How can I prove that two null spaces are equal by using the existence of an invertible operator?

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Suppose $W$ is finite dimensional and $T_1, T_2 in L(V,W).$ Prove that null($T_1$) $=$ null($T_2$) if and only if there exists and invertable operator $Sin L(W)$ such that $T_1 = Scirc T_2$.

I'm quite sure I have to make use of the rank-nullity theorem here. I started with:

$ dim(v) = dim(null(Scirc T_2)) + dim(Range(Scirc T_2)) $

$=dim(null(T_1)) + dim(Range(S circ T_2)) $

$=dim(null(S circ T_2) = dim(Range(nullT_1))$

$=dim(nullT_1) + dim(Range(T_1)) $

$= dim(nullT_1), + dim(Range(S circ T_2))$

$=dim(Range(T_1)) = dim(Range(S circ T_2))$

...I'm not sure how to take this further and put everything together.

edited 2 days ago

max_zorn

3,25861228

asked Nov 13 at 17:03

Jaigus

1948

What have you tried so far?
– jgon
Nov 13 at 17:06

Any time you're stuck on a proof, one place to start is to circle all the key terms and write down the definitions for them. PS it's "Invertible", not "invertable".
– Acccumulation
Nov 13 at 17:13

add a comment |

up vote
0
down vote

favorite

Suppose $W$ is finite dimensional and $T_1, T_2 in L(V,W).$ Prove that null($T_1$) $=$ null($T_2$) if and only if there exists and invertable operator $Sin L(W)$ such that $T_1 = Scirc T_2$.

I'm quite sure I have to make use of the rank-nullity theorem here. I started with:

$ dim(v) = dim(null(Scirc T_2)) + dim(Range(Scirc T_2)) $

$=dim(null(T_1)) + dim(Range(S circ T_2)) $

$=dim(null(S circ T_2) = dim(Range(nullT_1))$

$=dim(nullT_1) + dim(Range(T_1)) $

$= dim(nullT_1), + dim(Range(S circ T_2))$

$=dim(Range(T_1)) = dim(Range(S circ T_2))$

...I'm not sure how to take this further and put everything together.

edited 2 days ago

max_zorn

3,25861228

asked Nov 13 at 17:03

Jaigus

1948

What have you tried so far?
– jgon
Nov 13 at 17:06

Any time you're stuck on a proof, one place to start is to circle all the key terms and write down the definitions for them. PS it's "Invertible", not "invertable".
– Acccumulation
Nov 13 at 17:13

add a comment |

up vote
0
down vote

favorite

Suppose $W$ is finite dimensional and $T_1, T_2 in L(V,W).$ Prove that null($T_1$) $=$ null($T_2$) if and only if there exists and invertable operator $Sin L(W)$ such that $T_1 = Scirc T_2$.

I'm quite sure I have to make use of the rank-nullity theorem here. I started with:

$ dim(v) = dim(null(Scirc T_2)) + dim(Range(Scirc T_2)) $

$=dim(null(T_1)) + dim(Range(S circ T_2)) $

$=dim(null(S circ T_2) = dim(Range(nullT_1))$

$=dim(nullT_1) + dim(Range(T_1)) $

$= dim(nullT_1), + dim(Range(S circ T_2))$

$=dim(Range(T_1)) = dim(Range(S circ T_2))$

...I'm not sure how to take this further and put everything together.

edited 2 days ago

max_zorn

3,25861228

asked Nov 13 at 17:03

Jaigus

1948

Suppose $W$ is finite dimensional and $T_1, T_2 in L(V,W).$ Prove that null($T_1$) $=$ null($T_2$) if and only if there exists and invertable operator $Sin L(W)$ such that $T_1 = Scirc T_2$.

I'm quite sure I have to make use of the rank-nullity theorem here. I started with:

$ dim(v) = dim(null(Scirc T_2)) + dim(Range(Scirc T_2)) $

$=dim(null(T_1)) + dim(Range(S circ T_2)) $

$=dim(null(S circ T_2) = dim(Range(nullT_1))$

$=dim(nullT_1) + dim(Range(T_1)) $

$= dim(nullT_1), + dim(Range(S circ T_2))$

$=dim(Range(T_1)) = dim(Range(S circ T_2))$

...I'm not sure how to take this further and put everything together.

linear-algebra vector-spaces linear-transformations proof-explanation

edited 2 days ago

max_zorn

3,25861228

asked Nov 13 at 17:03

Jaigus

1948

edited 2 days ago

max_zorn

3,25861228

asked Nov 13 at 17:03

Jaigus

1948

edited 2 days ago

max_zorn

3,25861228

edited 2 days ago

max_zorn

3,25861228

edited 2 days ago

max_zorn

3,25861228

asked Nov 13 at 17:03

Jaigus

1948

asked Nov 13 at 17:03

Jaigus

1948

asked Nov 13 at 17:03

Jaigus

1948

What have you tried so far?
– jgon
Nov 13 at 17:06

Any time you're stuck on a proof, one place to start is to circle all the key terms and write down the definitions for them. PS it's "Invertible", not "invertable".
– Acccumulation
Nov 13 at 17:13

add a comment |

What have you tried so far?
– jgon
Nov 13 at 17:06

Any time you're stuck on a proof, one place to start is to circle all the key terms and write down the definitions for them. PS it's "Invertible", not "invertable".
– Acccumulation
Nov 13 at 17:13

What have you tried so far?
– jgon
Nov 13 at 17:06

Any time you're stuck on a proof, one place to start is to circle all the key terms and write down the definitions for them. PS it's "Invertible", not "invertable".
– Acccumulation
Nov 13 at 17:13

add a comment |

1 Answer
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We know that multiplication by an invertible matrix don't change the rank of a matrix,

$Longrightarrow rk(T_1) = rk(SoT_2) = rk(T_2)$

$Longrightarrow null(T_1) = dim(V)-rk(T_1)$

$=dim(V) - rk(T_2)$

$=null(T_2)$

$therefore RHSLongrightarrow LHS$

Hope it helps:)

edited Nov 13 at 17:40

answered Nov 13 at 17:22

Crazy for maths

4948

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

We know that multiplication by an invertible matrix don't change the rank of a matrix,

$Longrightarrow rk(T_1) = rk(SoT_2) = rk(T_2)$

$Longrightarrow null(T_1) = dim(V)-rk(T_1)$

$=dim(V) - rk(T_2)$

$=null(T_2)$

$therefore RHSLongrightarrow LHS$

Hope it helps:)

edited Nov 13 at 17:40

answered Nov 13 at 17:22

Crazy for maths

4948

add a comment |

up vote
0
down vote

We know that multiplication by an invertible matrix don't change the rank of a matrix,

$Longrightarrow rk(T_1) = rk(SoT_2) = rk(T_2)$

$Longrightarrow null(T_1) = dim(V)-rk(T_1)$

$=dim(V) - rk(T_2)$

$=null(T_2)$

$therefore RHSLongrightarrow LHS$

Hope it helps:)

edited Nov 13 at 17:40

answered Nov 13 at 17:22

Crazy for maths

4948

add a comment |

up vote
0
down vote

We know that multiplication by an invertible matrix don't change the rank of a matrix,

$Longrightarrow rk(T_1) = rk(SoT_2) = rk(T_2)$

$Longrightarrow null(T_1) = dim(V)-rk(T_1)$

$=dim(V) - rk(T_2)$

$=null(T_2)$

$therefore RHSLongrightarrow LHS$

Hope it helps:)

edited Nov 13 at 17:40

answered Nov 13 at 17:22

Crazy for maths

4948

We know that multiplication by an invertible matrix don't change the rank of a matrix,

$Longrightarrow rk(T_1) = rk(SoT_2) = rk(T_2)$

$Longrightarrow null(T_1) = dim(V)-rk(T_1)$

$=dim(V) - rk(T_2)$

$=null(T_2)$

$therefore RHSLongrightarrow LHS$

Hope it helps:)

edited Nov 13 at 17:40

answered Nov 13 at 17:22

Crazy for maths

4948

edited Nov 13 at 17:40

answered Nov 13 at 17:22

Crazy for maths

4948

answered Nov 13 at 17:22

Crazy for maths

4948

answered Nov 13 at 17:22

Crazy for maths

4948

add a comment |

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