Is not full rank matrix invertible?
$begingroup$
Problem
$A$ is a $4 times 4$ matrix. It is known that $text{rank}(A)=3$. Is matrix A invertible ?
Attempt to solve
$text{rank(A)}=3 implies det(A)=0$
which implies matrix is $textbf{not}$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $0$ and that some information is lost in this linear transformation.
Is my intuition behind this correct ?
linear-algebra matrices linear-transformations determinant matrix-rank
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
$endgroup$
add a comment |
$begingroup$
Problem
$A$ is a $4 times 4$ matrix. It is known that $text{rank}(A)=3$. Is matrix A invertible ?
Attempt to solve
$text{rank(A)}=3 implies det(A)=0$
which implies matrix is $textbf{not}$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $0$ and that some information is lost in this linear transformation.
Is my intuition behind this correct ?
linear-algebra matrices linear-transformations determinant matrix-rank
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
$endgroup$
$begingroup$
Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together.
$endgroup$
– Doug M
Dec 14 '18 at 16:14
add a comment |
$begingroup$
Problem
$A$ is a $4 times 4$ matrix. It is known that $text{rank}(A)=3$. Is matrix A invertible ?
Attempt to solve
$text{rank(A)}=3 implies det(A)=0$
which implies matrix is $textbf{not}$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $0$ and that some information is lost in this linear transformation.
Is my intuition behind this correct ?
linear-algebra matrices linear-transformations determinant matrix-rank
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
$endgroup$
Problem
$A$ is a $4 times 4$ matrix. It is known that $text{rank}(A)=3$. Is matrix A invertible ?
Attempt to solve
$text{rank(A)}=3 implies det(A)=0$
which implies matrix is $textbf{not}$ invertible. One dimension is lost during linear transformation if matrix is not full rank by definition. This implies determinant will be $0$ and that some information is lost in this linear transformation.
Is my intuition behind this correct ?
linear-algebra matrices linear-transformations determinant matrix-rank
linear-algebra matrices linear-transformations determinant matrix-rank
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.7k3939
15.7k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
1,019416
$begingroup$
Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together.
$endgroup$
– Doug M
Dec 14 '18 at 16:14
add a comment |
$begingroup$
Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together.
$endgroup$
– Doug M
Dec 14 '18 at 16:14
$begingroup$
Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together.
$endgroup$
– Doug M
Dec 14 '18 at 16:14
$begingroup$
Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together.
$endgroup$
– Doug M
Dec 14 '18 at 16:14
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $f:U to V$ we have $$mbox{rank}+mbox{nullity} = dim U$$
where the nullity is the dimension of the kernel, $ker f$.
A four-by-four matrix represents a linear map $f: U to V$ where $dim U = dim V = 4$. If the rank is three then $3+mbox{nullity}=4$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
$endgroup$
add a comment |
$begingroup$
Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.
The following properties are equivalent for a square matrix $A$:
$A$ has full rank
$A$ is invertible- the determinant of $A$ is non-zero
There are more, but the first two are sufficient to immediately draw the desired conclusion.
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
$endgroup$
add a comment |
$begingroup$
$rank(A) = 3 Rightarrow det(A) = 0$
needs to be proven, it is right though.
basically, you can say that:
$rank(A) < dim(A) Rightarrow det(A) = 0$
but it still needs to be proven.
an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
$endgroup$
add a comment |
$begingroup$
If $A$ and $B$ are matrices for which $AB$ makes sense, then
$$
operatorname{rank}(AB)lemin{operatorname{rank}(A),operatorname{rank}(B)}
$$
In particular, for every $B$, $operatorname{rank}(AB)leoperatorname{rank}(A)=3$. Can now $AB=I$?
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
$endgroup$
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%2f3039554%2fis-not-full-rank-matrix-invertible%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
$begingroup$
This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $f:U to V$ we have $$mbox{rank}+mbox{nullity} = dim U$$
where the nullity is the dimension of the kernel, $ker f$.
A four-by-four matrix represents a linear map $f: U to V$ where $dim U = dim V = 4$. If the rank is three then $3+mbox{nullity}=4$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
$endgroup$
add a comment |
$begingroup$
This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $f:U to V$ we have $$mbox{rank}+mbox{nullity} = dim U$$
where the nullity is the dimension of the kernel, $ker f$.
A four-by-four matrix represents a linear map $f: U to V$ where $dim U = dim V = 4$. If the rank is three then $3+mbox{nullity}=4$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
$endgroup$
add a comment |
$begingroup$
This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $f:U to V$ we have $$mbox{rank}+mbox{nullity} = dim U$$
where the nullity is the dimension of the kernel, $ker f$.
A four-by-four matrix represents a linear map $f: U to V$ where $dim U = dim V = 4$. If the rank is three then $3+mbox{nullity}=4$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
$endgroup$
This is exactly right. You're better off mentioning the rank-nullity theorem: for a linear map $f:U to V$ we have $$mbox{rank}+mbox{nullity} = dim U$$
where the nullity is the dimension of the kernel, $ker f$.
A four-by-four matrix represents a linear map $f: U to V$ where $dim U = dim V = 4$. If the rank is three then $3+mbox{nullity}=4$, i.e. there is a one-dimensional kernel. That means the map is not injective and has no inverse.
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
edited Dec 17 '18 at 19:30
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
26k32978
26k32978
add a comment |
add a comment |
$begingroup$
Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.
The following properties are equivalent for a square matrix $A$:
$A$ has full rank
$A$ is invertible- the determinant of $A$ is non-zero
There are more, but the first two are sufficient to immediately draw the desired conclusion.
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
$endgroup$
add a comment |
$begingroup$
Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.
The following properties are equivalent for a square matrix $A$:
$A$ has full rank
$A$ is invertible- the determinant of $A$ is non-zero
There are more, but the first two are sufficient to immediately draw the desired conclusion.
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
$endgroup$
add a comment |
$begingroup$
Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.
The following properties are equivalent for a square matrix $A$:
$A$ has full rank
$A$ is invertible- the determinant of $A$ is non-zero
There are more, but the first two are sufficient to immediately draw the desired conclusion.
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
$endgroup$
Your intuition seems fine. How you arrive at that conclusion depends on what properties you have seen, and/or which ones you are allowed to use.
The following properties are equivalent for a square matrix $A$:
$A$ has full rank
$A$ is invertible- the determinant of $A$ is non-zero
There are more, but the first two are sufficient to immediately draw the desired conclusion.
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
answered Dec 14 '18 at 16:15
StackTDStackTD
23.2k2153
23.2k2153
add a comment |
add a comment |
$begingroup$
$rank(A) = 3 Rightarrow det(A) = 0$
needs to be proven, it is right though.
basically, you can say that:
$rank(A) < dim(A) Rightarrow det(A) = 0$
but it still needs to be proven.
an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
$endgroup$
add a comment |
$begingroup$
$rank(A) = 3 Rightarrow det(A) = 0$
needs to be proven, it is right though.
basically, you can say that:
$rank(A) < dim(A) Rightarrow det(A) = 0$
but it still needs to be proven.
an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
$endgroup$
add a comment |
$begingroup$
$rank(A) = 3 Rightarrow det(A) = 0$
needs to be proven, it is right though.
basically, you can say that:
$rank(A) < dim(A) Rightarrow det(A) = 0$
but it still needs to be proven.
an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
$endgroup$
$rank(A) = 3 Rightarrow det(A) = 0$
needs to be proven, it is right though.
basically, you can say that:
$rank(A) < dim(A) Rightarrow det(A) = 0$
but it still needs to be proven.
an easy way to prove it is by showing that you will get a row of zeroes when trying to use raw reduction.
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
answered Dec 14 '18 at 16:19
matan anavimatan anavi
213
213
add a comment |
add a comment |
$begingroup$
If $A$ and $B$ are matrices for which $AB$ makes sense, then
$$
operatorname{rank}(AB)lemin{operatorname{rank}(A),operatorname{rank}(B)}
$$
In particular, for every $B$, $operatorname{rank}(AB)leoperatorname{rank}(A)=3$. Can now $AB=I$?
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
If $A$ and $B$ are matrices for which $AB$ makes sense, then
$$
operatorname{rank}(AB)lemin{operatorname{rank}(A),operatorname{rank}(B)}
$$
In particular, for every $B$, $operatorname{rank}(AB)leoperatorname{rank}(A)=3$. Can now $AB=I$?
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
$endgroup$
add a comment |
$begingroup$
If $A$ and $B$ are matrices for which $AB$ makes sense, then
$$
operatorname{rank}(AB)lemin{operatorname{rank}(A),operatorname{rank}(B)}
$$
In particular, for every $B$, $operatorname{rank}(AB)leoperatorname{rank}(A)=3$. Can now $AB=I$?
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
$endgroup$
If $A$ and $B$ are matrices for which $AB$ makes sense, then
$$
operatorname{rank}(AB)lemin{operatorname{rank}(A),operatorname{rank}(B)}
$$
In particular, for every $B$, $operatorname{rank}(AB)leoperatorname{rank}(A)=3$. Can now $AB=I$?
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
answered Dec 14 '18 at 17:08
egregegreg
183k1486205
183k1486205
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%2f3039554%2fis-not-full-rank-matrix-invertible%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
$begingroup$
Yes, you can even cut the lines regarding the determinant of A, and all of the rest of it still holds together.
$endgroup$
– Doug M
Dec 14 '18 at 16:14