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
15.6k3939
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
15.6k3939
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
15.6k3939
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
15.6k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.6k3939
asked Dec 14 '18 at 16:12
TukiTuki
1,019416
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.6k3939
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.6k3939
edited Dec 14 '18 at 21:47
Mostafa Ayaz
15.6k3939
15.6k3939
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
25.9k32978
$endgroup$
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
22.9k2152
$endgroup$
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 |
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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
25.9k32978
$endgroup$
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
25.9k32978
$endgroup$
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
25.9k32978
$endgroup$
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
25.9k32978
edited Dec 17 '18 at 19:30
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
25.9k32978
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
25.9k32978
answered Dec 14 '18 at 16:47
Fly by NightFly by Night
25.9k32978
25.9k32978
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
22.9k2152
$endgroup$
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
22.9k2152
$endgroup$
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
22.9k2152
$endgroup$
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
22.9k2152
answered Dec 14 '18 at 16:15
StackTDStackTD
22.9k2152
answered Dec 14 '18 at 16:15
StackTDStackTD
22.9k2152
answered Dec 14 '18 at 16:15
StackTDStackTD
22.9k2152
22.9k2152
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 |
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$begingroup$
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