Linear transformations - 2 opposite claims, solution attempt included












0












$begingroup$


Suppose $V, W$ are vector spaces and $ T: V to W $ is a linear transformation.



$v_1, v_2, ... , v_k in V$.



Prove or disprove:




  • If $span( v_1, v_2, ... , v_k) = V$, then $span(T(v_1), ... , T(v_k) = W$.


  • If $span(T(v_1), ... , T(v_k)) = W$, then $span( v_1, v_2, ... , v_k) = V$.



My solution goes:




  • The first claim is false. let $, V, W = mathbb R$, and $ T: mathbb R to mathbb R$. We define $T$ as $T(v) = 0$, and $v_1 = [1]$. Therefore, $span(v_1) =mathbb R$, but $span(T(v_1)) = span(0) neq mathbb R$.


  • Second claim: We can easily prove that if $(T(v_1),...,T(v_n))$ is linearly independent, then $(v_1,...,v_n)$ is linearly independent as well. Is this enough to conclude that since we know $span(T(v_1), ... , T(v_k)) = W$, then $T(v_1), ... , T(v_k)$ is linearly independent, therefore $ v_1, v_2, ... , v_k$ is linearly independent, and thus $span( v_1, v_2, ... , v_k) = V$?



Thanks in advance!






linear-algebra vector-spaces linear-transformations







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  • $begingroup$
    Changed it, the question remains the same. Cant figure out how to prove or disprove the second claim.
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:04










  • $begingroup$
    What if $dim Vgtdim W$?
    $endgroup$
    – amd
    Dec 13 '18 at 18:16
















0












$begingroup$


Suppose $V, W$ are vector spaces and $ T: V to W $ is a linear transformation.



$v_1, v_2, ... , v_k in V$.



Prove or disprove:




  • If $span( v_1, v_2, ... , v_k) = V$, then $span(T(v_1), ... , T(v_k) = W$.


  • If $span(T(v_1), ... , T(v_k)) = W$, then $span( v_1, v_2, ... , v_k) = V$.



My solution goes:




  • The first claim is false. let $, V, W = mathbb R$, and $ T: mathbb R to mathbb R$. We define $T$ as $T(v) = 0$, and $v_1 = [1]$. Therefore, $span(v_1) =mathbb R$, but $span(T(v_1)) = span(0) neq mathbb R$.


  • Second claim: We can easily prove that if $(T(v_1),...,T(v_n))$ is linearly independent, then $(v_1,...,v_n)$ is linearly independent as well. Is this enough to conclude that since we know $span(T(v_1), ... , T(v_k)) = W$, then $T(v_1), ... , T(v_k)$ is linearly independent, therefore $ v_1, v_2, ... , v_k$ is linearly independent, and thus $span( v_1, v_2, ... , v_k) = V$?



Thanks in advance!






linear-algebra vector-spaces linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Changed it, the question remains the same. Cant figure out how to prove or disprove the second claim.
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:04










  • $begingroup$
    What if $dim Vgtdim W$?
    $endgroup$
    – amd
    Dec 13 '18 at 18:16














0












0








0





$begingroup$


Suppose $V, W$ are vector spaces and $ T: V to W $ is a linear transformation.



$v_1, v_2, ... , v_k in V$.



Prove or disprove:




  • If $span( v_1, v_2, ... , v_k) = V$, then $span(T(v_1), ... , T(v_k) = W$.


  • If $span(T(v_1), ... , T(v_k)) = W$, then $span( v_1, v_2, ... , v_k) = V$.



My solution goes:




  • The first claim is false. let $, V, W = mathbb R$, and $ T: mathbb R to mathbb R$. We define $T$ as $T(v) = 0$, and $v_1 = [1]$. Therefore, $span(v_1) =mathbb R$, but $span(T(v_1)) = span(0) neq mathbb R$.


  • Second claim: We can easily prove that if $(T(v_1),...,T(v_n))$ is linearly independent, then $(v_1,...,v_n)$ is linearly independent as well. Is this enough to conclude that since we know $span(T(v_1), ... , T(v_k)) = W$, then $T(v_1), ... , T(v_k)$ is linearly independent, therefore $ v_1, v_2, ... , v_k$ is linearly independent, and thus $span( v_1, v_2, ... , v_k) = V$?



Thanks in advance!






linear-algebra vector-spaces linear-transformations







share|cite|improve this question











$endgroup$




Suppose $V, W$ are vector spaces and $ T: V to W $ is a linear transformation.



$v_1, v_2, ... , v_k in V$.



Prove or disprove:




  • If $span( v_1, v_2, ... , v_k) = V$, then $span(T(v_1), ... , T(v_k) = W$.


  • If $span(T(v_1), ... , T(v_k)) = W$, then $span( v_1, v_2, ... , v_k) = V$.



My solution goes:




  • The first claim is false. let $, V, W = mathbb R$, and $ T: mathbb R to mathbb R$. We define $T$ as $T(v) = 0$, and $v_1 = [1]$. Therefore, $span(v_1) =mathbb R$, but $span(T(v_1)) = span(0) neq mathbb R$.


  • Second claim: We can easily prove that if $(T(v_1),...,T(v_n))$ is linearly independent, then $(v_1,...,v_n)$ is linearly independent as well. Is this enough to conclude that since we know $span(T(v_1), ... , T(v_k)) = W$, then $T(v_1), ... , T(v_k)$ is linearly independent, therefore $ v_1, v_2, ... , v_k$ is linearly independent, and thus $span( v_1, v_2, ... , v_k) = V$?



Thanks in advance!






linear-algebra vector-spaces linear-transformations




linear-algebra vector-spaces linear-transformations






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 13 '18 at 18:04







Tegernako

















asked Dec 13 '18 at 16:57









TegernakoTegernako

877




877












  • $begingroup$
    Changed it, the question remains the same. Cant figure out how to prove or disprove the second claim.
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:04










  • $begingroup$
    What if $dim Vgtdim W$?
    $endgroup$
    – amd
    Dec 13 '18 at 18:16


















  • $begingroup$
    Changed it, the question remains the same. Cant figure out how to prove or disprove the second claim.
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:04










  • $begingroup$
    What if $dim Vgtdim W$?
    $endgroup$
    – amd
    Dec 13 '18 at 18:16
















$begingroup$
Changed it, the question remains the same. Cant figure out how to prove or disprove the second claim.
$endgroup$
– Tegernako
Dec 13 '18 at 18:04




$begingroup$
Changed it, the question remains the same. Cant figure out how to prove or disprove the second claim.
$endgroup$
– Tegernako
Dec 13 '18 at 18:04












$begingroup$
What if $dim Vgtdim W$?
$endgroup$
– amd
Dec 13 '18 at 18:16




$begingroup$
What if $dim Vgtdim W$?
$endgroup$
– amd
Dec 13 '18 at 18:16










1 Answer
1






active

oldest

votes


















1












$begingroup$

The first you did great. For the second consider $W={0}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is this a counter example?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:11










  • $begingroup$
    What I mean, if the empty set spans W, what could possibly be my T?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:59






  • 1




    $begingroup$
    @Tegernako: The point is that if $W={mathbf{0}}$ (so your $T$ is the zero map), then any collection $v_1,ldots,v_k$ will satisfy that $mathrm{span}(T(v_1),ldots,T(v_k)) = mathrm{span}(mathbf{0},mathbf{0},ldots,mathbf{0}) = mathrm{span}(mathbf{0}) = {mathbf{0}}=W$. The empty set is not the only set that spans $W$.
    $endgroup$
    – Arturo Magidin
    Dec 13 '18 at 18:12








  • 1




    $begingroup$
    @Tegernako $T: vmapsto 0$ for all $v$.
    $endgroup$
    – amd
    Dec 13 '18 at 18:17










  • $begingroup$
    I see. Basically every set will satisfy what we have, but not the 'then' part. Thanks a lot, cheers for the weekend!
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:37











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

The first you did great. For the second consider $W={0}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is this a counter example?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:11










  • $begingroup$
    What I mean, if the empty set spans W, what could possibly be my T?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:59






  • 1




    $begingroup$
    @Tegernako: The point is that if $W={mathbf{0}}$ (so your $T$ is the zero map), then any collection $v_1,ldots,v_k$ will satisfy that $mathrm{span}(T(v_1),ldots,T(v_k)) = mathrm{span}(mathbf{0},mathbf{0},ldots,mathbf{0}) = mathrm{span}(mathbf{0}) = {mathbf{0}}=W$. The empty set is not the only set that spans $W$.
    $endgroup$
    – Arturo Magidin
    Dec 13 '18 at 18:12








  • 1




    $begingroup$
    @Tegernako $T: vmapsto 0$ for all $v$.
    $endgroup$
    – amd
    Dec 13 '18 at 18:17










  • $begingroup$
    I see. Basically every set will satisfy what we have, but not the 'then' part. Thanks a lot, cheers for the weekend!
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:37
















1












$begingroup$

The first you did great. For the second consider $W={0}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How is this a counter example?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:11










  • $begingroup$
    What I mean, if the empty set spans W, what could possibly be my T?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:59






  • 1




    $begingroup$
    @Tegernako: The point is that if $W={mathbf{0}}$ (so your $T$ is the zero map), then any collection $v_1,ldots,v_k$ will satisfy that $mathrm{span}(T(v_1),ldots,T(v_k)) = mathrm{span}(mathbf{0},mathbf{0},ldots,mathbf{0}) = mathrm{span}(mathbf{0}) = {mathbf{0}}=W$. The empty set is not the only set that spans $W$.
    $endgroup$
    – Arturo Magidin
    Dec 13 '18 at 18:12








  • 1




    $begingroup$
    @Tegernako $T: vmapsto 0$ for all $v$.
    $endgroup$
    – amd
    Dec 13 '18 at 18:17










  • $begingroup$
    I see. Basically every set will satisfy what we have, but not the 'then' part. Thanks a lot, cheers for the weekend!
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:37














1












1








1





$begingroup$

The first you did great. For the second consider $W={0}$






share|cite|improve this answer









$endgroup$



The first you did great. For the second consider $W={0}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 13 '18 at 17:01









YankoYanko

6,5541728




6,5541728












  • $begingroup$
    How is this a counter example?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:11










  • $begingroup$
    What I mean, if the empty set spans W, what could possibly be my T?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:59






  • 1




    $begingroup$
    @Tegernako: The point is that if $W={mathbf{0}}$ (so your $T$ is the zero map), then any collection $v_1,ldots,v_k$ will satisfy that $mathrm{span}(T(v_1),ldots,T(v_k)) = mathrm{span}(mathbf{0},mathbf{0},ldots,mathbf{0}) = mathrm{span}(mathbf{0}) = {mathbf{0}}=W$. The empty set is not the only set that spans $W$.
    $endgroup$
    – Arturo Magidin
    Dec 13 '18 at 18:12








  • 1




    $begingroup$
    @Tegernako $T: vmapsto 0$ for all $v$.
    $endgroup$
    – amd
    Dec 13 '18 at 18:17










  • $begingroup$
    I see. Basically every set will satisfy what we have, but not the 'then' part. Thanks a lot, cheers for the weekend!
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:37


















  • $begingroup$
    How is this a counter example?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:11










  • $begingroup$
    What I mean, if the empty set spans W, what could possibly be my T?
    $endgroup$
    – Tegernako
    Dec 13 '18 at 17:59






  • 1




    $begingroup$
    @Tegernako: The point is that if $W={mathbf{0}}$ (so your $T$ is the zero map), then any collection $v_1,ldots,v_k$ will satisfy that $mathrm{span}(T(v_1),ldots,T(v_k)) = mathrm{span}(mathbf{0},mathbf{0},ldots,mathbf{0}) = mathrm{span}(mathbf{0}) = {mathbf{0}}=W$. The empty set is not the only set that spans $W$.
    $endgroup$
    – Arturo Magidin
    Dec 13 '18 at 18:12








  • 1




    $begingroup$
    @Tegernako $T: vmapsto 0$ for all $v$.
    $endgroup$
    – amd
    Dec 13 '18 at 18:17










  • $begingroup$
    I see. Basically every set will satisfy what we have, but not the 'then' part. Thanks a lot, cheers for the weekend!
    $endgroup$
    – Tegernako
    Dec 13 '18 at 18:37
















$begingroup$
How is this a counter example?
$endgroup$
– Tegernako
Dec 13 '18 at 17:11




$begingroup$
How is this a counter example?
$endgroup$
– Tegernako
Dec 13 '18 at 17:11












$begingroup$
What I mean, if the empty set spans W, what could possibly be my T?
$endgroup$
– Tegernako
Dec 13 '18 at 17:59




$begingroup$
What I mean, if the empty set spans W, what could possibly be my T?
$endgroup$
– Tegernako
Dec 13 '18 at 17:59




1




1




$begingroup$
@Tegernako: The point is that if $W={mathbf{0}}$ (so your $T$ is the zero map), then any collection $v_1,ldots,v_k$ will satisfy that $mathrm{span}(T(v_1),ldots,T(v_k)) = mathrm{span}(mathbf{0},mathbf{0},ldots,mathbf{0}) = mathrm{span}(mathbf{0}) = {mathbf{0}}=W$. The empty set is not the only set that spans $W$.
$endgroup$
– Arturo Magidin
Dec 13 '18 at 18:12






$begingroup$
@Tegernako: The point is that if $W={mathbf{0}}$ (so your $T$ is the zero map), then any collection $v_1,ldots,v_k$ will satisfy that $mathrm{span}(T(v_1),ldots,T(v_k)) = mathrm{span}(mathbf{0},mathbf{0},ldots,mathbf{0}) = mathrm{span}(mathbf{0}) = {mathbf{0}}=W$. The empty set is not the only set that spans $W$.
$endgroup$
– Arturo Magidin
Dec 13 '18 at 18:12






1




1




$begingroup$
@Tegernako $T: vmapsto 0$ for all $v$.
$endgroup$
– amd
Dec 13 '18 at 18:17




$begingroup$
@Tegernako $T: vmapsto 0$ for all $v$.
$endgroup$
– amd
Dec 13 '18 at 18:17












$begingroup$
I see. Basically every set will satisfy what we have, but not the 'then' part. Thanks a lot, cheers for the weekend!
$endgroup$
– Tegernako
Dec 13 '18 at 18:37




$begingroup$
I see. Basically every set will satisfy what we have, but not the 'then' part. Thanks a lot, cheers for the weekend!
$endgroup$
– Tegernako
Dec 13 '18 at 18:37


















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