Mixing
How can I find the rank of a matrix $A$ when the solution sets to $Ax=b$ is given?
$begingroup$
Suppose the complete set of solutions to $Ax = b$ where $b = [2, 4, 2]^T$ is given by
$$
x = [2, 0, 0]^T + c[1, 1, 0]^T + d[0, 0, 1]^T.
$$
This is from a YouToube Video of Gilbert Strang's lecture on Linear Algebra. He said that the rank of the matrix $A$ was $1$. I understand that $b$ is a column vector so its rank is $1$. How did the students identify the dimension of the null space of $A$ to be $2$? This is something that baffles me.
$$A = begin{bmatrix}1 & -1 & 0 \ 2 & -2 & 0 \ 1 & -1 & 0 end{bmatrix}$$
Firstly, why is it that if $[0, 0, 1]^T$ is in the null space, the third column is a zero vector.
Assuming $c = 1$, $d = 1,$
and $A$ is the matrix as proposed, in the first row of $B$, $1times 2 - 1 times 1 + 1 times 0$ gives $1$ instead of $2.$
linear-algebra
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
asked Jan 3 '17 at 14:03
acemineraceminer
230112
$endgroup$
add a comment |
$begingroup$
Suppose the complete set of solutions to $Ax = b$ where $b = [2, 4, 2]^T$ is given by
$$
x = [2, 0, 0]^T + c[1, 1, 0]^T + d[0, 0, 1]^T.
$$
This is from a YouToube Video of Gilbert Strang's lecture on Linear Algebra. He said that the rank of the matrix $A$ was $1$. I understand that $b$ is a column vector so its rank is $1$. How did the students identify the dimension of the null space of $A$ to be $2$? This is something that baffles me.
$$A = begin{bmatrix}1 & -1 & 0 \ 2 & -2 & 0 \ 1 & -1 & 0 end{bmatrix}$$
Firstly, why is it that if $[0, 0, 1]^T$ is in the null space, the third column is a zero vector.
Assuming $c = 1$, $d = 1,$
and $A$ is the matrix as proposed, in the first row of $B$, $1times 2 - 1 times 1 + 1 times 0$ gives $1$ instead of $2.$
linear-algebra
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
asked Jan 3 '17 at 14:03
acemineraceminer
230112
$endgroup$
$begingroup$
Your question is about the matrix $A$. Would you write it down in your post instead of giving the video link?
$endgroup$
– Jack
Jan 3 '17 at 14:28
$begingroup$
@Jack this is what is mentioned. The A matrix is not given
$endgroup$
– aceminer
Jan 3 '17 at 15:05
add a comment |
$begingroup$
Suppose the complete set of solutions to $Ax = b$ where $b = [2, 4, 2]^T$ is given by
$$
x = [2, 0, 0]^T + c[1, 1, 0]^T + d[0, 0, 1]^T.
$$
This is from a YouToube Video of Gilbert Strang's lecture on Linear Algebra. He said that the rank of the matrix $A$ was $1$. I understand that $b$ is a column vector so its rank is $1$. How did the students identify the dimension of the null space of $A$ to be $2$? This is something that baffles me.
$$A = begin{bmatrix}1 & -1 & 0 \ 2 & -2 & 0 \ 1 & -1 & 0 end{bmatrix}$$
Firstly, why is it that if $[0, 0, 1]^T$ is in the null space, the third column is a zero vector.
Assuming $c = 1$, $d = 1,$
and $A$ is the matrix as proposed, in the first row of $B$, $1times 2 - 1 times 1 + 1 times 0$ gives $1$ instead of $2.$
linear-algebra
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
asked Jan 3 '17 at 14:03
acemineraceminer
230112
$endgroup$
Suppose the complete set of solutions to $Ax = b$ where $b = [2, 4, 2]^T$ is given by
$$
x = [2, 0, 0]^T + c[1, 1, 0]^T + d[0, 0, 1]^T.
$$
This is from a YouToube Video of Gilbert Strang's lecture on Linear Algebra. He said that the rank of the matrix $A$ was $1$. I understand that $b$ is a column vector so its rank is $1$. How did the students identify the dimension of the null space of $A$ to be $2$? This is something that baffles me.
$$A = begin{bmatrix}1 & -1 & 0 \ 2 & -2 & 0 \ 1 & -1 & 0 end{bmatrix}$$
Firstly, why is it that if $[0, 0, 1]^T$ is in the null space, the third column is a zero vector.
Assuming $c = 1$, $d = 1,$
and $A$ is the matrix as proposed, in the first row of $B$, $1times 2 - 1 times 1 + 1 times 0$ gives $1$ instead of $2.$
linear-algebra
linear-algebra
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
asked Jan 3 '17 at 14:03
acemineraceminer
230112
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
asked Jan 3 '17 at 14:03
acemineraceminer
230112
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
edited Jan 4 '17 at 13:04
Antoni Parellada
3,11421341
3,11421341
asked Jan 3 '17 at 14:03
acemineraceminer
230112
asked Jan 3 '17 at 14:03
acemineraceminer
230112
asked Jan 3 '17 at 14:03
acemineraceminer
230112
230112
$begingroup$
Your question is about the matrix $A$. Would you write it down in your post instead of giving the video link?
$endgroup$
– Jack
Jan 3 '17 at 14:28
$begingroup$
@Jack this is what is mentioned. The A matrix is not given
$endgroup$
– aceminer
Jan 3 '17 at 15:05
add a comment |
$begingroup$
Your question is about the matrix $A$. Would you write it down in your post instead of giving the video link?
$endgroup$
– Jack
Jan 3 '17 at 14:28
$begingroup$
@Jack this is what is mentioned. The A matrix is not given
$endgroup$
– aceminer
Jan 3 '17 at 15:05
$begingroup$
Your question is about the matrix $A$. Would you write it down in your post instead of giving the video link?
$endgroup$
– Jack
Jan 3 '17 at 14:28
$begingroup$
Your question is about the matrix $A$. Would you write it down in your post instead of giving the video link?
$endgroup$
– Jack
Jan 3 '17 at 14:28
$begingroup$
@Jack this is what is mentioned. The A matrix is not given
$endgroup$
– aceminer
Jan 3 '17 at 15:05
$begingroup$
@Jack this is what is mentioned. The A matrix is not given
$endgroup$
– aceminer
Jan 3 '17 at 15:05
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
I wonder if the easiest way to look at this is a follows:
$[2,0,0]^top$ is a particular solution to $Ax=b$. Why? Because it is true that it doesn't matter what $c$ and $d$ values we pick: for any $c$ and $d$ we get a vector, $x = [2, 0, 0]^top + c[1, 1, 0]^top + d[0, 0, 1]^top$ that satisfies $Ax=b$. Therefore, we can pick $c=0$ and $d=0$, and we end up with $x = [2, 0, 0]^top.$
This is tantamount to saying that $c[1, 1, 0]^top + d[0, 0, 1]^top$ are really non-contributory. Or in other words, that they are in the null space of $A.$ If a system of three equations with three unknowns can be solved in more than one way is because the system is under-determined, and there are free variables.
$c[1, 1, 0]^top + d[0, 0, 1]^top$ represents all linear combinations of two linearly independent vectors in the null space, and any vector space (or subspace) observes closure under addition and scalar multiplication. Hence, $c[1, 1, 0]^top + d[0, 0, 1]^top=0.$
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
$endgroup$
$begingroup$
Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question
$endgroup$
– aceminer
Jan 4 '17 at 12:53
2
$begingroup$
$[color{blue}{0},color{red}{0},1]^top$ is in the null space of $A$ because if you take $color{blue}{0}times begin{bmatrix}1\2\1end{bmatrix}+color{red}{0}times begin{bmatrix}-1\-2\-1end{bmatrix}+1times begin{bmatrix}0\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:09
1
$begingroup$
$Ax$ where in this case $x=[color{blue}{0},color{red}{0},1]^top=begin{bmatrix}color{blue}{0}\color{red}{0}\1end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $color{blue}{0}times text{ first column of A} + color{red}{0}times text{ second column of A} +1times text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $bf 0$ vector will result in the $bf 0$ vector.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:16
1
$begingroup$
I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:21
1
$begingroup$
Look up how matrix vector multiplication works and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get?
$endgroup$
– Piotr Benedysiuk
Jan 4 '17 at 13:21
|
show 5 more comments
$begingroup$
From the rank-nullity theorem, the following relation satisfies for an $m times n$ matrix.
$$text{rank}(A)+text{nullity}(A)=n$$
The dimension of the null space is called the nullity of A.
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
$endgroup$
$begingroup$
Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1?
$endgroup$
– aceminer
Jan 3 '17 at 14:16
$begingroup$
See here.
$endgroup$
– projectilemotion
Jan 3 '17 at 15:51
add a comment |
$begingroup$
The homogeneous equation $A x = 0$ has as set of solutions $mathcal{O}$ the set of differences of the solutions of the original equation (as $A x = A y = b$ implies $A (x - y) = 0$, and conversely if $Ax = b$ and $A u = 0$ then $A (x+u) = b$), so this set is
$$
mathcal{O} = { a,[1, 1, 0]^T + b,[0, 0, 1]^T : a, b in mathbb{R}},
$$
a $2$-dimensional space.
Therefore the nullity of $A$ is $2$ and the rank is $3 - 2 = 1$.
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
$endgroup$
$begingroup$
how did you derive the last set of a and b? It seems likes there's a huge leap for me
$endgroup$
– aceminer
Jan 3 '17 at 23:24
$begingroup$
Just take two of your solutions, and their difference.
$endgroup$
– Andreas Caranti
Jan 4 '17 at 7:18
add a comment |
$begingroup$
Watch the video from 10:38 again.
First of all he points out that the size of the matrix $A$ is $3times 3$. Secondly, the assumption of the complete sets of solutions tells you that
$$
Null(A)=2.
$$
Now the rank-nullity theorem tells you the rank of $A$ is given by
$$
Rank(A)=3-Null(A)=1.
$$
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
$endgroup$
$begingroup$
at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions
$endgroup$
– aceminer
Jan 3 '17 at 23:22
$begingroup$
@aceminer: I have edited my answer.
$endgroup$
– Jack
Jan 3 '17 at 23:31
$begingroup$
At 10:53, he said that "and I give you the 'complete solutions'".
$endgroup$
– Jack
Jan 3 '17 at 23:32
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%2f2081927%2fhow-can-i-find-the-rank-of-a-matrix-a-when-the-solution-sets-to-ax-b-is-give%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I wonder if the easiest way to look at this is a follows:
$[2,0,0]^top$ is a particular solution to $Ax=b$. Why? Because it is true that it doesn't matter what $c$ and $d$ values we pick: for any $c$ and $d$ we get a vector, $x = [2, 0, 0]^top + c[1, 1, 0]^top + d[0, 0, 1]^top$ that satisfies $Ax=b$. Therefore, we can pick $c=0$ and $d=0$, and we end up with $x = [2, 0, 0]^top.$
This is tantamount to saying that $c[1, 1, 0]^top + d[0, 0, 1]^top$ are really non-contributory. Or in other words, that they are in the null space of $A.$ If a system of three equations with three unknowns can be solved in more than one way is because the system is under-determined, and there are free variables.
$c[1, 1, 0]^top + d[0, 0, 1]^top$ represents all linear combinations of two linearly independent vectors in the null space, and any vector space (or subspace) observes closure under addition and scalar multiplication. Hence, $c[1, 1, 0]^top + d[0, 0, 1]^top=0.$
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
$endgroup$
$begingroup$
Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question
$endgroup$
– aceminer
Jan 4 '17 at 12:53
2
$begingroup$
$[color{blue}{0},color{red}{0},1]^top$ is in the null space of $A$ because if you take $color{blue}{0}times begin{bmatrix}1\2\1end{bmatrix}+color{red}{0}times begin{bmatrix}-1\-2\-1end{bmatrix}+1times begin{bmatrix}0\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:09
1
$begingroup$
$Ax$ where in this case $x=[color{blue}{0},color{red}{0},1]^top=begin{bmatrix}color{blue}{0}\color{red}{0}\1end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $color{blue}{0}times text{ first column of A} + color{red}{0}times text{ second column of A} +1times text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $bf 0$ vector will result in the $bf 0$ vector.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:16
1
$begingroup$
I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:21
1
$begingroup$
Look up how matrix vector multiplication works and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get?
$endgroup$
– Piotr Benedysiuk
Jan 4 '17 at 13:21
|
show 5 more comments
$begingroup$
I wonder if the easiest way to look at this is a follows:
$[2,0,0]^top$ is a particular solution to $Ax=b$. Why? Because it is true that it doesn't matter what $c$ and $d$ values we pick: for any $c$ and $d$ we get a vector, $x = [2, 0, 0]^top + c[1, 1, 0]^top + d[0, 0, 1]^top$ that satisfies $Ax=b$. Therefore, we can pick $c=0$ and $d=0$, and we end up with $x = [2, 0, 0]^top.$
This is tantamount to saying that $c[1, 1, 0]^top + d[0, 0, 1]^top$ are really non-contributory. Or in other words, that they are in the null space of $A.$ If a system of three equations with three unknowns can be solved in more than one way is because the system is under-determined, and there are free variables.
$c[1, 1, 0]^top + d[0, 0, 1]^top$ represents all linear combinations of two linearly independent vectors in the null space, and any vector space (or subspace) observes closure under addition and scalar multiplication. Hence, $c[1, 1, 0]^top + d[0, 0, 1]^top=0.$
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
$endgroup$
$begingroup$
Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question
$endgroup$
– aceminer
Jan 4 '17 at 12:53
2
$begingroup$
$[color{blue}{0},color{red}{0},1]^top$ is in the null space of $A$ because if you take $color{blue}{0}times begin{bmatrix}1\2\1end{bmatrix}+color{red}{0}times begin{bmatrix}-1\-2\-1end{bmatrix}+1times begin{bmatrix}0\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:09
1
$begingroup$
$Ax$ where in this case $x=[color{blue}{0},color{red}{0},1]^top=begin{bmatrix}color{blue}{0}\color{red}{0}\1end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $color{blue}{0}times text{ first column of A} + color{red}{0}times text{ second column of A} +1times text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $bf 0$ vector will result in the $bf 0$ vector.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:16
1
$begingroup$
I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:21
1
$begingroup$
Look up how matrix vector multiplication works and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get?
$endgroup$
– Piotr Benedysiuk
Jan 4 '17 at 13:21
|
show 5 more comments
$begingroup$
I wonder if the easiest way to look at this is a follows:
$[2,0,0]^top$ is a particular solution to $Ax=b$. Why? Because it is true that it doesn't matter what $c$ and $d$ values we pick: for any $c$ and $d$ we get a vector, $x = [2, 0, 0]^top + c[1, 1, 0]^top + d[0, 0, 1]^top$ that satisfies $Ax=b$. Therefore, we can pick $c=0$ and $d=0$, and we end up with $x = [2, 0, 0]^top.$
This is tantamount to saying that $c[1, 1, 0]^top + d[0, 0, 1]^top$ are really non-contributory. Or in other words, that they are in the null space of $A.$ If a system of three equations with three unknowns can be solved in more than one way is because the system is under-determined, and there are free variables.
$c[1, 1, 0]^top + d[0, 0, 1]^top$ represents all linear combinations of two linearly independent vectors in the null space, and any vector space (or subspace) observes closure under addition and scalar multiplication. Hence, $c[1, 1, 0]^top + d[0, 0, 1]^top=0.$
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
$endgroup$
I wonder if the easiest way to look at this is a follows:
$[2,0,0]^top$ is a particular solution to $Ax=b$. Why? Because it is true that it doesn't matter what $c$ and $d$ values we pick: for any $c$ and $d$ we get a vector, $x = [2, 0, 0]^top + c[1, 1, 0]^top + d[0, 0, 1]^top$ that satisfies $Ax=b$. Therefore, we can pick $c=0$ and $d=0$, and we end up with $x = [2, 0, 0]^top.$
This is tantamount to saying that $c[1, 1, 0]^top + d[0, 0, 1]^top$ are really non-contributory. Or in other words, that they are in the null space of $A.$ If a system of three equations with three unknowns can be solved in more than one way is because the system is under-determined, and there are free variables.
$c[1, 1, 0]^top + d[0, 0, 1]^top$ represents all linear combinations of two linearly independent vectors in the null space, and any vector space (or subspace) observes closure under addition and scalar multiplication. Hence, $c[1, 1, 0]^top + d[0, 0, 1]^top=0.$
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
edited Jan 4 '17 at 1:38
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
answered Jan 4 '17 at 1:06
Antoni ParelladaAntoni Parellada
3,11421341
3,11421341
$begingroup$
Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question
$endgroup$
– aceminer
Jan 4 '17 at 12:53
2
$begingroup$
$[color{blue}{0},color{red}{0},1]^top$ is in the null space of $A$ because if you take $color{blue}{0}times begin{bmatrix}1\2\1end{bmatrix}+color{red}{0}times begin{bmatrix}-1\-2\-1end{bmatrix}+1times begin{bmatrix}0\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:09
1
$begingroup$
$Ax$ where in this case $x=[color{blue}{0},color{red}{0},1]^top=begin{bmatrix}color{blue}{0}\color{red}{0}\1end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $color{blue}{0}times text{ first column of A} + color{red}{0}times text{ second column of A} +1times text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $bf 0$ vector will result in the $bf 0$ vector.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:16
1
$begingroup$
I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:21
1
$begingroup$
Look up how matrix vector multiplication works and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get?
$endgroup$
– Piotr Benedysiuk
Jan 4 '17 at 13:21
|
show 5 more comments
$begingroup$
Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question
$endgroup$
– aceminer
Jan 4 '17 at 12:53
2
$begingroup$
$[color{blue}{0},color{red}{0},1]^top$ is in the null space of $A$ because if you take $color{blue}{0}times begin{bmatrix}1\2\1end{bmatrix}+color{red}{0}times begin{bmatrix}-1\-2\-1end{bmatrix}+1times begin{bmatrix}0\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:09
1
$begingroup$
$Ax$ where in this case $x=[color{blue}{0},color{red}{0},1]^top=begin{bmatrix}color{blue}{0}\color{red}{0}\1end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $color{blue}{0}times text{ first column of A} + color{red}{0}times text{ second column of A} +1times text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $bf 0$ vector will result in the $bf 0$ vector.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:16
1
$begingroup$
I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:21
1
$begingroup$
Look up how matrix vector multiplication works and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get?
$endgroup$
– Piotr Benedysiuk
Jan 4 '17 at 13:21
$begingroup$
Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question
$endgroup$
– aceminer
Jan 4 '17 at 12:53
$begingroup$
Hi, Thanks for answer. I modified my question to further expand on this question. Could you kindly help me out with this question
$endgroup$
– aceminer
Jan 4 '17 at 12:53
2
2
$begingroup$
$[color{blue}{0},color{red}{0},1]^top$ is in the null space of $A$ because if you take $color{blue}{0}times begin{bmatrix}1\2\1end{bmatrix}+color{red}{0}times begin{bmatrix}-1\-2\-1end{bmatrix}+1times begin{bmatrix}0\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:09
$begingroup$
$[color{blue}{0},color{red}{0},1]^top$ is in the null space of $A$ because if you take $color{blue}{0}times begin{bmatrix}1\2\1end{bmatrix}+color{red}{0}times begin{bmatrix}-1\-2\-1end{bmatrix}+1times begin{bmatrix}0\0\0end{bmatrix}=begin{bmatrix}0\0\0end{bmatrix}$
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:09
1
1
$begingroup$
$Ax$ where in this case $x=[color{blue}{0},color{red}{0},1]^top=begin{bmatrix}color{blue}{0}\color{red}{0}\1end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $color{blue}{0}times text{ first column of A} + color{red}{0}times text{ second column of A} +1times text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $bf 0$ vector will result in the $bf 0$ vector.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:16
$begingroup$
$Ax$ where in this case $x=[color{blue}{0},color{red}{0},1]^top=begin{bmatrix}color{blue}{0}\color{red}{0}\1end{bmatrix}$ is calculated (and that was the idea behind using colors) as taking $color{blue}{0}times text{ first column of A} + color{red}{0}times text{ second column of A} +1times text{ third column of A}.$ Now the two color zeros would nullify whatever first and second columns you could place in $A$, and as for the third entry of $x$ (in this case $1$), it doesn't even matter - any entry times the $bf 0$ vector will result in the $bf 0$ vector.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:16
1
1
$begingroup$
I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:21
$begingroup$
I think you are very close to seeing it clearly, but you may want to go to the first lectures of Professor Strang, where he talks about column and row interpretations of $Ax$.
$endgroup$
– Antoni Parellada
Jan 4 '17 at 13:21
1
1
$begingroup$
Look up how matrix vector multiplication works and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get?
$endgroup$
– Piotr Benedysiuk
Jan 4 '17 at 13:21
$begingroup$
Look up how matrix vector multiplication works and try multiplying $A$ by $[0,0,1]^T.$ Then do the same with $[1,1,0]^T.$ What do you get?
$endgroup$
– Piotr Benedysiuk
Jan 4 '17 at 13:21
|
show 5 more comments
$begingroup$
From the rank-nullity theorem, the following relation satisfies for an $m times n$ matrix.
$$text{rank}(A)+text{nullity}(A)=n$$
The dimension of the null space is called the nullity of A.
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
$endgroup$
$begingroup$
Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1?
$endgroup$
– aceminer
Jan 3 '17 at 14:16
$begingroup$
See here.
$endgroup$
– projectilemotion
Jan 3 '17 at 15:51
add a comment |
$begingroup$
From the rank-nullity theorem, the following relation satisfies for an $m times n$ matrix.
$$text{rank}(A)+text{nullity}(A)=n$$
The dimension of the null space is called the nullity of A.
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
$endgroup$
$begingroup$
Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1?
$endgroup$
– aceminer
Jan 3 '17 at 14:16
$begingroup$
See here.
$endgroup$
– projectilemotion
Jan 3 '17 at 15:51
add a comment |
$begingroup$
From the rank-nullity theorem, the following relation satisfies for an $m times n$ matrix.
$$text{rank}(A)+text{nullity}(A)=n$$
The dimension of the null space is called the nullity of A.
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
$endgroup$
From the rank-nullity theorem, the following relation satisfies for an $m times n$ matrix.
$$text{rank}(A)+text{nullity}(A)=n$$
The dimension of the null space is called the nullity of A.
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
answered Jan 3 '17 at 14:15
projectilemotionprojectilemotion
11.4k62241
11.4k62241
$begingroup$
Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1?
$endgroup$
– aceminer
Jan 3 '17 at 14:16
$begingroup$
See here.
$endgroup$
– projectilemotion
Jan 3 '17 at 15:51
add a comment |
$begingroup$
Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1?
$endgroup$
– aceminer
Jan 3 '17 at 14:16
$begingroup$
See here.
$endgroup$
– projectilemotion
Jan 3 '17 at 15:51
$begingroup$
Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1?
$endgroup$
– aceminer
Jan 3 '17 at 14:16
$begingroup$
Yes I understand this but where is the rank or rather the nullity? How did he get the rank of the matrix A to be 1?
$endgroup$
– aceminer
Jan 3 '17 at 14:16
$begingroup$
See here.
$endgroup$
– projectilemotion
Jan 3 '17 at 15:51
$begingroup$
See here.
$endgroup$
– projectilemotion
Jan 3 '17 at 15:51
add a comment |
$begingroup$
The homogeneous equation $A x = 0$ has as set of solutions $mathcal{O}$ the set of differences of the solutions of the original equation (as $A x = A y = b$ implies $A (x - y) = 0$, and conversely if $Ax = b$ and $A u = 0$ then $A (x+u) = b$), so this set is
$$
mathcal{O} = { a,[1, 1, 0]^T + b,[0, 0, 1]^T : a, b in mathbb{R}},
$$
a $2$-dimensional space.
Therefore the nullity of $A$ is $2$ and the rank is $3 - 2 = 1$.
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
$endgroup$
$begingroup$
how did you derive the last set of a and b? It seems likes there's a huge leap for me
$endgroup$
– aceminer
Jan 3 '17 at 23:24
$begingroup$
Just take two of your solutions, and their difference.
$endgroup$
– Andreas Caranti
Jan 4 '17 at 7:18
add a comment |
$begingroup$
The homogeneous equation $A x = 0$ has as set of solutions $mathcal{O}$ the set of differences of the solutions of the original equation (as $A x = A y = b$ implies $A (x - y) = 0$, and conversely if $Ax = b$ and $A u = 0$ then $A (x+u) = b$), so this set is
$$
mathcal{O} = { a,[1, 1, 0]^T + b,[0, 0, 1]^T : a, b in mathbb{R}},
$$
a $2$-dimensional space.
Therefore the nullity of $A$ is $2$ and the rank is $3 - 2 = 1$.
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
$endgroup$
$begingroup$
how did you derive the last set of a and b? It seems likes there's a huge leap for me
$endgroup$
– aceminer
Jan 3 '17 at 23:24
$begingroup$
Just take two of your solutions, and their difference.
$endgroup$
– Andreas Caranti
Jan 4 '17 at 7:18
add a comment |
$begingroup$
The homogeneous equation $A x = 0$ has as set of solutions $mathcal{O}$ the set of differences of the solutions of the original equation (as $A x = A y = b$ implies $A (x - y) = 0$, and conversely if $Ax = b$ and $A u = 0$ then $A (x+u) = b$), so this set is
$$
mathcal{O} = { a,[1, 1, 0]^T + b,[0, 0, 1]^T : a, b in mathbb{R}},
$$
a $2$-dimensional space.
Therefore the nullity of $A$ is $2$ and the rank is $3 - 2 = 1$.
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
$endgroup$
The homogeneous equation $A x = 0$ has as set of solutions $mathcal{O}$ the set of differences of the solutions of the original equation (as $A x = A y = b$ implies $A (x - y) = 0$, and conversely if $Ax = b$ and $A u = 0$ then $A (x+u) = b$), so this set is
$$
mathcal{O} = { a,[1, 1, 0]^T + b,[0, 0, 1]^T : a, b in mathbb{R}},
$$
a $2$-dimensional space.
Therefore the nullity of $A$ is $2$ and the rank is $3 - 2 = 1$.
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
edited Jan 3 '17 at 15:55
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
answered Jan 3 '17 at 15:49
Andreas CarantiAndreas Caranti
57.2k34497
57.2k34497
$begingroup$
how did you derive the last set of a and b? It seems likes there's a huge leap for me
$endgroup$
– aceminer
Jan 3 '17 at 23:24
$begingroup$
Just take two of your solutions, and their difference.
$endgroup$
– Andreas Caranti
Jan 4 '17 at 7:18
add a comment |
$begingroup$
how did you derive the last set of a and b? It seems likes there's a huge leap for me
$endgroup$
– aceminer
Jan 3 '17 at 23:24
$begingroup$
Just take two of your solutions, and their difference.
$endgroup$
– Andreas Caranti
Jan 4 '17 at 7:18
$begingroup$
how did you derive the last set of a and b? It seems likes there's a huge leap for me
$endgroup$
– aceminer
Jan 3 '17 at 23:24
$begingroup$
how did you derive the last set of a and b? It seems likes there's a huge leap for me
$endgroup$
– aceminer
Jan 3 '17 at 23:24
$begingroup$
Just take two of your solutions, and their difference.
$endgroup$
– Andreas Caranti
Jan 4 '17 at 7:18
$begingroup$
Just take two of your solutions, and their difference.
$endgroup$
– Andreas Caranti
Jan 4 '17 at 7:18
add a comment |
$begingroup$
Watch the video from 10:38 again.
First of all he points out that the size of the matrix $A$ is $3times 3$. Secondly, the assumption of the complete sets of solutions tells you that
$$
Null(A)=2.
$$
Now the rank-nullity theorem tells you the rank of $A$ is given by
$$
Rank(A)=3-Null(A)=1.
$$
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
$endgroup$
$begingroup$
at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions
$endgroup$
– aceminer
Jan 3 '17 at 23:22
$begingroup$
@aceminer: I have edited my answer.
$endgroup$
– Jack
Jan 3 '17 at 23:31
$begingroup$
At 10:53, he said that "and I give you the 'complete solutions'".
$endgroup$
– Jack
Jan 3 '17 at 23:32
add a comment |
$begingroup$
Watch the video from 10:38 again.
First of all he points out that the size of the matrix $A$ is $3times 3$. Secondly, the assumption of the complete sets of solutions tells you that
$$
Null(A)=2.
$$
Now the rank-nullity theorem tells you the rank of $A$ is given by
$$
Rank(A)=3-Null(A)=1.
$$
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
$endgroup$
$begingroup$
at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions
$endgroup$
– aceminer
Jan 3 '17 at 23:22
$begingroup$
@aceminer: I have edited my answer.
$endgroup$
– Jack
Jan 3 '17 at 23:31
$begingroup$
At 10:53, he said that "and I give you the 'complete solutions'".
$endgroup$
– Jack
Jan 3 '17 at 23:32
add a comment |
$begingroup$
Watch the video from 10:38 again.
First of all he points out that the size of the matrix $A$ is $3times 3$. Secondly, the assumption of the complete sets of solutions tells you that
$$
Null(A)=2.
$$
Now the rank-nullity theorem tells you the rank of $A$ is given by
$$
Rank(A)=3-Null(A)=1.
$$
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
$endgroup$
Watch the video from 10:38 again.
First of all he points out that the size of the matrix $A$ is $3times 3$. Secondly, the assumption of the complete sets of solutions tells you that
$$
Null(A)=2.
$$
Now the rank-nullity theorem tells you the rank of $A$ is given by
$$
Rank(A)=3-Null(A)=1.
$$
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
edited Jan 3 '17 at 23:30
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
answered Jan 3 '17 at 15:48
JackJack
27.7k1784204
27.7k1784204
$begingroup$
at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions
$endgroup$
– aceminer
Jan 3 '17 at 23:22
$begingroup$
@aceminer: I have edited my answer.
$endgroup$
– Jack
Jan 3 '17 at 23:31
$begingroup$
At 10:53, he said that "and I give you the 'complete solutions'".
$endgroup$
– Jack
Jan 3 '17 at 23:32
add a comment |
$begingroup$
at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions
$endgroup$
– aceminer
Jan 3 '17 at 23:22
$begingroup$
@aceminer: I have edited my answer.
$endgroup$
– Jack
Jan 3 '17 at 23:31
$begingroup$
At 10:53, he said that "and I give you the 'complete solutions'".
$endgroup$
– Jack
Jan 3 '17 at 23:32
$begingroup$
at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions
$endgroup$
– aceminer
Jan 3 '17 at 23:22
$begingroup$
at which min mark should I watch? I started the video and it loaded from the start. And what is this assumption of complete set of solutions
$endgroup$
– aceminer
Jan 3 '17 at 23:22
$begingroup$
@aceminer: I have edited my answer.
$endgroup$
– Jack
Jan 3 '17 at 23:31
$begingroup$
@aceminer: I have edited my answer.
$endgroup$
– Jack
Jan 3 '17 at 23:31
$begingroup$
At 10:53, he said that "and I give you the 'complete solutions'".
$endgroup$
– Jack
Jan 3 '17 at 23:32
$begingroup$
At 10:53, he said that "and I give you the 'complete solutions'".
$endgroup$
– Jack
Jan 3 '17 at 23:32
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%2f2081927%2fhow-can-i-find-the-rank-of-a-matrix-a-when-the-solution-sets-to-ax-b-is-give%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
$begingroup$
Your question is about the matrix $A$. Would you write it down in your post instead of giving the video link?
$endgroup$
– Jack
Jan 3 '17 at 14:28
$begingroup$
@Jack this is what is mentioned. The A matrix is not given
$endgroup$
– aceminer
Jan 3 '17 at 15:05