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?
linear-transformations
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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?
linear-transformations
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
add a comment |Â
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?
linear-transformations
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
linear-transformations
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
add a comment |Â
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
add a comment |Â
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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