Proofs with linear maps
$begingroup$
For my linear algebra course at university I was given this task:
Let $f :Bbb R^n to Bbb R^m$ be a linear map. Let further $A := (v_1, ldots , v_k)$ be a family of linearly
independent vectors in $Bbb R^n$ and let $B := (f(v_1), ldots, f(v_k))$ be their images. Also denote by $M subset Bbb R_n$ an arbitrary subset of $Bbb R^n$ such that $operatorname{Span}(M) = Bbb R^n$. Show that
a) $f$ injective $implies$ The family B is linearly independent,
b) $f$ surjective $implies$ $operatorname{Span}(f(M)) =Bbb R^m$.
My problem is that I can't imagine a useful approach for this task and I would be very thankful for some hints how to start my proofs.
Thank you guys!
linear-algebra proof-writing linear-transformations
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
$endgroup$
add a comment |
$begingroup$
For my linear algebra course at university I was given this task:
Let $f :Bbb R^n to Bbb R^m$ be a linear map. Let further $A := (v_1, ldots , v_k)$ be a family of linearly
independent vectors in $Bbb R^n$ and let $B := (f(v_1), ldots, f(v_k))$ be their images. Also denote by $M subset Bbb R_n$ an arbitrary subset of $Bbb R^n$ such that $operatorname{Span}(M) = Bbb R^n$. Show that
a) $f$ injective $implies$ The family B is linearly independent,
b) $f$ surjective $implies$ $operatorname{Span}(f(M)) =Bbb R^m$.
My problem is that I can't imagine a useful approach for this task and I would be very thankful for some hints how to start my proofs.
Thank you guys!
linear-algebra proof-writing linear-transformations
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
$endgroup$
1
$begingroup$
You have used theproof-verificationtag. Which proof do you want was to verify whether it is correct or not?
$endgroup$
– José Carlos Santos
Nov 24 '18 at 12:26
add a comment |
$begingroup$
For my linear algebra course at university I was given this task:
Let $f :Bbb R^n to Bbb R^m$ be a linear map. Let further $A := (v_1, ldots , v_k)$ be a family of linearly
independent vectors in $Bbb R^n$ and let $B := (f(v_1), ldots, f(v_k))$ be their images. Also denote by $M subset Bbb R_n$ an arbitrary subset of $Bbb R^n$ such that $operatorname{Span}(M) = Bbb R^n$. Show that
a) $f$ injective $implies$ The family B is linearly independent,
b) $f$ surjective $implies$ $operatorname{Span}(f(M)) =Bbb R^m$.
My problem is that I can't imagine a useful approach for this task and I would be very thankful for some hints how to start my proofs.
Thank you guys!
linear-algebra proof-writing linear-transformations
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
$endgroup$
For my linear algebra course at university I was given this task:
Let $f :Bbb R^n to Bbb R^m$ be a linear map. Let further $A := (v_1, ldots , v_k)$ be a family of linearly
independent vectors in $Bbb R^n$ and let $B := (f(v_1), ldots, f(v_k))$ be their images. Also denote by $M subset Bbb R_n$ an arbitrary subset of $Bbb R^n$ such that $operatorname{Span}(M) = Bbb R^n$. Show that
a) $f$ injective $implies$ The family B is linearly independent,
b) $f$ surjective $implies$ $operatorname{Span}(f(M)) =Bbb R^m$.
My problem is that I can't imagine a useful approach for this task and I would be very thankful for some hints how to start my proofs.
Thank you guys!
linear-algebra proof-writing linear-transformations
linear-algebra proof-writing linear-transformations
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
edited Nov 24 '18 at 14:39
Mathmeeeeen
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
asked Nov 24 '18 at 12:11
MathmeeeeenMathmeeeeen
133
133
1
$begingroup$
You have used theproof-verificationtag. Which proof do you want was to verify whether it is correct or not?
$endgroup$
– José Carlos Santos
Nov 24 '18 at 12:26
add a comment |
1
$begingroup$
You have used theproof-verificationtag. Which proof do you want was to verify whether it is correct or not?
$endgroup$
– José Carlos Santos
Nov 24 '18 at 12:26
1
1
$begingroup$
You have used the
proof-verification tag. Which proof do you want was to verify whether it is correct or not?$endgroup$
– José Carlos Santos
Nov 24 '18 at 12:26
$begingroup$
You have used the
proof-verification tag. Which proof do you want was to verify whether it is correct or not?$endgroup$
– José Carlos Santos
Nov 24 '18 at 12:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
a) If there are $alpha_1,ldots,alpha_k in mathbb{R}$ such that
$$sumlimits_{i=0}^kalpha_if(v_i) = 0,$$
then, since $f$ is linear, $$fleft(sumlimits_{i=0}^kalpha_iv_iright) = 0,$$
and since $f$ is injective, $$sumlimits_{i=0}^kalpha_iv_i = 0,$$
but the $v_i$ all lie in $A$, so are linearly independent, so no such $alpha_i$ exist, hence $B$ is linearly independent.
b)If there is some $xinmathbb{R}^msetminus mathop{mathrm{Span}}(f(M)),$ then, since $f$ is surjective, there is some $y in mathbb{R}^n$ such that $f(y) = x$. But since $M$ spans $mathbb{R}^n$, there are some $m_1,ldots,m_kin M$ and some $alpha_1,ldots,alpha_kinmathbb{R}$ such that $$y = sumlimits_{i=0}^kalpha_im_i.$$
But then begin{align*}x = f(y) &= fleft(sumlimits_{i=0}^kalpha_im_iright)
\&=sumlimits_{i=0}^kalpha_if(m_i)&mbox{since }fmbox{ is linear}
\&inmathop{mathrm{Span}}{f(m_1),ldots,f(m_k)}
\&subseteqmathop{mathrm{Span}}(f(M))&mbox{since }m_i in Mmbox{ for all }i.end{align*}
This contradicts our definition of $x$, so no such $x$ exists, so $mathrm{Span}(f(M)) = mathbb{R}^m$.
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
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add a comment |
$begingroup$
For (a) show at first, that $f$ is injective iff its kernel is trivial. Now assume $0=sumlambda_i f(v_i)$. As $f$ is linear we have $0=fbigl(sumlambda_i v_i)bigr)$, hence $sumlambda_i v_i=0$.
For (b) assume that $f(M)$ doesn't span $mathbb R^m$.
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
$begingroup$
Unless I'm missing something, that's terrible advice for (b).
$endgroup$
– Aweygan
Nov 24 '18 at 18:04
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
a) If there are $alpha_1,ldots,alpha_k in mathbb{R}$ such that
$$sumlimits_{i=0}^kalpha_if(v_i) = 0,$$
then, since $f$ is linear, $$fleft(sumlimits_{i=0}^kalpha_iv_iright) = 0,$$
and since $f$ is injective, $$sumlimits_{i=0}^kalpha_iv_i = 0,$$
but the $v_i$ all lie in $A$, so are linearly independent, so no such $alpha_i$ exist, hence $B$ is linearly independent.
b)If there is some $xinmathbb{R}^msetminus mathop{mathrm{Span}}(f(M)),$ then, since $f$ is surjective, there is some $y in mathbb{R}^n$ such that $f(y) = x$. But since $M$ spans $mathbb{R}^n$, there are some $m_1,ldots,m_kin M$ and some $alpha_1,ldots,alpha_kinmathbb{R}$ such that $$y = sumlimits_{i=0}^kalpha_im_i.$$
But then begin{align*}x = f(y) &= fleft(sumlimits_{i=0}^kalpha_im_iright)
\&=sumlimits_{i=0}^kalpha_if(m_i)&mbox{since }fmbox{ is linear}
\&inmathop{mathrm{Span}}{f(m_1),ldots,f(m_k)}
\&subseteqmathop{mathrm{Span}}(f(M))&mbox{since }m_i in Mmbox{ for all }i.end{align*}
This contradicts our definition of $x$, so no such $x$ exists, so $mathrm{Span}(f(M)) = mathbb{R}^m$.
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
$endgroup$
add a comment |
$begingroup$
a) If there are $alpha_1,ldots,alpha_k in mathbb{R}$ such that
$$sumlimits_{i=0}^kalpha_if(v_i) = 0,$$
then, since $f$ is linear, $$fleft(sumlimits_{i=0}^kalpha_iv_iright) = 0,$$
and since $f$ is injective, $$sumlimits_{i=0}^kalpha_iv_i = 0,$$
but the $v_i$ all lie in $A$, so are linearly independent, so no such $alpha_i$ exist, hence $B$ is linearly independent.
b)If there is some $xinmathbb{R}^msetminus mathop{mathrm{Span}}(f(M)),$ then, since $f$ is surjective, there is some $y in mathbb{R}^n$ such that $f(y) = x$. But since $M$ spans $mathbb{R}^n$, there are some $m_1,ldots,m_kin M$ and some $alpha_1,ldots,alpha_kinmathbb{R}$ such that $$y = sumlimits_{i=0}^kalpha_im_i.$$
But then begin{align*}x = f(y) &= fleft(sumlimits_{i=0}^kalpha_im_iright)
\&=sumlimits_{i=0}^kalpha_if(m_i)&mbox{since }fmbox{ is linear}
\&inmathop{mathrm{Span}}{f(m_1),ldots,f(m_k)}
\&subseteqmathop{mathrm{Span}}(f(M))&mbox{since }m_i in Mmbox{ for all }i.end{align*}
This contradicts our definition of $x$, so no such $x$ exists, so $mathrm{Span}(f(M)) = mathbb{R}^m$.
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
$endgroup$
add a comment |
$begingroup$
a) If there are $alpha_1,ldots,alpha_k in mathbb{R}$ such that
$$sumlimits_{i=0}^kalpha_if(v_i) = 0,$$
then, since $f$ is linear, $$fleft(sumlimits_{i=0}^kalpha_iv_iright) = 0,$$
and since $f$ is injective, $$sumlimits_{i=0}^kalpha_iv_i = 0,$$
but the $v_i$ all lie in $A$, so are linearly independent, so no such $alpha_i$ exist, hence $B$ is linearly independent.
b)If there is some $xinmathbb{R}^msetminus mathop{mathrm{Span}}(f(M)),$ then, since $f$ is surjective, there is some $y in mathbb{R}^n$ such that $f(y) = x$. But since $M$ spans $mathbb{R}^n$, there are some $m_1,ldots,m_kin M$ and some $alpha_1,ldots,alpha_kinmathbb{R}$ such that $$y = sumlimits_{i=0}^kalpha_im_i.$$
But then begin{align*}x = f(y) &= fleft(sumlimits_{i=0}^kalpha_im_iright)
\&=sumlimits_{i=0}^kalpha_if(m_i)&mbox{since }fmbox{ is linear}
\&inmathop{mathrm{Span}}{f(m_1),ldots,f(m_k)}
\&subseteqmathop{mathrm{Span}}(f(M))&mbox{since }m_i in Mmbox{ for all }i.end{align*}
This contradicts our definition of $x$, so no such $x$ exists, so $mathrm{Span}(f(M)) = mathbb{R}^m$.
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
$endgroup$
a) If there are $alpha_1,ldots,alpha_k in mathbb{R}$ such that
$$sumlimits_{i=0}^kalpha_if(v_i) = 0,$$
then, since $f$ is linear, $$fleft(sumlimits_{i=0}^kalpha_iv_iright) = 0,$$
and since $f$ is injective, $$sumlimits_{i=0}^kalpha_iv_i = 0,$$
but the $v_i$ all lie in $A$, so are linearly independent, so no such $alpha_i$ exist, hence $B$ is linearly independent.
b)If there is some $xinmathbb{R}^msetminus mathop{mathrm{Span}}(f(M)),$ then, since $f$ is surjective, there is some $y in mathbb{R}^n$ such that $f(y) = x$. But since $M$ spans $mathbb{R}^n$, there are some $m_1,ldots,m_kin M$ and some $alpha_1,ldots,alpha_kinmathbb{R}$ such that $$y = sumlimits_{i=0}^kalpha_im_i.$$
But then begin{align*}x = f(y) &= fleft(sumlimits_{i=0}^kalpha_im_iright)
\&=sumlimits_{i=0}^kalpha_if(m_i)&mbox{since }fmbox{ is linear}
\&inmathop{mathrm{Span}}{f(m_1),ldots,f(m_k)}
\&subseteqmathop{mathrm{Span}}(f(M))&mbox{since }m_i in Mmbox{ for all }i.end{align*}
This contradicts our definition of $x$, so no such $x$ exists, so $mathrm{Span}(f(M)) = mathbb{R}^m$.
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
answered Nov 24 '18 at 12:40
user3482749user3482749
3,722417
3,722417
add a comment |
add a comment |
$begingroup$
For (a) show at first, that $f$ is injective iff its kernel is trivial. Now assume $0=sumlambda_i f(v_i)$. As $f$ is linear we have $0=fbigl(sumlambda_i v_i)bigr)$, hence $sumlambda_i v_i=0$.
For (b) assume that $f(M)$ doesn't span $mathbb R^m$.
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
$begingroup$
Unless I'm missing something, that's terrible advice for (b).
$endgroup$
– Aweygan
Nov 24 '18 at 18:04
add a comment |
$begingroup$
For (a) show at first, that $f$ is injective iff its kernel is trivial. Now assume $0=sumlambda_i f(v_i)$. As $f$ is linear we have $0=fbigl(sumlambda_i v_i)bigr)$, hence $sumlambda_i v_i=0$.
For (b) assume that $f(M)$ doesn't span $mathbb R^m$.
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
$begingroup$
Unless I'm missing something, that's terrible advice for (b).
$endgroup$
– Aweygan
Nov 24 '18 at 18:04
add a comment |
$begingroup$
For (a) show at first, that $f$ is injective iff its kernel is trivial. Now assume $0=sumlambda_i f(v_i)$. As $f$ is linear we have $0=fbigl(sumlambda_i v_i)bigr)$, hence $sumlambda_i v_i=0$.
For (b) assume that $f(M)$ doesn't span $mathbb R^m$.
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
$endgroup$
For (a) show at first, that $f$ is injective iff its kernel is trivial. Now assume $0=sumlambda_i f(v_i)$. As $f$ is linear we have $0=fbigl(sumlambda_i v_i)bigr)$, hence $sumlambda_i v_i=0$.
For (b) assume that $f(M)$ doesn't span $mathbb R^m$.
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
answered Nov 24 '18 at 12:31
Michael HoppeMichael Hoppe
10.8k31834
10.8k31834
$begingroup$
Unless I'm missing something, that's terrible advice for (b).
$endgroup$
– Aweygan
Nov 24 '18 at 18:04
add a comment |
$begingroup$
Unless I'm missing something, that's terrible advice for (b).
$endgroup$
– Aweygan
Nov 24 '18 at 18:04
$begingroup$
Unless I'm missing something, that's terrible advice for (b).
$endgroup$
– Aweygan
Nov 24 '18 at 18:04
$begingroup$
Unless I'm missing something, that's terrible advice for (b).
$endgroup$
– Aweygan
Nov 24 '18 at 18:04
add a comment |
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$begingroup$
You have used the
proof-verificationtag. Which proof do you want was to verify whether it is correct or not?$endgroup$
– José Carlos Santos
Nov 24 '18 at 12:26