Dimension of the vector space spanned by ${e,f,g,..}$ equals the maximal number of linearly independent...
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I am reading Gelfand's Lectures on Linear Algebra.
He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.
He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.
My approach:
Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.
Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.
Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.
Is this argument correct?
linear-algebra
asked Nov 18 at 22:23
trynalearn
631314
add a comment |
up vote
0
down vote
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I am reading Gelfand's Lectures on Linear Algebra.
He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.
He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.
My approach:
Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.
Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.
Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.
Is this argument correct?
linear-algebra
asked Nov 18 at 22:23
trynalearn
631314
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading Gelfand's Lectures on Linear Algebra.
He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.
He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.
My approach:
Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.
Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.
Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.
Is this argument correct?
linear-algebra
asked Nov 18 at 22:23
trynalearn
631314
I am reading Gelfand's Lectures on Linear Algebra.
He defines a vector space to have dimension $n$ if there exists a linearly independent set of $n$ vectors, and if every set of $n+1$ vectors is linearly dependent.
He asks to prove that the dimension of the vector space generated by ${e,f,g,..}$ equals the maximal number of linearly independent vectors in ${e,f,g...}$.
My approach:
Let $N$ be the maximal number of linearly independent vectors in ${e,f,g..}$.
Then since ${e,f,g..}$ contains $N$ linearly independent vectors its dimension must be greater than or equal to $N$.
Suppose it were strictly greater than $N$. Then there must exist a linearly independent set of $N+1$ vectors, contradicting the fact that the maximal number of linearly independent vectors among ${e,f,g..}$ was $N$.
Is this argument correct?
linear-algebra
linear-algebra
asked Nov 18 at 22:23
trynalearn
631314
asked Nov 18 at 22:23
trynalearn
631314
asked Nov 18 at 22:23
trynalearn
631314
asked Nov 18 at 22:23
trynalearn
631314
asked Nov 18 at 22:23
trynalearn
631314
631314
add a comment |
add a comment |
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