Need some help for clarifying an example on infinite-dimensional vector space
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In the book Linear Algebra Done Right, example 2.14 said that $P_m(mathbb{F})$ is a finite-dimensional vector space for each non-negative integer $m$.
$P_m(mathbb{F})$ is defined as "For $m$ a nonnegative integer, $P_m(mathbb{F})$ denotes the set of all polynomials with coefficients in $mathbb{F}$ and degree at most $m$"
Then in example 2.16, it showed that $P(mathbb{F})$ is infinite-dimensional.
Consider any list of elements of $P(mathbb{F})$. Let $m$ denote the highest degree of the polynomials inn the list. Then every polynomial in the span of this list has degree at most $m$. Thus $z^{m+1}$ is not in the span of our list. Hence no list spans $P(mathbb{F})$. Thus $P(mathbb{F})$ is infinite dimensional.
I am a bit confused on why $P_m(mathbb{F})$ is finite dimensional, and $P(mathbb{F})$ is infinite dimensional.
Is it because $P_m(mathbb{F})$ is just talking about polynomials with degree at most $m$, while $P(mathbb{F})$ is talking about polynomials with degree that can be higher than $m$?
linear-algebra
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
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add a comment |
$begingroup$
In the book Linear Algebra Done Right, example 2.14 said that $P_m(mathbb{F})$ is a finite-dimensional vector space for each non-negative integer $m$.
$P_m(mathbb{F})$ is defined as "For $m$ a nonnegative integer, $P_m(mathbb{F})$ denotes the set of all polynomials with coefficients in $mathbb{F}$ and degree at most $m$"
Then in example 2.16, it showed that $P(mathbb{F})$ is infinite-dimensional.
Consider any list of elements of $P(mathbb{F})$. Let $m$ denote the highest degree of the polynomials inn the list. Then every polynomial in the span of this list has degree at most $m$. Thus $z^{m+1}$ is not in the span of our list. Hence no list spans $P(mathbb{F})$. Thus $P(mathbb{F})$ is infinite dimensional.
I am a bit confused on why $P_m(mathbb{F})$ is finite dimensional, and $P(mathbb{F})$ is infinite dimensional.
Is it because $P_m(mathbb{F})$ is just talking about polynomials with degree at most $m$, while $P(mathbb{F})$ is talking about polynomials with degree that can be higher than $m$?
linear-algebra
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
$endgroup$
2
$begingroup$
Yes that's the reason. In $P(mathbb{F})$ the degree is unbounded. You have polynomials of all degrees.
$endgroup$
– mouthetics
Dec 29 '18 at 22:51
3
$begingroup$
I love when people answer their own question in the question itself :)
$endgroup$
– James Groon
Dec 29 '18 at 22:52
add a comment |
$begingroup$
In the book Linear Algebra Done Right, example 2.14 said that $P_m(mathbb{F})$ is a finite-dimensional vector space for each non-negative integer $m$.
$P_m(mathbb{F})$ is defined as "For $m$ a nonnegative integer, $P_m(mathbb{F})$ denotes the set of all polynomials with coefficients in $mathbb{F}$ and degree at most $m$"
Then in example 2.16, it showed that $P(mathbb{F})$ is infinite-dimensional.
Consider any list of elements of $P(mathbb{F})$. Let $m$ denote the highest degree of the polynomials inn the list. Then every polynomial in the span of this list has degree at most $m$. Thus $z^{m+1}$ is not in the span of our list. Hence no list spans $P(mathbb{F})$. Thus $P(mathbb{F})$ is infinite dimensional.
I am a bit confused on why $P_m(mathbb{F})$ is finite dimensional, and $P(mathbb{F})$ is infinite dimensional.
Is it because $P_m(mathbb{F})$ is just talking about polynomials with degree at most $m$, while $P(mathbb{F})$ is talking about polynomials with degree that can be higher than $m$?
linear-algebra
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
$endgroup$
In the book Linear Algebra Done Right, example 2.14 said that $P_m(mathbb{F})$ is a finite-dimensional vector space for each non-negative integer $m$.
$P_m(mathbb{F})$ is defined as "For $m$ a nonnegative integer, $P_m(mathbb{F})$ denotes the set of all polynomials with coefficients in $mathbb{F}$ and degree at most $m$"
Then in example 2.16, it showed that $P(mathbb{F})$ is infinite-dimensional.
Consider any list of elements of $P(mathbb{F})$. Let $m$ denote the highest degree of the polynomials inn the list. Then every polynomial in the span of this list has degree at most $m$. Thus $z^{m+1}$ is not in the span of our list. Hence no list spans $P(mathbb{F})$. Thus $P(mathbb{F})$ is infinite dimensional.
I am a bit confused on why $P_m(mathbb{F})$ is finite dimensional, and $P(mathbb{F})$ is infinite dimensional.
Is it because $P_m(mathbb{F})$ is just talking about polynomials with degree at most $m$, while $P(mathbb{F})$ is talking about polynomials with degree that can be higher than $m$?
linear-algebra
linear-algebra
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
asked Dec 29 '18 at 22:46
JOHN JOHN
3969
3969
2
$begingroup$
Yes that's the reason. In $P(mathbb{F})$ the degree is unbounded. You have polynomials of all degrees.
$endgroup$
– mouthetics
Dec 29 '18 at 22:51
3
$begingroup$
I love when people answer their own question in the question itself :)
$endgroup$
– James Groon
Dec 29 '18 at 22:52
add a comment |
2
$begingroup$
Yes that's the reason. In $P(mathbb{F})$ the degree is unbounded. You have polynomials of all degrees.
$endgroup$
– mouthetics
Dec 29 '18 at 22:51
3
$begingroup$
I love when people answer their own question in the question itself :)
$endgroup$
– James Groon
Dec 29 '18 at 22:52
2
2
$begingroup$
Yes that's the reason. In $P(mathbb{F})$ the degree is unbounded. You have polynomials of all degrees.
$endgroup$
– mouthetics
Dec 29 '18 at 22:51
$begingroup$
Yes that's the reason. In $P(mathbb{F})$ the degree is unbounded. You have polynomials of all degrees.
$endgroup$
– mouthetics
Dec 29 '18 at 22:51
3
3
$begingroup$
I love when people answer their own question in the question itself :)
$endgroup$
– James Groon
Dec 29 '18 at 22:52
$begingroup$
I love when people answer their own question in the question itself :)
$endgroup$
– James Groon
Dec 29 '18 at 22:52
add a comment |
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$begingroup$
Yes that's the reason. In $P(mathbb{F})$ the degree is unbounded. You have polynomials of all degrees.
$endgroup$
– mouthetics
Dec 29 '18 at 22:51
3
$begingroup$
I love when people answer their own question in the question itself :)
$endgroup$
– James Groon
Dec 29 '18 at 22:52