1\0} + c_3pmatrix{1\1\2\2} + c_4pmatrix{-1\1\4\1} = pmatrix{0\0\0\0}$$ only has the trivial solution $c_1 = c_2 = c_3 = c_4 = 0$. The left-side can be rewritten to yield $$pmatrix{c_1 + c_2 + c_3 - c_4\ c_1 + c_3 + c_4\c_1 + c_2 + 2c_3 + 4c_4\ c_1 + 2c_3 + c_4} = pmatrix{0\0\0\0},$$ which we can write as a linear system $$pmatrix{1 & 1 & 1 & -1 & 0\ 1 & 0 & 1 & 1 & 0\1 & 1 & 2 & 4 & 0\1 & 0 & 2 & 1 & 0},$$ from which you can use elementary row operations to find the reduced-row echelon form to determine the solution.
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
To decide if the four vectors $(1,1,1,1)^T, (1,0,1,0)^T,(1,1,2,2)^T,(-1,1,4,1)^T$, you need to show that $$c_1pmatrix{1\1\1\1} + c_2pmatrix{1\0\1\0} + c_3pmatrix{1\1\2\2} + c_4pmatrix{-1\1\4\1} = pmatrix{0\0\0\0}$$ only has the trivial solution $c_1 = c_2 = c_3 = c_4 = 0$. The left-side can be rewritten to yield $$pmatrix{c_1 + c_2 + c_3 - c_4\ c_1 + c_3 + c_4\c_1 + c_2 + 2c_3 + 4c_4\ c_1 + 2c_3 + c_4} = pmatrix{0\0\0\0},$$ which we can write as a linear system $$pmatrix{1 & 1 & 1 & -1 & 0\ 1 & 0 & 1 & 1 & 0\1 & 1 & 2 & 4 & 0\1 & 0 & 2 & 1 & 0},$$ from which you can use elementary row operations to find the reduced-row echelon form to determine the solution.
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
answered Nov 21 '18 at 18:53
Decaf-MathDecaf-Math
3,209825
3,209825
add a comment |
add a comment |
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1
Are you familiar with the definition of linear independence?
– Anurag A
Nov 21 '18 at 18:45
To test for linear independence of $n$ vectors in $Bbb R^n$, you can simply form a matrix using those vectors and find the determinant of it. If the determinant is zero then they are dependent. If the determinant is nonzero then they are independent. This is an application of a more general result about independence of vectors and the rank of a matrix formed from those vectors, often seen via row reduction.
– JMoravitz
Nov 21 '18 at 19:17