Linear dependent, and linear independent, linear Algebra
$begingroup$
Given vector v = [
1
2
], write some vector w that is linearly dependent with v, and some other
vector r that is linearly independent of v.
linear-algebra
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
$endgroup$
add a comment |
$begingroup$
Given vector v = [
1
2
], write some vector w that is linearly dependent with v, and some other
vector r that is linearly independent of v.
linear-algebra
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
$endgroup$
1
$begingroup$
Welcome to MSE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
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– platty
Dec 1 '18 at 7:14
add a comment |
$begingroup$
Given vector v = [
1
2
], write some vector w that is linearly dependent with v, and some other
vector r that is linearly independent of v.
linear-algebra
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
$endgroup$
Given vector v = [
1
2
], write some vector w that is linearly dependent with v, and some other
vector r that is linearly independent of v.
linear-algebra
linear-algebra
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
asked Dec 1 '18 at 7:12
Yosy bYosy b
1
1
1
$begingroup$
Welcome to MSE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– platty
Dec 1 '18 at 7:14
add a comment |
1
$begingroup$
Welcome to MSE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– platty
Dec 1 '18 at 7:14
1
1
$begingroup$
Welcome to MSE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– platty
Dec 1 '18 at 7:14
$begingroup$
Welcome to MSE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
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– platty
Dec 1 '18 at 7:14
add a comment |
1 Answer
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By definition, $n$ vectors are linearly dependent if there exist constants/scalars $c_1, c_2, dots, c_n $ s.t. $c_1 alpha_1 + c_2 alpha_2 + dots c_n alpha_n = 0$ where $c_i's$ are not all zero, i.e. there exists some non-zero $c_i$.
So, for $n = 2$ we want $c_1 alpha_1 + c_2 alpha_2 = 0$. Atleast one of $c_1$ and $c_2$ is non-zero, so divide the equation by that non-zero scalar to get $alpha_1 = c alpha_2$
And, by definition vectors that are not linearly dependent are linearly independent.
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
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add a comment |
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$begingroup$
By definition, $n$ vectors are linearly dependent if there exist constants/scalars $c_1, c_2, dots, c_n $ s.t. $c_1 alpha_1 + c_2 alpha_2 + dots c_n alpha_n = 0$ where $c_i's$ are not all zero, i.e. there exists some non-zero $c_i$.
So, for $n = 2$ we want $c_1 alpha_1 + c_2 alpha_2 = 0$. Atleast one of $c_1$ and $c_2$ is non-zero, so divide the equation by that non-zero scalar to get $alpha_1 = c alpha_2$
And, by definition vectors that are not linearly dependent are linearly independent.
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
$endgroup$
add a comment |
$begingroup$
By definition, $n$ vectors are linearly dependent if there exist constants/scalars $c_1, c_2, dots, c_n $ s.t. $c_1 alpha_1 + c_2 alpha_2 + dots c_n alpha_n = 0$ where $c_i's$ are not all zero, i.e. there exists some non-zero $c_i$.
So, for $n = 2$ we want $c_1 alpha_1 + c_2 alpha_2 = 0$. Atleast one of $c_1$ and $c_2$ is non-zero, so divide the equation by that non-zero scalar to get $alpha_1 = c alpha_2$
And, by definition vectors that are not linearly dependent are linearly independent.
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
$endgroup$
add a comment |
$begingroup$
By definition, $n$ vectors are linearly dependent if there exist constants/scalars $c_1, c_2, dots, c_n $ s.t. $c_1 alpha_1 + c_2 alpha_2 + dots c_n alpha_n = 0$ where $c_i's$ are not all zero, i.e. there exists some non-zero $c_i$.
So, for $n = 2$ we want $c_1 alpha_1 + c_2 alpha_2 = 0$. Atleast one of $c_1$ and $c_2$ is non-zero, so divide the equation by that non-zero scalar to get $alpha_1 = c alpha_2$
And, by definition vectors that are not linearly dependent are linearly independent.
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
$endgroup$
By definition, $n$ vectors are linearly dependent if there exist constants/scalars $c_1, c_2, dots, c_n $ s.t. $c_1 alpha_1 + c_2 alpha_2 + dots c_n alpha_n = 0$ where $c_i's$ are not all zero, i.e. there exists some non-zero $c_i$.
So, for $n = 2$ we want $c_1 alpha_1 + c_2 alpha_2 = 0$. Atleast one of $c_1$ and $c_2$ is non-zero, so divide the equation by that non-zero scalar to get $alpha_1 = c alpha_2$
And, by definition vectors that are not linearly dependent are linearly independent.
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
answered Dec 1 '18 at 7:23
ab123ab123
1,654423
1,654423
add a comment |
add a comment |
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$begingroup$
Welcome to MSE! You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an edit): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance.
$endgroup$
– platty
Dec 1 '18 at 7:14