Proof of $dim ker (f circ g) le dim ker (f) + dim ker (g)$ The 2019 Stack Overflow Developer Survey Results Are InWhy is $dim(V/bigcap_i=1^nker(f_i))leq n$ for linear functionals $f_i$?$dim(mboxim(f)) = dim(U)$ and $dim(ker(g)) = dim(V)-dim(U)$$ker(T) subseteqker(S)$ implies the exist some $H$ s.t. $Hcirc T=S$$V$ be a vector space , $T:V to V$ be a linear operator , then is $ker (T) cap R(T) cong R(T)/R(T^2) $?$T^2=O$ and if $dim V=2$, $T^2=O$ ($Tneq O$), then $ker T=mathrmIm ; T$$renewcommandImtextIm$Prove that $dim(ker(T))=dim(ker(T^2))$ if $dim(Im(T))=dim(Im(T^2))$$dim(kervarphicapkerpsi)=n-2$ proofRank-Nullity Theorem: Struggling to Reconcile $rm Rank(A) +rm Nul(A) = n$ and $dim(V) = dim(rm Im(f)) +dim(rm ker(f))$.Prove/Disprove: $S$ not invertible and $dim(ker(T)+dim(ker(S)geqdim(V)$Linear transformations $f, g:Vto V$ with properties $f circ f = g circ g = 0_V$ and $f circ g + g circ f = 1_V$ imply $dim V$ is even?
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Proof of $dim ker (f circ g) le dim ker (f) + dim ker (g)$
The 2019 Stack Overflow Developer Survey Results Are InWhy is $dim(V/bigcap_i=1^nker(f_i))leq n$ for linear functionals $f_i$?$dim(mboxim(f)) = dim(U)$ and $dim(ker(g)) = dim(V)-dim(U)$$ker(T) subseteqker(S)$ implies the exist some $H$ s.t. $Hcirc T=S$$V$ be a vector space , $T:V to V$ be a linear operator , then is $ker (T) cap R(T) cong R(T)/R(T^2) $?$T^2=O$ and if $dim V=2$, $T^2=O$ ($Tneq O$), then $ker T=mathrmIm ; T$$renewcommandImtextIm$Prove that $dim(ker(T))=dim(ker(T^2))$ if $dim(Im(T))=dim(Im(T^2))$$dim(kervarphicapkerpsi)=n-2$ proofRank-Nullity Theorem: Struggling to Reconcile $rm Rank(A) +rm Nul(A) = n$ and $dim(V) = dim(rm Im(f)) +dim(rm ker(f))$.Prove/Disprove: $S$ not invertible and $dim(ker(T)+dim(ker(S)geqdim(V)$Linear transformations $f, g:Vto V$ with properties $f circ f = g circ g = 0_V$ and $f circ g + g circ f = 1_V$ imply $dim V$ is even?
$begingroup$
In my lecture notes I have an inequality for any linear transformations $f: X to X$, $g: X to X$ for any finite-dimensional vector space $X$:
$$dim ker (f circ g) le dim ker (f) + dim ker (g)$$
Can anyone help me with proofing it?
Of course I know basic equality: $dim ker f + dim im f = dim X$, but I do not know any inequality, which involves dimension of transpormations, therefore I do not even know where to start.
Thanks in advance
linear-algebra vector-spaces linear-transformations
asked Jan 10 '16 at 22:00
AndyAndy
1606
$endgroup$
add a comment |
$begingroup$
In my lecture notes I have an inequality for any linear transformations $f: X to X$, $g: X to X$ for any finite-dimensional vector space $X$:
$$dim ker (f circ g) le dim ker (f) + dim ker (g)$$
Can anyone help me with proofing it?
Of course I know basic equality: $dim ker f + dim im f = dim X$, but I do not know any inequality, which involves dimension of transpormations, therefore I do not even know where to start.
Thanks in advance
linear-algebra vector-spaces linear-transformations
asked Jan 10 '16 at 22:00
AndyAndy
1606
$endgroup$
$begingroup$
Can you write expression for $ker(f circ g) $?
$endgroup$
– shilov
Jan 10 '16 at 22:03
add a comment |
$begingroup$
In my lecture notes I have an inequality for any linear transformations $f: X to X$, $g: X to X$ for any finite-dimensional vector space $X$:
$$dim ker (f circ g) le dim ker (f) + dim ker (g)$$
Can anyone help me with proofing it?
Of course I know basic equality: $dim ker f + dim im f = dim X$, but I do not know any inequality, which involves dimension of transpormations, therefore I do not even know where to start.
Thanks in advance
linear-algebra vector-spaces linear-transformations
asked Jan 10 '16 at 22:00
AndyAndy
1606
$endgroup$
In my lecture notes I have an inequality for any linear transformations $f: X to X$, $g: X to X$ for any finite-dimensional vector space $X$:
$$dim ker (f circ g) le dim ker (f) + dim ker (g)$$
Can anyone help me with proofing it?
Of course I know basic equality: $dim ker f + dim im f = dim X$, but I do not know any inequality, which involves dimension of transpormations, therefore I do not even know where to start.
Thanks in advance
linear-algebra vector-spaces linear-transformations
linear-algebra vector-spaces linear-transformations
asked Jan 10 '16 at 22:00
AndyAndy
1606
asked Jan 10 '16 at 22:00
AndyAndy
1606
asked Jan 10 '16 at 22:00
AndyAndy
1606
asked Jan 10 '16 at 22:00
AndyAndy
1606
asked Jan 10 '16 at 22:00
AndyAndy
1606
1606
$begingroup$
Can you write expression for $ker(f circ g) $?
$endgroup$
– shilov
Jan 10 '16 at 22:03
add a comment |
$begingroup$
Can you write expression for $ker(f circ g) $?
$endgroup$
– shilov
Jan 10 '16 at 22:03
$begingroup$
Can you write expression for $ker(f circ g) $?
$endgroup$
– shilov
Jan 10 '16 at 22:03
$begingroup$
Can you write expression for $ker(f circ g) $?
$endgroup$
– shilov
Jan 10 '16 at 22:03
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $g:Vto W$ and $f:Wto X$. So, $fcirc g:Vto X$. Note that $ker(fcirc g)=g^-1(ker f)$. Let $V=g^-1(ker f)$. This is a subspace of $V$, and $g(V')subseteqker f$. Note that $ker gsubseteq V'$.
$$operatornamedimV=operatornamedim(ker g)+dim(g(V')),$$
and
$$dim(g(V'))=dim(ker f|_g(V'))+dim(f(g(V')).$$
Since $V'=ker(fcirc g)$, $dim(f(g(V'))=0$, and $dim(f|_g(V'))=dim(ker fcap g(V'))ledimker f$, we get the desired inquality.
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
$endgroup$
$begingroup$
Shouln't it be $g^-1(ker f)$.? Also, where is $f$ described?
$endgroup$
– Bernard
Jan 10 '16 at 22:13
$begingroup$
Yes, you're right. Silly me.
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:14
$begingroup$
You're welcome!
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:16
$begingroup$
Ok, at the first look I though that I understand, but now I am not sure, how do we know $ker(f) = g(V)$? (so that later we can say $ker(f) = im(g)$ since $g(V) =im (g)$)
$endgroup$
– Andy
Jan 11 '16 at 2:30
$begingroup$
You're right, I was a bit careless. I'll fix my answer.
$endgroup$
– Tim Raczkowski
Jan 11 '16 at 3:29
add a comment |
$begingroup$
I know this is an old post but I think based on Tim Raczkowski's idea, we can construct a more succinct proof:
Again, let $~g: V to W,~ f: W to X$ with $ V' = textker(fg) subseteq V $.
The important point to note is that $ V' = textker(fg) = g^-1(textkerf) $ so that $ g(V') subseteq textker(f) $. Hence, $textdim(g(V')) leq textnullity(f)$. By the Dimension Theorem, $$ textdim(V') = textdim(textker~ g|_V') + textdim(textIm~g|_V').$$
Now, $textdim(textIm~g|_V') = textdim(g(V'))$ and from above, $textdim(g(V')) leq textnullity(f)$.
Thus, $$ textdim(V') leq textdim(textker~ g|_V') + textnullity(f). $$
Since $ textker~ g subseteq V'$, $ textker~g|_V' = textker ~g $ which implies $textdim(textker~g|_V' ) = textdim(textker ~g)$.
Finally, $$ textnullity~fg = textdim(V') leq textnullity~g + textnullity~f $$
as needed.
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Let $g:Vto W$ and $f:Wto X$. So, $fcirc g:Vto X$. Note that $ker(fcirc g)=g^-1(ker f)$. Let $V=g^-1(ker f)$. This is a subspace of $V$, and $g(V')subseteqker f$. Note that $ker gsubseteq V'$.
$$operatornamedimV=operatornamedim(ker g)+dim(g(V')),$$
and
$$dim(g(V'))=dim(ker f|_g(V'))+dim(f(g(V')).$$
Since $V'=ker(fcirc g)$, $dim(f(g(V'))=0$, and $dim(f|_g(V'))=dim(ker fcap g(V'))ledimker f$, we get the desired inquality.
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
$endgroup$
$begingroup$
Shouln't it be $g^-1(ker f)$.? Also, where is $f$ described?
$endgroup$
– Bernard
Jan 10 '16 at 22:13
$begingroup$
Yes, you're right. Silly me.
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:14
$begingroup$
You're welcome!
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:16
$begingroup$
Ok, at the first look I though that I understand, but now I am not sure, how do we know $ker(f) = g(V)$? (so that later we can say $ker(f) = im(g)$ since $g(V) =im (g)$)
$endgroup$
– Andy
Jan 11 '16 at 2:30
$begingroup$
You're right, I was a bit careless. I'll fix my answer.
$endgroup$
– Tim Raczkowski
Jan 11 '16 at 3:29
add a comment |
$begingroup$
Let $g:Vto W$ and $f:Wto X$. So, $fcirc g:Vto X$. Note that $ker(fcirc g)=g^-1(ker f)$. Let $V=g^-1(ker f)$. This is a subspace of $V$, and $g(V')subseteqker f$. Note that $ker gsubseteq V'$.
$$operatornamedimV=operatornamedim(ker g)+dim(g(V')),$$
and
$$dim(g(V'))=dim(ker f|_g(V'))+dim(f(g(V')).$$
Since $V'=ker(fcirc g)$, $dim(f(g(V'))=0$, and $dim(f|_g(V'))=dim(ker fcap g(V'))ledimker f$, we get the desired inquality.
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
$endgroup$
$begingroup$
Shouln't it be $g^-1(ker f)$.? Also, where is $f$ described?
$endgroup$
– Bernard
Jan 10 '16 at 22:13
$begingroup$
Yes, you're right. Silly me.
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:14
$begingroup$
You're welcome!
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:16
$begingroup$
Ok, at the first look I though that I understand, but now I am not sure, how do we know $ker(f) = g(V)$? (so that later we can say $ker(f) = im(g)$ since $g(V) =im (g)$)
$endgroup$
– Andy
Jan 11 '16 at 2:30
$begingroup$
You're right, I was a bit careless. I'll fix my answer.
$endgroup$
– Tim Raczkowski
Jan 11 '16 at 3:29
add a comment |
$begingroup$
Let $g:Vto W$ and $f:Wto X$. So, $fcirc g:Vto X$. Note that $ker(fcirc g)=g^-1(ker f)$. Let $V=g^-1(ker f)$. This is a subspace of $V$, and $g(V')subseteqker f$. Note that $ker gsubseteq V'$.
$$operatornamedimV=operatornamedim(ker g)+dim(g(V')),$$
and
$$dim(g(V'))=dim(ker f|_g(V'))+dim(f(g(V')).$$
Since $V'=ker(fcirc g)$, $dim(f(g(V'))=0$, and $dim(f|_g(V'))=dim(ker fcap g(V'))ledimker f$, we get the desired inquality.
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
$endgroup$
Let $g:Vto W$ and $f:Wto X$. So, $fcirc g:Vto X$. Note that $ker(fcirc g)=g^-1(ker f)$. Let $V=g^-1(ker f)$. This is a subspace of $V$, and $g(V')subseteqker f$. Note that $ker gsubseteq V'$.
$$operatornamedimV=operatornamedim(ker g)+dim(g(V')),$$
and
$$dim(g(V'))=dim(ker f|_g(V'))+dim(f(g(V')).$$
Since $V'=ker(fcirc g)$, $dim(f(g(V'))=0$, and $dim(f|_g(V'))=dim(ker fcap g(V'))ledimker f$, we get the desired inquality.
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
edited Jan 11 '16 at 5:06
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
answered Jan 10 '16 at 22:07
Tim RaczkowskiTim Raczkowski
17.4k21244
17.4k21244
$begingroup$
Shouln't it be $g^-1(ker f)$.? Also, where is $f$ described?
$endgroup$
– Bernard
Jan 10 '16 at 22:13
$begingroup$
Yes, you're right. Silly me.
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:14
$begingroup$
You're welcome!
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:16
$begingroup$
Ok, at the first look I though that I understand, but now I am not sure, how do we know $ker(f) = g(V)$? (so that later we can say $ker(f) = im(g)$ since $g(V) =im (g)$)
$endgroup$
– Andy
Jan 11 '16 at 2:30
$begingroup$
You're right, I was a bit careless. I'll fix my answer.
$endgroup$
– Tim Raczkowski
Jan 11 '16 at 3:29
add a comment |
$begingroup$
Shouln't it be $g^-1(ker f)$.? Also, where is $f$ described?
$endgroup$
– Bernard
Jan 10 '16 at 22:13
$begingroup$
Yes, you're right. Silly me.
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:14
$begingroup$
You're welcome!
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:16
$begingroup$
Ok, at the first look I though that I understand, but now I am not sure, how do we know $ker(f) = g(V)$? (so that later we can say $ker(f) = im(g)$ since $g(V) =im (g)$)
$endgroup$
– Andy
Jan 11 '16 at 2:30
$begingroup$
You're right, I was a bit careless. I'll fix my answer.
$endgroup$
– Tim Raczkowski
Jan 11 '16 at 3:29
$begingroup$
Shouln't it be $g^-1(ker f)$.? Also, where is $f$ described?
$endgroup$
– Bernard
Jan 10 '16 at 22:13
$begingroup$
Shouln't it be $g^-1(ker f)$.? Also, where is $f$ described?
$endgroup$
– Bernard
Jan 10 '16 at 22:13
$begingroup$
Yes, you're right. Silly me.
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:14
$begingroup$
Yes, you're right. Silly me.
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:14
$begingroup$
You're welcome!
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:16
$begingroup$
You're welcome!
$endgroup$
– Tim Raczkowski
Jan 10 '16 at 22:16
$begingroup$
Ok, at the first look I though that I understand, but now I am not sure, how do we know $ker(f) = g(V)$? (so that later we can say $ker(f) = im(g)$ since $g(V) =im (g)$)
$endgroup$
– Andy
Jan 11 '16 at 2:30
$begingroup$
Ok, at the first look I though that I understand, but now I am not sure, how do we know $ker(f) = g(V)$? (so that later we can say $ker(f) = im(g)$ since $g(V) =im (g)$)
$endgroup$
– Andy
Jan 11 '16 at 2:30
$begingroup$
You're right, I was a bit careless. I'll fix my answer.
$endgroup$
– Tim Raczkowski
Jan 11 '16 at 3:29
$begingroup$
You're right, I was a bit careless. I'll fix my answer.
$endgroup$
– Tim Raczkowski
Jan 11 '16 at 3:29
add a comment |
$begingroup$
I know this is an old post but I think based on Tim Raczkowski's idea, we can construct a more succinct proof:
Again, let $~g: V to W,~ f: W to X$ with $ V' = textker(fg) subseteq V $.
The important point to note is that $ V' = textker(fg) = g^-1(textkerf) $ so that $ g(V') subseteq textker(f) $. Hence, $textdim(g(V')) leq textnullity(f)$. By the Dimension Theorem, $$ textdim(V') = textdim(textker~ g|_V') + textdim(textIm~g|_V').$$
Now, $textdim(textIm~g|_V') = textdim(g(V'))$ and from above, $textdim(g(V')) leq textnullity(f)$.
Thus, $$ textdim(V') leq textdim(textker~ g|_V') + textnullity(f). $$
Since $ textker~ g subseteq V'$, $ textker~g|_V' = textker ~g $ which implies $textdim(textker~g|_V' ) = textdim(textker ~g)$.
Finally, $$ textnullity~fg = textdim(V') leq textnullity~g + textnullity~f $$
as needed.
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
$endgroup$
add a comment |
$begingroup$
I know this is an old post but I think based on Tim Raczkowski's idea, we can construct a more succinct proof:
Again, let $~g: V to W,~ f: W to X$ with $ V' = textker(fg) subseteq V $.
The important point to note is that $ V' = textker(fg) = g^-1(textkerf) $ so that $ g(V') subseteq textker(f) $. Hence, $textdim(g(V')) leq textnullity(f)$. By the Dimension Theorem, $$ textdim(V') = textdim(textker~ g|_V') + textdim(textIm~g|_V').$$
Now, $textdim(textIm~g|_V') = textdim(g(V'))$ and from above, $textdim(g(V')) leq textnullity(f)$.
Thus, $$ textdim(V') leq textdim(textker~ g|_V') + textnullity(f). $$
Since $ textker~ g subseteq V'$, $ textker~g|_V' = textker ~g $ which implies $textdim(textker~g|_V' ) = textdim(textker ~g)$.
Finally, $$ textnullity~fg = textdim(V') leq textnullity~g + textnullity~f $$
as needed.
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
$endgroup$
add a comment |
$begingroup$
I know this is an old post but I think based on Tim Raczkowski's idea, we can construct a more succinct proof:
Again, let $~g: V to W,~ f: W to X$ with $ V' = textker(fg) subseteq V $.
The important point to note is that $ V' = textker(fg) = g^-1(textkerf) $ so that $ g(V') subseteq textker(f) $. Hence, $textdim(g(V')) leq textnullity(f)$. By the Dimension Theorem, $$ textdim(V') = textdim(textker~ g|_V') + textdim(textIm~g|_V').$$
Now, $textdim(textIm~g|_V') = textdim(g(V'))$ and from above, $textdim(g(V')) leq textnullity(f)$.
Thus, $$ textdim(V') leq textdim(textker~ g|_V') + textnullity(f). $$
Since $ textker~ g subseteq V'$, $ textker~g|_V' = textker ~g $ which implies $textdim(textker~g|_V' ) = textdim(textker ~g)$.
Finally, $$ textnullity~fg = textdim(V') leq textnullity~g + textnullity~f $$
as needed.
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
$endgroup$
I know this is an old post but I think based on Tim Raczkowski's idea, we can construct a more succinct proof:
Again, let $~g: V to W,~ f: W to X$ with $ V' = textker(fg) subseteq V $.
The important point to note is that $ V' = textker(fg) = g^-1(textkerf) $ so that $ g(V') subseteq textker(f) $. Hence, $textdim(g(V')) leq textnullity(f)$. By the Dimension Theorem, $$ textdim(V') = textdim(textker~ g|_V') + textdim(textIm~g|_V').$$
Now, $textdim(textIm~g|_V') = textdim(g(V'))$ and from above, $textdim(g(V')) leq textnullity(f)$.
Thus, $$ textdim(V') leq textdim(textker~ g|_V') + textnullity(f). $$
Since $ textker~ g subseteq V'$, $ textker~g|_V' = textker ~g $ which implies $textdim(textker~g|_V' ) = textdim(textker ~g)$.
Finally, $$ textnullity~fg = textdim(V') leq textnullity~g + textnullity~f $$
as needed.
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
answered Mar 22 at 19:27
Khoa taKhoa ta
23719
23719
add a comment |
add a comment |
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$begingroup$
Can you write expression for $ker(f circ g) $?
$endgroup$
– shilov
Jan 10 '16 at 22:03