Bounds for the rank of the sum of two linear maps
$begingroup$
The following is Exercise 3.15 from the German textbook Lineare Algebra by Hans-Joachim Kowalsky and Gerhard O. Michler:
Let $varphi$ and $psi$ two linear maps from a finite-dimensional vector space $V$ to a finite-dimensional vector space $W$ with $dim{varphi(V)} = m$ and $dim{psi(V)} = n$. Show that $$|m-n| leq operatorname{rk}(varphi + psi) leq m + n.$$
Progress and Approaches:
- The second inequality can be shown by using the fact that the image of $varphi + psi$ is contained in the sum of the images of $varphi$ and $psi$ and the dimension formula for subspaces.
- However, the other direction does not turn to work as smoothly. I tried to consider the kernels of $varphi$, $psi$ and $varphi+psi$ with the hope that I could apply the Rank-Nullity-Theorem and the dimension formula from before. But it turns out that there is no relationship between these.
I also do not know how to acquire the absolute value of $m-n$ in the end. This must mean that one needs to assume that one image is larger than the other. But I do not see how this is supposed to work.
Could you please help me with this exercise? Thank you in advance!
linear-algebra inequality vector-spaces linear-transformations
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
$endgroup$
add a comment |
$begingroup$
The following is Exercise 3.15 from the German textbook Lineare Algebra by Hans-Joachim Kowalsky and Gerhard O. Michler:
Let $varphi$ and $psi$ two linear maps from a finite-dimensional vector space $V$ to a finite-dimensional vector space $W$ with $dim{varphi(V)} = m$ and $dim{psi(V)} = n$. Show that $$|m-n| leq operatorname{rk}(varphi + psi) leq m + n.$$
Progress and Approaches:
- The second inequality can be shown by using the fact that the image of $varphi + psi$ is contained in the sum of the images of $varphi$ and $psi$ and the dimension formula for subspaces.
- However, the other direction does not turn to work as smoothly. I tried to consider the kernels of $varphi$, $psi$ and $varphi+psi$ with the hope that I could apply the Rank-Nullity-Theorem and the dimension formula from before. But it turns out that there is no relationship between these.
I also do not know how to acquire the absolute value of $m-n$ in the end. This must mean that one needs to assume that one image is larger than the other. But I do not see how this is supposed to work.
Could you please help me with this exercise? Thank you in advance!
linear-algebra inequality vector-spaces linear-transformations
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
$endgroup$
add a comment |
$begingroup$
The following is Exercise 3.15 from the German textbook Lineare Algebra by Hans-Joachim Kowalsky and Gerhard O. Michler:
Let $varphi$ and $psi$ two linear maps from a finite-dimensional vector space $V$ to a finite-dimensional vector space $W$ with $dim{varphi(V)} = m$ and $dim{psi(V)} = n$. Show that $$|m-n| leq operatorname{rk}(varphi + psi) leq m + n.$$
Progress and Approaches:
- The second inequality can be shown by using the fact that the image of $varphi + psi$ is contained in the sum of the images of $varphi$ and $psi$ and the dimension formula for subspaces.
- However, the other direction does not turn to work as smoothly. I tried to consider the kernels of $varphi$, $psi$ and $varphi+psi$ with the hope that I could apply the Rank-Nullity-Theorem and the dimension formula from before. But it turns out that there is no relationship between these.
I also do not know how to acquire the absolute value of $m-n$ in the end. This must mean that one needs to assume that one image is larger than the other. But I do not see how this is supposed to work.
Could you please help me with this exercise? Thank you in advance!
linear-algebra inequality vector-spaces linear-transformations
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
$endgroup$
The following is Exercise 3.15 from the German textbook Lineare Algebra by Hans-Joachim Kowalsky and Gerhard O. Michler:
Let $varphi$ and $psi$ two linear maps from a finite-dimensional vector space $V$ to a finite-dimensional vector space $W$ with $dim{varphi(V)} = m$ and $dim{psi(V)} = n$. Show that $$|m-n| leq operatorname{rk}(varphi + psi) leq m + n.$$
Progress and Approaches:
- The second inequality can be shown by using the fact that the image of $varphi + psi$ is contained in the sum of the images of $varphi$ and $psi$ and the dimension formula for subspaces.
- However, the other direction does not turn to work as smoothly. I tried to consider the kernels of $varphi$, $psi$ and $varphi+psi$ with the hope that I could apply the Rank-Nullity-Theorem and the dimension formula from before. But it turns out that there is no relationship between these.
I also do not know how to acquire the absolute value of $m-n$ in the end. This must mean that one needs to assume that one image is larger than the other. But I do not see how this is supposed to work.
Could you please help me with this exercise? Thank you in advance!
linear-algebra inequality vector-spaces linear-transformations
linear-algebra inequality vector-spaces linear-transformations
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
edited Dec 9 '18 at 7:28
Ben Millwood
11.3k32049
11.3k32049
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
asked Dec 9 '18 at 6:18
DiglettDiglett
9701521
9701521
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
$newcommand{rk}{operatorname{rk}} newcommand{im}{operatorname{im}}$
Note that $rk varphi = rk (- varphi)$, since $im varphi = - im varphi$. Now, by the item that you have already solved,
$$
rk psi = rk (psi + varphi -varphi) leq rk(psi+varphi) + rk (-varphi) = rk(psi+varphi) + rk varphi.
$$
and so $rk psi - rk varphi leq rk (psi + varphi)$. Interchanging the roles of $psi$ and $varphi$ we get that $rk varphi - rk psi leq rk psi + varphi$ and so
$$
|m-n| = |rk psi - rk varphi| leq rk (psi + varphi).
$$
edited Dec 9 '18 at 7:04
Tengu
2,63211021
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
votes
$begingroup$
$newcommand{rk}{operatorname{rk}} newcommand{im}{operatorname{im}}$
Note that $rk varphi = rk (- varphi)$, since $im varphi = - im varphi$. Now, by the item that you have already solved,
$$
rk psi = rk (psi + varphi -varphi) leq rk(psi+varphi) + rk (-varphi) = rk(psi+varphi) + rk varphi.
$$
and so $rk psi - rk varphi leq rk (psi + varphi)$. Interchanging the roles of $psi$ and $varphi$ we get that $rk varphi - rk psi leq rk psi + varphi$ and so
$$
|m-n| = |rk psi - rk varphi| leq rk (psi + varphi).
$$
edited Dec 9 '18 at 7:04
Tengu
2,63211021
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
$endgroup$
add a comment |
$begingroup$
$newcommand{rk}{operatorname{rk}} newcommand{im}{operatorname{im}}$
Note that $rk varphi = rk (- varphi)$, since $im varphi = - im varphi$. Now, by the item that you have already solved,
$$
rk psi = rk (psi + varphi -varphi) leq rk(psi+varphi) + rk (-varphi) = rk(psi+varphi) + rk varphi.
$$
and so $rk psi - rk varphi leq rk (psi + varphi)$. Interchanging the roles of $psi$ and $varphi$ we get that $rk varphi - rk psi leq rk psi + varphi$ and so
$$
|m-n| = |rk psi - rk varphi| leq rk (psi + varphi).
$$
edited Dec 9 '18 at 7:04
Tengu
2,63211021
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
$endgroup$
add a comment |
$begingroup$
$newcommand{rk}{operatorname{rk}} newcommand{im}{operatorname{im}}$
Note that $rk varphi = rk (- varphi)$, since $im varphi = - im varphi$. Now, by the item that you have already solved,
$$
rk psi = rk (psi + varphi -varphi) leq rk(psi+varphi) + rk (-varphi) = rk(psi+varphi) + rk varphi.
$$
and so $rk psi - rk varphi leq rk (psi + varphi)$. Interchanging the roles of $psi$ and $varphi$ we get that $rk varphi - rk psi leq rk psi + varphi$ and so
$$
|m-n| = |rk psi - rk varphi| leq rk (psi + varphi).
$$
edited Dec 9 '18 at 7:04
Tengu
2,63211021
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
$endgroup$
$newcommand{rk}{operatorname{rk}} newcommand{im}{operatorname{im}}$
Note that $rk varphi = rk (- varphi)$, since $im varphi = - im varphi$. Now, by the item that you have already solved,
$$
rk psi = rk (psi + varphi -varphi) leq rk(psi+varphi) + rk (-varphi) = rk(psi+varphi) + rk varphi.
$$
and so $rk psi - rk varphi leq rk (psi + varphi)$. Interchanging the roles of $psi$ and $varphi$ we get that $rk varphi - rk psi leq rk psi + varphi$ and so
$$
|m-n| = |rk psi - rk varphi| leq rk (psi + varphi).
$$
edited Dec 9 '18 at 7:04
Tengu
2,63211021
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
edited Dec 9 '18 at 7:04
Tengu
2,63211021
edited Dec 9 '18 at 7:04
Tengu
2,63211021
edited Dec 9 '18 at 7:04
Tengu
2,63211021
2,63211021
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
answered Dec 9 '18 at 6:59
Guido A.Guido A.
7,3621730
7,3621730
add a comment |
add a comment |
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