If $T : V to V$ is linear transformation that satisfies $Tcirc T=0$, which statement is true?
If $T: V to V$ is a linear transformation of vector space $V$ and $Tcirc T=0$, then:
a) $ker T subseteq operatorname{image} T$
b) $operatorname{image} T subseteq ker T$
c) $T = 0$
d) $T$ is a nonsingular linear transformation.
My Try: As $T(T(x))=0$ that means $operatorname{image} T subseteq ker T$.
But kernel and image are always disjoint, except $0$ is always common in them.
That means the only possibility is $T(x)=0$ for every $xin V$. So it is a zero linear transformation.
But on the answer sheet, option (d) is given as correct.
linear-algebra linear-transformations
edited Nov 26 at 10:35
Travis
59.5k767145
asked Nov 26 at 10:22
user532616
254
add a comment |
If $T: V to V$ is a linear transformation of vector space $V$ and $Tcirc T=0$, then:
a) $ker T subseteq operatorname{image} T$
b) $operatorname{image} T subseteq ker T$
c) $T = 0$
d) $T$ is a nonsingular linear transformation.
My Try: As $T(T(x))=0$ that means $operatorname{image} T subseteq ker T$.
But kernel and image are always disjoint, except $0$ is always common in them.
That means the only possibility is $T(x)=0$ for every $xin V$. So it is a zero linear transformation.
But on the answer sheet, option (d) is given as correct.
linear-algebra linear-transformations
edited Nov 26 at 10:35
Travis
59.5k767145
asked Nov 26 at 10:22
user532616
254
The correct answer is b. Your claim that kernel and image are always disjoint is false, take e.g. $begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$. Not sure why they put d as $T$ is always singular (unless $V$ is zero-dimensional I guess.)
– hunter
Nov 26 at 10:45
add a comment |
If $T: V to V$ is a linear transformation of vector space $V$ and $Tcirc T=0$, then:
a) $ker T subseteq operatorname{image} T$
b) $operatorname{image} T subseteq ker T$
c) $T = 0$
d) $T$ is a nonsingular linear transformation.
My Try: As $T(T(x))=0$ that means $operatorname{image} T subseteq ker T$.
But kernel and image are always disjoint, except $0$ is always common in them.
That means the only possibility is $T(x)=0$ for every $xin V$. So it is a zero linear transformation.
But on the answer sheet, option (d) is given as correct.
linear-algebra linear-transformations
edited Nov 26 at 10:35
Travis
59.5k767145
asked Nov 26 at 10:22
user532616
254
If $T: V to V$ is a linear transformation of vector space $V$ and $Tcirc T=0$, then:
a) $ker T subseteq operatorname{image} T$
b) $operatorname{image} T subseteq ker T$
c) $T = 0$
d) $T$ is a nonsingular linear transformation.
My Try: As $T(T(x))=0$ that means $operatorname{image} T subseteq ker T$.
But kernel and image are always disjoint, except $0$ is always common in them.
That means the only possibility is $T(x)=0$ for every $xin V$. So it is a zero linear transformation.
But on the answer sheet, option (d) is given as correct.
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Nov 26 at 10:35
Travis
59.5k767145
asked Nov 26 at 10:22
user532616
254
edited Nov 26 at 10:35
Travis
59.5k767145
asked Nov 26 at 10:22
user532616
254
edited Nov 26 at 10:35
Travis
59.5k767145
edited Nov 26 at 10:35
Travis
59.5k767145
edited Nov 26 at 10:35
Travis
59.5k767145
59.5k767145
asked Nov 26 at 10:22
user532616
254
asked Nov 26 at 10:22
user532616
254
asked Nov 26 at 10:22
user532616
254
254
The correct answer is b. Your claim that kernel and image are always disjoint is false, take e.g. $begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$. Not sure why they put d as $T$ is always singular (unless $V$ is zero-dimensional I guess.)
– hunter
Nov 26 at 10:45
add a comment |
The correct answer is b. Your claim that kernel and image are always disjoint is false, take e.g. $begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$. Not sure why they put d as $T$ is always singular (unless $V$ is zero-dimensional I guess.)
– hunter
Nov 26 at 10:45
The correct answer is b. Your claim that kernel and image are always disjoint is false, take e.g. $begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$. Not sure why they put d as $T$ is always singular (unless $V$ is zero-dimensional I guess.)
– hunter
Nov 26 at 10:45
The correct answer is b. Your claim that kernel and image are always disjoint is false, take e.g. $begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$. Not sure why they put d as $T$ is always singular (unless $V$ is zero-dimensional I guess.)
– hunter
Nov 26 at 10:45
add a comment |
1 Answer
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votes
The image and kernel are not always transverse; consider for example $$T : V to V, qquad T(v, w) := (w, 0) .$$
Certainly (d) cannot be correct, as it implies $det (T circ T) = (det T)^2 neq 0$.
answered Nov 26 at 10:31
Travis
59.5k767145
Is my explanation is correct? Which is correct option (b) or (c)
– user532616
Nov 26 at 10:52
I got it, range nd kernel are not disjoint so option b is correct
– user532616
Nov 26 at 11:06
add a comment |
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1 Answer
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1 Answer
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votes
The image and kernel are not always transverse; consider for example $$T : V to V, qquad T(v, w) := (w, 0) .$$
Certainly (d) cannot be correct, as it implies $det (T circ T) = (det T)^2 neq 0$.
answered Nov 26 at 10:31
Travis
59.5k767145
Is my explanation is correct? Which is correct option (b) or (c)
– user532616
Nov 26 at 10:52
I got it, range nd kernel are not disjoint so option b is correct
– user532616
Nov 26 at 11:06
add a comment |
The image and kernel are not always transverse; consider for example $$T : V to V, qquad T(v, w) := (w, 0) .$$
Certainly (d) cannot be correct, as it implies $det (T circ T) = (det T)^2 neq 0$.
answered Nov 26 at 10:31
Travis
59.5k767145
Is my explanation is correct? Which is correct option (b) or (c)
– user532616
Nov 26 at 10:52
I got it, range nd kernel are not disjoint so option b is correct
– user532616
Nov 26 at 11:06
add a comment |
The image and kernel are not always transverse; consider for example $$T : V to V, qquad T(v, w) := (w, 0) .$$
Certainly (d) cannot be correct, as it implies $det (T circ T) = (det T)^2 neq 0$.
answered Nov 26 at 10:31
Travis
59.5k767145
The image and kernel are not always transverse; consider for example $$T : V to V, qquad T(v, w) := (w, 0) .$$
Certainly (d) cannot be correct, as it implies $det (T circ T) = (det T)^2 neq 0$.
answered Nov 26 at 10:31
Travis
59.5k767145
answered Nov 26 at 10:31
Travis
59.5k767145
answered Nov 26 at 10:31
Travis
59.5k767145
answered Nov 26 at 10:31
Travis
59.5k767145
59.5k767145
Is my explanation is correct? Which is correct option (b) or (c)
– user532616
Nov 26 at 10:52
I got it, range nd kernel are not disjoint so option b is correct
– user532616
Nov 26 at 11:06
add a comment |
Is my explanation is correct? Which is correct option (b) or (c)
– user532616
Nov 26 at 10:52
I got it, range nd kernel are not disjoint so option b is correct
– user532616
Nov 26 at 11:06
Is my explanation is correct? Which is correct option (b) or (c)
– user532616
Nov 26 at 10:52
Is my explanation is correct? Which is correct option (b) or (c)
– user532616
Nov 26 at 10:52
I got it, range nd kernel are not disjoint so option b is correct
– user532616
Nov 26 at 11:06
I got it, range nd kernel are not disjoint so option b is correct
– user532616
Nov 26 at 11:06
add a comment |
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The correct answer is b. Your claim that kernel and image are always disjoint is false, take e.g. $begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}$. Not sure why they put d as $T$ is always singular (unless $V$ is zero-dimensional I guess.)
– hunter
Nov 26 at 10:45