Null Space of Sum of Two Matrices is a subset or supset of null space of one
$begingroup$
Could anyone explain how either of these can be proven? I don't see how either of these statements by themselves can be true, much less how to prove them.
$$N(A+B)⊂N(A)$$
$$N(A+B)⊃N(A)$$
linear-algebra matrices
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
$endgroup$
add a comment |
$begingroup$
Could anyone explain how either of these can be proven? I don't see how either of these statements by themselves can be true, much less how to prove them.
$$N(A+B)⊂N(A)$$
$$N(A+B)⊃N(A)$$
linear-algebra matrices
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
$endgroup$
$begingroup$
Suppose $x in N(A+B)$. This implies $(A+B)x = 0 implies Ax + Bx = 0$. Is it necessarily true that $Ax = 0$? What about the other way around, if we have $Ax = 0$, then must it be true that $(A+B)x$? In fact I don't believe either of your statements, but I do buy that $N(A) cap N(B) subset N(A+B)$.
$endgroup$
– TrostAft
Dec 10 '18 at 7:49
add a comment |
$begingroup$
Could anyone explain how either of these can be proven? I don't see how either of these statements by themselves can be true, much less how to prove them.
$$N(A+B)⊂N(A)$$
$$N(A+B)⊃N(A)$$
linear-algebra matrices
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
$endgroup$
Could anyone explain how either of these can be proven? I don't see how either of these statements by themselves can be true, much less how to prove them.
$$N(A+B)⊂N(A)$$
$$N(A+B)⊃N(A)$$
linear-algebra matrices
linear-algebra matrices
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
edited Dec 10 '18 at 7:51
Asaf Karagila♦
303k32429760
303k32429760
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
asked Dec 10 '18 at 7:42
Grant LeechGrant Leech
11
11
$begingroup$
Suppose $x in N(A+B)$. This implies $(A+B)x = 0 implies Ax + Bx = 0$. Is it necessarily true that $Ax = 0$? What about the other way around, if we have $Ax = 0$, then must it be true that $(A+B)x$? In fact I don't believe either of your statements, but I do buy that $N(A) cap N(B) subset N(A+B)$.
$endgroup$
– TrostAft
Dec 10 '18 at 7:49
add a comment |
$begingroup$
Suppose $x in N(A+B)$. This implies $(A+B)x = 0 implies Ax + Bx = 0$. Is it necessarily true that $Ax = 0$? What about the other way around, if we have $Ax = 0$, then must it be true that $(A+B)x$? In fact I don't believe either of your statements, but I do buy that $N(A) cap N(B) subset N(A+B)$.
$endgroup$
– TrostAft
Dec 10 '18 at 7:49
$begingroup$
Suppose $x in N(A+B)$. This implies $(A+B)x = 0 implies Ax + Bx = 0$. Is it necessarily true that $Ax = 0$? What about the other way around, if we have $Ax = 0$, then must it be true that $(A+B)x$? In fact I don't believe either of your statements, but I do buy that $N(A) cap N(B) subset N(A+B)$.
$endgroup$
– TrostAft
Dec 10 '18 at 7:49
$begingroup$
Suppose $x in N(A+B)$. This implies $(A+B)x = 0 implies Ax + Bx = 0$. Is it necessarily true that $Ax = 0$? What about the other way around, if we have $Ax = 0$, then must it be true that $(A+B)x$? In fact I don't believe either of your statements, but I do buy that $N(A) cap N(B) subset N(A+B)$.
$endgroup$
– TrostAft
Dec 10 '18 at 7:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First one false when $A=-B=I$. the second one is false when $A=0$ and $B=I$.
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
$begingroup$
For the first case, I think $A =I$ .
$endgroup$
– AnyAD
Dec 10 '18 at 7:51
add a comment |
$begingroup$
They will/may be true for special choice/s of $A,B $ but they are not true in general as shown in the other answer.
For the case that this is true, you'd prove the first statement fir example by taking $xin N (A+B)$ and showing that this implies also that $xin N(A)$.
That is $(A+B)x=0implies Ax=0$.
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033597%2fnull-space-of-sum-of-two-matrices-is-a-subset-or-supset-of-null-space-of-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First one false when $A=-B=I$. the second one is false when $A=0$ and $B=I$.
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
$begingroup$
For the first case, I think $A =I$ .
$endgroup$
– AnyAD
Dec 10 '18 at 7:51
add a comment |
$begingroup$
First one false when $A=-B=I$. the second one is false when $A=0$ and $B=I$.
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
$begingroup$
For the first case, I think $A =I$ .
$endgroup$
– AnyAD
Dec 10 '18 at 7:51
add a comment |
$begingroup$
First one false when $A=-B=I$. the second one is false when $A=0$ and $B=I$.
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
$endgroup$
First one false when $A=-B=I$. the second one is false when $A=0$ and $B=I$.
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
answered Dec 10 '18 at 7:47
Kavi Rama MurthyKavi Rama Murthy
55.7k42158
55.7k42158
$begingroup$
For the first case, I think $A =I$ .
$endgroup$
– AnyAD
Dec 10 '18 at 7:51
add a comment |
$begingroup$
For the first case, I think $A =I$ .
$endgroup$
– AnyAD
Dec 10 '18 at 7:51
$begingroup$
For the first case, I think $A =I$ .
$endgroup$
– AnyAD
Dec 10 '18 at 7:51
$begingroup$
For the first case, I think $A =I$ .
$endgroup$
– AnyAD
Dec 10 '18 at 7:51
add a comment |
$begingroup$
They will/may be true for special choice/s of $A,B $ but they are not true in general as shown in the other answer.
For the case that this is true, you'd prove the first statement fir example by taking $xin N (A+B)$ and showing that this implies also that $xin N(A)$.
That is $(A+B)x=0implies Ax=0$.
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
$endgroup$
add a comment |
$begingroup$
They will/may be true for special choice/s of $A,B $ but they are not true in general as shown in the other answer.
For the case that this is true, you'd prove the first statement fir example by taking $xin N (A+B)$ and showing that this implies also that $xin N(A)$.
That is $(A+B)x=0implies Ax=0$.
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
$endgroup$
add a comment |
$begingroup$
They will/may be true for special choice/s of $A,B $ but they are not true in general as shown in the other answer.
For the case that this is true, you'd prove the first statement fir example by taking $xin N (A+B)$ and showing that this implies also that $xin N(A)$.
That is $(A+B)x=0implies Ax=0$.
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
$endgroup$
They will/may be true for special choice/s of $A,B $ but they are not true in general as shown in the other answer.
For the case that this is true, you'd prove the first statement fir example by taking $xin N (A+B)$ and showing that this implies also that $xin N(A)$.
That is $(A+B)x=0implies Ax=0$.
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
edited Dec 10 '18 at 8:12
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
answered Dec 10 '18 at 7:55
AnyADAnyAD
2,098812
2,098812
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3033597%2fnull-space-of-sum-of-two-matrices-is-a-subset-or-supset-of-null-space-of-one%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Suppose $x in N(A+B)$. This implies $(A+B)x = 0 implies Ax + Bx = 0$. Is it necessarily true that $Ax = 0$? What about the other way around, if we have $Ax = 0$, then must it be true that $(A+B)x$? In fact I don't believe either of your statements, but I do buy that $N(A) cap N(B) subset N(A+B)$.
$endgroup$
– TrostAft
Dec 10 '18 at 7:49