How can I make a matrix form when $BbbR^3 to BbbR^4$ is one-to-one?

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Problem : if $T: BbbR^3to BbbR^4$ is one-to-one, describe the possible
echelon forms of the standard matrix for a linear transformation $T$.




My solution is below.



To be one-to-one every column vectors must be independent, it means every columns must have pivot position. So I can form like this.
$$
beginpmatrix
a&*&*\
0&b&*\
0&0&c\
0&0&0
endpmatrix
$$



Am I right? If not, how can I approach this problem?










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  • 2




    Yes, $T$ is injective if and only if $operatornamerank(T) = 3$, which is true if and only if it is similar to a matrix of the above form.
    – mechanodroid
    Sep 10 at 11:15










  • @mechanodroid I didn't learn about "injective" and "$textrank(T)$" yet, can you explain?
    – baeharam
    Sep 10 at 11:20







  • 1




    Injective means one-to-one. Rank is the number of linearly independent columns (or rows) of $T$.
    – mechanodroid
    Sep 10 at 11:23










  • Thx for the explanation.
    – baeharam
    Sep 10 at 11:27










  • You need nonzero $a$, $b$ and $c$.
    – amd
    Sep 10 at 23:36














up vote
2
down vote

favorite













Problem : if $T: BbbR^3to BbbR^4$ is one-to-one, describe the possible
echelon forms of the standard matrix for a linear transformation $T$.




My solution is below.



To be one-to-one every column vectors must be independent, it means every columns must have pivot position. So I can form like this.
$$
beginpmatrix
a&*&*\
0&b&*\
0&0&c\
0&0&0
endpmatrix
$$



Am I right? If not, how can I approach this problem?










share|cite|improve this question



















  • 2




    Yes, $T$ is injective if and only if $operatornamerank(T) = 3$, which is true if and only if it is similar to a matrix of the above form.
    – mechanodroid
    Sep 10 at 11:15










  • @mechanodroid I didn't learn about "injective" and "$textrank(T)$" yet, can you explain?
    – baeharam
    Sep 10 at 11:20







  • 1




    Injective means one-to-one. Rank is the number of linearly independent columns (or rows) of $T$.
    – mechanodroid
    Sep 10 at 11:23










  • Thx for the explanation.
    – baeharam
    Sep 10 at 11:27










  • You need nonzero $a$, $b$ and $c$.
    – amd
    Sep 10 at 23:36












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Problem : if $T: BbbR^3to BbbR^4$ is one-to-one, describe the possible
echelon forms of the standard matrix for a linear transformation $T$.




My solution is below.



To be one-to-one every column vectors must be independent, it means every columns must have pivot position. So I can form like this.
$$
beginpmatrix
a&*&*\
0&b&*\
0&0&c\
0&0&0
endpmatrix
$$



Am I right? If not, how can I approach this problem?










share|cite|improve this question
















Problem : if $T: BbbR^3to BbbR^4$ is one-to-one, describe the possible
echelon forms of the standard matrix for a linear transformation $T$.




My solution is below.



To be one-to-one every column vectors must be independent, it means every columns must have pivot position. So I can form like this.
$$
beginpmatrix
a&*&*\
0&b&*\
0&0&c\
0&0&0
endpmatrix
$$



Am I right? If not, how can I approach this problem?







linear-transformations






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share|cite|improve this question













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edited Sep 10 at 11:16









mechanodroid

24.7k62245




24.7k62245










asked Sep 10 at 11:04









baeharam

396




396







  • 2




    Yes, $T$ is injective if and only if $operatornamerank(T) = 3$, which is true if and only if it is similar to a matrix of the above form.
    – mechanodroid
    Sep 10 at 11:15










  • @mechanodroid I didn't learn about "injective" and "$textrank(T)$" yet, can you explain?
    – baeharam
    Sep 10 at 11:20







  • 1




    Injective means one-to-one. Rank is the number of linearly independent columns (or rows) of $T$.
    – mechanodroid
    Sep 10 at 11:23










  • Thx for the explanation.
    – baeharam
    Sep 10 at 11:27










  • You need nonzero $a$, $b$ and $c$.
    – amd
    Sep 10 at 23:36












  • 2




    Yes, $T$ is injective if and only if $operatornamerank(T) = 3$, which is true if and only if it is similar to a matrix of the above form.
    – mechanodroid
    Sep 10 at 11:15










  • @mechanodroid I didn't learn about "injective" and "$textrank(T)$" yet, can you explain?
    – baeharam
    Sep 10 at 11:20







  • 1




    Injective means one-to-one. Rank is the number of linearly independent columns (or rows) of $T$.
    – mechanodroid
    Sep 10 at 11:23










  • Thx for the explanation.
    – baeharam
    Sep 10 at 11:27










  • You need nonzero $a$, $b$ and $c$.
    – amd
    Sep 10 at 23:36







2




2




Yes, $T$ is injective if and only if $operatornamerank(T) = 3$, which is true if and only if it is similar to a matrix of the above form.
– mechanodroid
Sep 10 at 11:15




Yes, $T$ is injective if and only if $operatornamerank(T) = 3$, which is true if and only if it is similar to a matrix of the above form.
– mechanodroid
Sep 10 at 11:15












@mechanodroid I didn't learn about "injective" and "$textrank(T)$" yet, can you explain?
– baeharam
Sep 10 at 11:20





@mechanodroid I didn't learn about "injective" and "$textrank(T)$" yet, can you explain?
– baeharam
Sep 10 at 11:20





1




1




Injective means one-to-one. Rank is the number of linearly independent columns (or rows) of $T$.
– mechanodroid
Sep 10 at 11:23




Injective means one-to-one. Rank is the number of linearly independent columns (or rows) of $T$.
– mechanodroid
Sep 10 at 11:23












Thx for the explanation.
– baeharam
Sep 10 at 11:27




Thx for the explanation.
– baeharam
Sep 10 at 11:27












You need nonzero $a$, $b$ and $c$.
– amd
Sep 10 at 23:36




You need nonzero $a$, $b$ and $c$.
– amd
Sep 10 at 23:36















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