Mixing
Let $v_1$, $v_2$, $v_3$ be a basis of the $mathbb{R}$-vector space $mathbb{R}^3$
$begingroup$
I'm not that good at math and would be very happy if you could give me some hints and so on
So my task is:
Let ${v_1, v_2, v_3}$ be a basis of the $mathbb{R}$-vector space $mathbb{R}^3$
show that ${v_1 + v_2, v_1 + v_3, v_2 + v_3}$ is also a base of $mathbb{R}^3$
let $a, b, c in mathbb{R}$, so that $a ≠ b ≠ c$; show that ${v_1 + v_2 + v_3, av_1 + bv_2 + cv_3, a^2v_1 + b^2v_2 + c^2v_3}$ is also a base of $mathbb{R}^3$
I will be gratefully for every help!
linear-algebra vector-spaces real-numbers invariant-subspace
edited Jan 13 at 11:26
egreg
181k1485203
$endgroup$
|
show 9 more comments
$begingroup$
I'm not that good at math and would be very happy if you could give me some hints and so on
So my task is:
Let ${v_1, v_2, v_3}$ be a basis of the $mathbb{R}$-vector space $mathbb{R}^3$
show that ${v_1 + v_2, v_1 + v_3, v_2 + v_3}$ is also a base of $mathbb{R}^3$
let $a, b, c in mathbb{R}$, so that $a ≠ b ≠ c$; show that ${v_1 + v_2 + v_3, av_1 + bv_2 + cv_3, a^2v_1 + b^2v_2 + c^2v_3}$ is also a base of $mathbb{R}^3$
I will be gratefully for every help!
linear-algebra vector-spaces real-numbers invariant-subspace
edited Jan 13 at 11:26
egreg
181k1485203
$endgroup$
1
$begingroup$
Do you know the definition of basis?
$endgroup$
– Jerry
Jan 13 at 10:53
$begingroup$
@Jerry yeah, let V be a K-vector space. Then it means U ⊆ V a base of V, If U is a generator of V and linearly independent. The empty set ∅ is the base of the zero space {0}
$endgroup$
– user634297
Jan 13 at 11:08
$begingroup$
Did you study determinants?
$endgroup$
– Berci
Jan 13 at 11:17
$begingroup$
@Berci No, Not yet but I looked now a YouTube video how to calculate determinants, and the first matrix is then -2 correct ?
$endgroup$
– user634297
Jan 13 at 11:31
$begingroup$
Yes, it's $-2$ for that. The determinant of a 3x3 matrix is the 3d (signed) volume of the parallelepiped spanned by its column vectors, and if those really span 3d, then the volume is nonzero, while if the vectors lie in a plane, the volume is 0.
$endgroup$
– Berci
Jan 13 at 11:34
|
show 9 more comments
$begingroup$
I'm not that good at math and would be very happy if you could give me some hints and so on
So my task is:
Let ${v_1, v_2, v_3}$ be a basis of the $mathbb{R}$-vector space $mathbb{R}^3$
show that ${v_1 + v_2, v_1 + v_3, v_2 + v_3}$ is also a base of $mathbb{R}^3$
let $a, b, c in mathbb{R}$, so that $a ≠ b ≠ c$; show that ${v_1 + v_2 + v_3, av_1 + bv_2 + cv_3, a^2v_1 + b^2v_2 + c^2v_3}$ is also a base of $mathbb{R}^3$
I will be gratefully for every help!
linear-algebra vector-spaces real-numbers invariant-subspace
edited Jan 13 at 11:26
egreg
181k1485203
$endgroup$
I'm not that good at math and would be very happy if you could give me some hints and so on
So my task is:
Let ${v_1, v_2, v_3}$ be a basis of the $mathbb{R}$-vector space $mathbb{R}^3$
show that ${v_1 + v_2, v_1 + v_3, v_2 + v_3}$ is also a base of $mathbb{R}^3$
let $a, b, c in mathbb{R}$, so that $a ≠ b ≠ c$; show that ${v_1 + v_2 + v_3, av_1 + bv_2 + cv_3, a^2v_1 + b^2v_2 + c^2v_3}$ is also a base of $mathbb{R}^3$
I will be gratefully for every help!
linear-algebra vector-spaces real-numbers invariant-subspace
linear-algebra vector-spaces real-numbers invariant-subspace
edited Jan 13 at 11:26
egreg
181k1485203
edited Jan 13 at 11:26
egreg
181k1485203
edited Jan 13 at 11:26
egreg
181k1485203
edited Jan 13 at 11:26
egreg
181k1485203
edited Jan 13 at 11:26
egreg
181k1485203
181k1485203
asked Jan 13 at 10:39
user634297
1
$begingroup$
Do you know the definition of basis?
$endgroup$
– Jerry
Jan 13 at 10:53
$begingroup$
@Jerry yeah, let V be a K-vector space. Then it means U ⊆ V a base of V, If U is a generator of V and linearly independent. The empty set ∅ is the base of the zero space {0}
$endgroup$
– user634297
Jan 13 at 11:08
$begingroup$
Did you study determinants?
$endgroup$
– Berci
Jan 13 at 11:17
$begingroup$
@Berci No, Not yet but I looked now a YouTube video how to calculate determinants, and the first matrix is then -2 correct ?
$endgroup$
– user634297
Jan 13 at 11:31
$begingroup$
Yes, it's $-2$ for that. The determinant of a 3x3 matrix is the 3d (signed) volume of the parallelepiped spanned by its column vectors, and if those really span 3d, then the volume is nonzero, while if the vectors lie in a plane, the volume is 0.
$endgroup$
– Berci
Jan 13 at 11:34
|
show 9 more comments
1
$begingroup$
Do you know the definition of basis?
$endgroup$
– Jerry
Jan 13 at 10:53
$begingroup$
@Jerry yeah, let V be a K-vector space. Then it means U ⊆ V a base of V, If U is a generator of V and linearly independent. The empty set ∅ is the base of the zero space {0}
$endgroup$
– user634297
Jan 13 at 11:08
$begingroup$
Did you study determinants?
$endgroup$
– Berci
Jan 13 at 11:17
$begingroup$
@Berci No, Not yet but I looked now a YouTube video how to calculate determinants, and the first matrix is then -2 correct ?
$endgroup$
– user634297
Jan 13 at 11:31
$begingroup$
Yes, it's $-2$ for that. The determinant of a 3x3 matrix is the 3d (signed) volume of the parallelepiped spanned by its column vectors, and if those really span 3d, then the volume is nonzero, while if the vectors lie in a plane, the volume is 0.
$endgroup$
– Berci
Jan 13 at 11:34
1
1
$begingroup$
Do you know the definition of basis?
$endgroup$
– Jerry
Jan 13 at 10:53
$begingroup$
Do you know the definition of basis?
$endgroup$
– Jerry
Jan 13 at 10:53
$begingroup$
@Jerry yeah, let V be a K-vector space. Then it means U ⊆ V a base of V, If U is a generator of V and linearly independent. The empty set ∅ is the base of the zero space {0}
$endgroup$
– user634297
Jan 13 at 11:08
$begingroup$
@Jerry yeah, let V be a K-vector space. Then it means U ⊆ V a base of V, If U is a generator of V and linearly independent. The empty set ∅ is the base of the zero space {0}
$endgroup$
– user634297
Jan 13 at 11:08
$begingroup$
Did you study determinants?
$endgroup$
– Berci
Jan 13 at 11:17
$begingroup$
Did you study determinants?
$endgroup$
– Berci
Jan 13 at 11:17
$begingroup$
@Berci No, Not yet but I looked now a YouTube video how to calculate determinants, and the first matrix is then -2 correct ?
$endgroup$
– user634297
Jan 13 at 11:31
$begingroup$
@Berci No, Not yet but I looked now a YouTube video how to calculate determinants, and the first matrix is then -2 correct ?
$endgroup$
– user634297
Jan 13 at 11:31
$begingroup$
Yes, it's $-2$ for that. The determinant of a 3x3 matrix is the 3d (signed) volume of the parallelepiped spanned by its column vectors, and if those really span 3d, then the volume is nonzero, while if the vectors lie in a plane, the volume is 0.
$endgroup$
– Berci
Jan 13 at 11:34
$begingroup$
Yes, it's $-2$ for that. The determinant of a 3x3 matrix is the 3d (signed) volume of the parallelepiped spanned by its column vectors, and if those really span 3d, then the volume is nonzero, while if the vectors lie in a plane, the volume is 0.
$endgroup$
– Berci
Jan 13 at 11:34
|
show 9 more comments
1 Answer
1
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oldest
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$begingroup$
A systematic way to solve it would be to arrange the given vectors in a matrix, with respect to the given basis $(v_1,v_2,v_3)$.
It's usually done by writing the coordinates of the vectors (w.r.t the given basis!) in the columns of the matrix.
Since we have $v_1+v_2={bf1}cdot v_1+{bf1}cdot v_2+{bf0}cdot v_3$, its coordinates w.r.t basis $(v_1,v_2,v_3)$ are $(1,1,0)$, and we will use it as a column. Similarly taken coordinates for the other two vectors, for part 1, we get the matrix
$$pmatrix{1&1&0\1&0&1\0&1&1}$$
For part 2., the matrix is
$$pmatrix{1&a&a^2\1&b&b^2\1&c&c^2}$$
To conclude that the columns of a square matrix form a basis, all we have to check is that its determinant is nonzero.
answered Jan 13 at 11:12
BerciBerci
60.8k23673
$endgroup$
add a comment |
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1 Answer
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$begingroup$
A systematic way to solve it would be to arrange the given vectors in a matrix, with respect to the given basis $(v_1,v_2,v_3)$.
It's usually done by writing the coordinates of the vectors (w.r.t the given basis!) in the columns of the matrix.
Since we have $v_1+v_2={bf1}cdot v_1+{bf1}cdot v_2+{bf0}cdot v_3$, its coordinates w.r.t basis $(v_1,v_2,v_3)$ are $(1,1,0)$, and we will use it as a column. Similarly taken coordinates for the other two vectors, for part 1, we get the matrix
$$pmatrix{1&1&0\1&0&1\0&1&1}$$
For part 2., the matrix is
$$pmatrix{1&a&a^2\1&b&b^2\1&c&c^2}$$
To conclude that the columns of a square matrix form a basis, all we have to check is that its determinant is nonzero.
answered Jan 13 at 11:12
BerciBerci
60.8k23673
$endgroup$
add a comment |
$begingroup$
A systematic way to solve it would be to arrange the given vectors in a matrix, with respect to the given basis $(v_1,v_2,v_3)$.
It's usually done by writing the coordinates of the vectors (w.r.t the given basis!) in the columns of the matrix.
Since we have $v_1+v_2={bf1}cdot v_1+{bf1}cdot v_2+{bf0}cdot v_3$, its coordinates w.r.t basis $(v_1,v_2,v_3)$ are $(1,1,0)$, and we will use it as a column. Similarly taken coordinates for the other two vectors, for part 1, we get the matrix
$$pmatrix{1&1&0\1&0&1\0&1&1}$$
For part 2., the matrix is
$$pmatrix{1&a&a^2\1&b&b^2\1&c&c^2}$$
To conclude that the columns of a square matrix form a basis, all we have to check is that its determinant is nonzero.
answered Jan 13 at 11:12
BerciBerci
60.8k23673
$endgroup$
add a comment |
$begingroup$
A systematic way to solve it would be to arrange the given vectors in a matrix, with respect to the given basis $(v_1,v_2,v_3)$.
It's usually done by writing the coordinates of the vectors (w.r.t the given basis!) in the columns of the matrix.
Since we have $v_1+v_2={bf1}cdot v_1+{bf1}cdot v_2+{bf0}cdot v_3$, its coordinates w.r.t basis $(v_1,v_2,v_3)$ are $(1,1,0)$, and we will use it as a column. Similarly taken coordinates for the other two vectors, for part 1, we get the matrix
$$pmatrix{1&1&0\1&0&1\0&1&1}$$
For part 2., the matrix is
$$pmatrix{1&a&a^2\1&b&b^2\1&c&c^2}$$
To conclude that the columns of a square matrix form a basis, all we have to check is that its determinant is nonzero.
answered Jan 13 at 11:12
BerciBerci
60.8k23673
$endgroup$
A systematic way to solve it would be to arrange the given vectors in a matrix, with respect to the given basis $(v_1,v_2,v_3)$.
It's usually done by writing the coordinates of the vectors (w.r.t the given basis!) in the columns of the matrix.
Since we have $v_1+v_2={bf1}cdot v_1+{bf1}cdot v_2+{bf0}cdot v_3$, its coordinates w.r.t basis $(v_1,v_2,v_3)$ are $(1,1,0)$, and we will use it as a column. Similarly taken coordinates for the other two vectors, for part 1, we get the matrix
$$pmatrix{1&1&0\1&0&1\0&1&1}$$
For part 2., the matrix is
$$pmatrix{1&a&a^2\1&b&b^2\1&c&c^2}$$
To conclude that the columns of a square matrix form a basis, all we have to check is that its determinant is nonzero.
answered Jan 13 at 11:12
BerciBerci
60.8k23673
answered Jan 13 at 11:12
BerciBerci
60.8k23673
answered Jan 13 at 11:12
BerciBerci
60.8k23673
answered Jan 13 at 11:12
BerciBerci
60.8k23673
60.8k23673
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1
$begingroup$
Do you know the definition of basis?
$endgroup$
– Jerry
Jan 13 at 10:53
$begingroup$
@Jerry yeah, let V be a K-vector space. Then it means U ⊆ V a base of V, If U is a generator of V and linearly independent. The empty set ∅ is the base of the zero space {0}
$endgroup$
– user634297
Jan 13 at 11:08
$begingroup$
Did you study determinants?
$endgroup$
– Berci
Jan 13 at 11:17
$begingroup$
@Berci No, Not yet but I looked now a YouTube video how to calculate determinants, and the first matrix is then -2 correct ?
$endgroup$
– user634297
Jan 13 at 11:31
$begingroup$
Yes, it's $-2$ for that. The determinant of a 3x3 matrix is the 3d (signed) volume of the parallelepiped spanned by its column vectors, and if those really span 3d, then the volume is nonzero, while if the vectors lie in a plane, the volume is 0.
$endgroup$
– Berci
Jan 13 at 11:34