Mixing
Dimensions of linear transformation in relation to kernel
$begingroup$
I am new to linear algebra, and I just wanted to doublecheck my understanding of the following:
$T: R^n to R^m$ is a linear transformation.
True or false?
If $n>m$, then $operatorname{Ker}T neq {0}$.
This statement is correct, because if $n>m$, then per definition $T$ is not injective. If $T$ is not injective, then $ operatorname{Ker}Tneq {0}$.
Is my understanding correct?
Thank you!
linear-algebra linear-transformations
edited Jan 23 at 15:01
Bernard
122k741116
asked Jan 23 at 14:53
daltadalta
1238
$endgroup$
add a comment |
$begingroup$
I am new to linear algebra, and I just wanted to doublecheck my understanding of the following:
$T: R^n to R^m$ is a linear transformation.
True or false?
If $n>m$, then $operatorname{Ker}T neq {0}$.
This statement is correct, because if $n>m$, then per definition $T$ is not injective. If $T$ is not injective, then $ operatorname{Ker}Tneq {0}$.
Is my understanding correct?
Thank you!
linear-algebra linear-transformations
edited Jan 23 at 15:01
Bernard
122k741116
asked Jan 23 at 14:53
daltadalta
1238
$endgroup$
$begingroup$
Per definition of what?
$endgroup$
– Bernard
Jan 23 at 15:02
$begingroup$
@bernard If the original (sub)space has more dimensions than the destination one, then not all elements can be transformed in an exclusive manner. Perhaps the word "per definition" is not appropriate here, but that is the gist of what I meant.
$endgroup$
– dalta
Jan 23 at 15:06
$begingroup$
I agree with, but it is only an informal way of seeing things. It is more rigourous to use the rank-nullity formula.
$endgroup$
– Bernard
Jan 23 at 15:10
$begingroup$
@Bernard Thanks for the feedback: like I said, I am new to this, and I haven't studied the rank-nullity formula yet. However, I will look it up later tonight. But basically the statement is "true"?
$endgroup$
– dalta
Jan 23 at 15:19
$begingroup$
Intuitively, it is true, yes.
$endgroup$
– Bernard
Jan 23 at 15:24
add a comment |
$begingroup$
I am new to linear algebra, and I just wanted to doublecheck my understanding of the following:
$T: R^n to R^m$ is a linear transformation.
True or false?
If $n>m$, then $operatorname{Ker}T neq {0}$.
This statement is correct, because if $n>m$, then per definition $T$ is not injective. If $T$ is not injective, then $ operatorname{Ker}Tneq {0}$.
Is my understanding correct?
Thank you!
linear-algebra linear-transformations
edited Jan 23 at 15:01
Bernard
122k741116
asked Jan 23 at 14:53
daltadalta
1238
$endgroup$
I am new to linear algebra, and I just wanted to doublecheck my understanding of the following:
$T: R^n to R^m$ is a linear transformation.
True or false?
If $n>m$, then $operatorname{Ker}T neq {0}$.
This statement is correct, because if $n>m$, then per definition $T$ is not injective. If $T$ is not injective, then $ operatorname{Ker}Tneq {0}$.
Is my understanding correct?
Thank you!
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 23 at 15:01
Bernard
122k741116
asked Jan 23 at 14:53
daltadalta
1238
edited Jan 23 at 15:01
Bernard
122k741116
asked Jan 23 at 14:53
daltadalta
1238
edited Jan 23 at 15:01
Bernard
122k741116
edited Jan 23 at 15:01
Bernard
122k741116
edited Jan 23 at 15:01
Bernard
122k741116
122k741116
asked Jan 23 at 14:53
daltadalta
1238
asked Jan 23 at 14:53
daltadalta
1238
asked Jan 23 at 14:53
daltadalta
1238
1238
$begingroup$
Per definition of what?
$endgroup$
– Bernard
Jan 23 at 15:02
$begingroup$
@bernard If the original (sub)space has more dimensions than the destination one, then not all elements can be transformed in an exclusive manner. Perhaps the word "per definition" is not appropriate here, but that is the gist of what I meant.
$endgroup$
– dalta
Jan 23 at 15:06
$begingroup$
I agree with, but it is only an informal way of seeing things. It is more rigourous to use the rank-nullity formula.
$endgroup$
– Bernard
Jan 23 at 15:10
$begingroup$
@Bernard Thanks for the feedback: like I said, I am new to this, and I haven't studied the rank-nullity formula yet. However, I will look it up later tonight. But basically the statement is "true"?
$endgroup$
– dalta
Jan 23 at 15:19
$begingroup$
Intuitively, it is true, yes.
$endgroup$
– Bernard
Jan 23 at 15:24
add a comment |
$begingroup$
Per definition of what?
$endgroup$
– Bernard
Jan 23 at 15:02
$begingroup$
@bernard If the original (sub)space has more dimensions than the destination one, then not all elements can be transformed in an exclusive manner. Perhaps the word "per definition" is not appropriate here, but that is the gist of what I meant.
$endgroup$
– dalta
Jan 23 at 15:06
$begingroup$
I agree with, but it is only an informal way of seeing things. It is more rigourous to use the rank-nullity formula.
$endgroup$
– Bernard
Jan 23 at 15:10
$begingroup$
@Bernard Thanks for the feedback: like I said, I am new to this, and I haven't studied the rank-nullity formula yet. However, I will look it up later tonight. But basically the statement is "true"?
$endgroup$
– dalta
Jan 23 at 15:19
$begingroup$
Intuitively, it is true, yes.
$endgroup$
– Bernard
Jan 23 at 15:24
$begingroup$
Per definition of what?
$endgroup$
– Bernard
Jan 23 at 15:02
$begingroup$
Per definition of what?
$endgroup$
– Bernard
Jan 23 at 15:02
$begingroup$
@bernard If the original (sub)space has more dimensions than the destination one, then not all elements can be transformed in an exclusive manner. Perhaps the word "per definition" is not appropriate here, but that is the gist of what I meant.
$endgroup$
– dalta
Jan 23 at 15:06
$begingroup$
@bernard If the original (sub)space has more dimensions than the destination one, then not all elements can be transformed in an exclusive manner. Perhaps the word "per definition" is not appropriate here, but that is the gist of what I meant.
$endgroup$
– dalta
Jan 23 at 15:06
$begingroup$
I agree with, but it is only an informal way of seeing things. It is more rigourous to use the rank-nullity formula.
$endgroup$
– Bernard
Jan 23 at 15:10
$begingroup$
I agree with, but it is only an informal way of seeing things. It is more rigourous to use the rank-nullity formula.
$endgroup$
– Bernard
Jan 23 at 15:10
$begingroup$
@Bernard Thanks for the feedback: like I said, I am new to this, and I haven't studied the rank-nullity formula yet. However, I will look it up later tonight. But basically the statement is "true"?
$endgroup$
– dalta
Jan 23 at 15:19
$begingroup$
@Bernard Thanks for the feedback: like I said, I am new to this, and I haven't studied the rank-nullity formula yet. However, I will look it up later tonight. But basically the statement is "true"?
$endgroup$
– dalta
Jan 23 at 15:19
$begingroup$
Intuitively, it is true, yes.
$endgroup$
– Bernard
Jan 23 at 15:24
$begingroup$
Intuitively, it is true, yes.
$endgroup$
– Bernard
Jan 23 at 15:24
add a comment |
1 Answer
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$begingroup$
It is true.
The rank nullity theorem says $operatorname{rank}T+operatorname{nullity}T=n$.
On the other hand, $operatorname{rank}Tle mlt n$. $therefore operatorname{nullity}Tgt0$.
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
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add a comment |
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$begingroup$
It is true.
The rank nullity theorem says $operatorname{rank}T+operatorname{nullity}T=n$.
On the other hand, $operatorname{rank}Tle mlt n$. $therefore operatorname{nullity}Tgt0$.
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
It is true.
The rank nullity theorem says $operatorname{rank}T+operatorname{nullity}T=n$.
On the other hand, $operatorname{rank}Tle mlt n$. $therefore operatorname{nullity}Tgt0$.
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
It is true.
The rank nullity theorem says $operatorname{rank}T+operatorname{nullity}T=n$.
On the other hand, $operatorname{rank}Tle mlt n$. $therefore operatorname{nullity}Tgt0$.
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
$endgroup$
It is true.
The rank nullity theorem says $operatorname{rank}T+operatorname{nullity}T=n$.
On the other hand, $operatorname{rank}Tle mlt n$. $therefore operatorname{nullity}Tgt0$.
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
answered Jan 23 at 18:01
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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$begingroup$
Per definition of what?
$endgroup$
– Bernard
Jan 23 at 15:02
$begingroup$
@bernard If the original (sub)space has more dimensions than the destination one, then not all elements can be transformed in an exclusive manner. Perhaps the word "per definition" is not appropriate here, but that is the gist of what I meant.
$endgroup$
– dalta
Jan 23 at 15:06
$begingroup$
I agree with, but it is only an informal way of seeing things. It is more rigourous to use the rank-nullity formula.
$endgroup$
– Bernard
Jan 23 at 15:10
$begingroup$
@Bernard Thanks for the feedback: like I said, I am new to this, and I haven't studied the rank-nullity formula yet. However, I will look it up later tonight. But basically the statement is "true"?
$endgroup$
– dalta
Jan 23 at 15:19
$begingroup$
Intuitively, it is true, yes.
$endgroup$
– Bernard
Jan 23 at 15:24