Does an n x n matrix A have a ker(A) = 0 no matter what even if A represents linear transformation?Matrix of Linear Transformation Involving Derivatives?Matrix of T, a linear transformation when Im T = Ker TFinding the transformation matrix of this linear map.Prove the size of a linear transformation matrixRepresenting a Linear Transformation as a MatrixWhat does vector mean in Linear Algebra?Basis for Ker(T), Range(T) of Linear Transformation without the MatrixFind $mathrmKer(T)$ and $mathrmIm(T)$ of the following linear transformation with basesKernel and image of a matrix converting linear transformationWhat formula can make a non-linear transformation into a matrix transformation?
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Does an n x n matrix A have a ker(A) = 0 no matter what even if A represents linear transformation?
Matrix of Linear Transformation Involving Derivatives?Matrix of T, a linear transformation when Im T = Ker TFinding the transformation matrix of this linear map.Prove the size of a linear transformation matrixRepresenting a Linear Transformation as a MatrixWhat does vector mean in Linear Algebra?Basis for Ker(T), Range(T) of Linear Transformation without the MatrixFind $mathrmKer(T)$ and $mathrmIm(T)$ of the following linear transformation with basesKernel and image of a matrix converting linear transformationWhat formula can make a non-linear transformation into a matrix transformation?
$begingroup$
In my textbook there is a theorem that says that because n x n matrices are inversible they have a bunch of proprieties. One of them that is the that Ker A = 0.
However I was doing a practice problem where there was n x n matrix representing a linear transformation and when I tried to solve Ax = 0 it didn't give me that Ker A = 0 and it was the correct answer.
Is there some sort of further conditon for nxn matrix to automatically have a Ker A = 0?
linear-algebra
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In my textbook there is a theorem that says that because n x n matrices are inversible they have a bunch of proprieties. One of them that is the that Ker A = 0.
However I was doing a practice problem where there was n x n matrix representing a linear transformation and when I tried to solve Ax = 0 it didn't give me that Ker A = 0 and it was the correct answer.
Is there some sort of further conditon for nxn matrix to automatically have a Ker A = 0?
linear-algebra
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
$0 leq dim(ker(A)) leq n$, depending on how many linearly independent columns there are.
$endgroup$
– Hyperion
Mar 28 at 15:06
add a comment |
$begingroup$
In my textbook there is a theorem that says that because n x n matrices are inversible they have a bunch of proprieties. One of them that is the that Ker A = 0.
However I was doing a practice problem where there was n x n matrix representing a linear transformation and when I tried to solve Ax = 0 it didn't give me that Ker A = 0 and it was the correct answer.
Is there some sort of further conditon for nxn matrix to automatically have a Ker A = 0?
linear-algebra
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In my textbook there is a theorem that says that because n x n matrices are inversible they have a bunch of proprieties. One of them that is the that Ker A = 0.
However I was doing a practice problem where there was n x n matrix representing a linear transformation and when I tried to solve Ax = 0 it didn't give me that Ker A = 0 and it was the correct answer.
Is there some sort of further conditon for nxn matrix to automatically have a Ker A = 0?
linear-algebra
linear-algebra
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
asked Mar 28 at 14:54
Dr.StoneDr.Stone
335
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
$0 leq dim(ker(A)) leq n$, depending on how many linearly independent columns there are.
$endgroup$
– Hyperion
Mar 28 at 15:06
add a comment |
1
$begingroup$
$0 leq dim(ker(A)) leq n$, depending on how many linearly independent columns there are.
$endgroup$
– Hyperion
Mar 28 at 15:06
1
1
$begingroup$
$0 leq dim(ker(A)) leq n$, depending on how many linearly independent columns there are.
$endgroup$
– Hyperion
Mar 28 at 15:06
$begingroup$
$0 leq dim(ker(A)) leq n$, depending on how many linearly independent columns there are.
$endgroup$
– Hyperion
Mar 28 at 15:06
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
because n x n matrices are inversible
and
But doesn't the fact that A is n x n matrix implies that it is invertible?
Here's the problem: not all $ntimes n$-matrices are invertible.
However: if an $ntimes n$-matrix is invertible, then its kernel contains only the zero vector and vice versa; so those two properties are equivalent. This also means that if an $ntimes n$-matrix is not invertible, the kernel will contain more than only the zero vector (and vice versa!).
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
For a $n times n$ matrix $A,$ $text Ker (A) = 0 $ if and only if $A$ is invertible.
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
$endgroup$
$begingroup$
But doesn't the fact that A is n x n matrix implies that it is invertible?
$endgroup$
– Dr.Stone
Mar 28 at 15:01
1
$begingroup$
Nope. Take $n=2.$ Consider the $2 times 2$ matrix $$A=beginpmatrix 1 & 0 \ 0 & 0 endpmatrix.$$ Is $A$ invertible?
$endgroup$
– Dbchatto67
Mar 28 at 15:03
$begingroup$
Ah ok, I see now.
$endgroup$
– Dr.Stone
Mar 28 at 15:05
add a comment |
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2 Answers
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2 Answers
2
active
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active
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votes
$begingroup$
because n x n matrices are inversible
and
But doesn't the fact that A is n x n matrix implies that it is invertible?
Here's the problem: not all $ntimes n$-matrices are invertible.
However: if an $ntimes n$-matrix is invertible, then its kernel contains only the zero vector and vice versa; so those two properties are equivalent. This also means that if an $ntimes n$-matrix is not invertible, the kernel will contain more than only the zero vector (and vice versa!).
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
because n x n matrices are inversible
and
But doesn't the fact that A is n x n matrix implies that it is invertible?
Here's the problem: not all $ntimes n$-matrices are invertible.
However: if an $ntimes n$-matrix is invertible, then its kernel contains only the zero vector and vice versa; so those two properties are equivalent. This also means that if an $ntimes n$-matrix is not invertible, the kernel will contain more than only the zero vector (and vice versa!).
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
$endgroup$
add a comment |
$begingroup$
because n x n matrices are inversible
and
But doesn't the fact that A is n x n matrix implies that it is invertible?
Here's the problem: not all $ntimes n$-matrices are invertible.
However: if an $ntimes n$-matrix is invertible, then its kernel contains only the zero vector and vice versa; so those two properties are equivalent. This also means that if an $ntimes n$-matrix is not invertible, the kernel will contain more than only the zero vector (and vice versa!).
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
$endgroup$
because n x n matrices are inversible
and
But doesn't the fact that A is n x n matrix implies that it is invertible?
Here's the problem: not all $ntimes n$-matrices are invertible.
However: if an $ntimes n$-matrix is invertible, then its kernel contains only the zero vector and vice versa; so those two properties are equivalent. This also means that if an $ntimes n$-matrix is not invertible, the kernel will contain more than only the zero vector (and vice versa!).
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
answered Mar 28 at 15:04
StackTDStackTD
24.3k2254
24.3k2254
add a comment |
add a comment |
$begingroup$
For a $n times n$ matrix $A,$ $text Ker (A) = 0 $ if and only if $A$ is invertible.
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
$endgroup$
$begingroup$
But doesn't the fact that A is n x n matrix implies that it is invertible?
$endgroup$
– Dr.Stone
Mar 28 at 15:01
1
$begingroup$
Nope. Take $n=2.$ Consider the $2 times 2$ matrix $$A=beginpmatrix 1 & 0 \ 0 & 0 endpmatrix.$$ Is $A$ invertible?
$endgroup$
– Dbchatto67
Mar 28 at 15:03
$begingroup$
Ah ok, I see now.
$endgroup$
– Dr.Stone
Mar 28 at 15:05
add a comment |
$begingroup$
For a $n times n$ matrix $A,$ $text Ker (A) = 0 $ if and only if $A$ is invertible.
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
$endgroup$
$begingroup$
But doesn't the fact that A is n x n matrix implies that it is invertible?
$endgroup$
– Dr.Stone
Mar 28 at 15:01
1
$begingroup$
Nope. Take $n=2.$ Consider the $2 times 2$ matrix $$A=beginpmatrix 1 & 0 \ 0 & 0 endpmatrix.$$ Is $A$ invertible?
$endgroup$
– Dbchatto67
Mar 28 at 15:03
$begingroup$
Ah ok, I see now.
$endgroup$
– Dr.Stone
Mar 28 at 15:05
add a comment |
$begingroup$
For a $n times n$ matrix $A,$ $text Ker (A) = 0 $ if and only if $A$ is invertible.
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
$endgroup$
For a $n times n$ matrix $A,$ $text Ker (A) = 0 $ if and only if $A$ is invertible.
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
answered Mar 28 at 14:56
Dbchatto67Dbchatto67
2,445522
2,445522
$begingroup$
But doesn't the fact that A is n x n matrix implies that it is invertible?
$endgroup$
– Dr.Stone
Mar 28 at 15:01
1
$begingroup$
Nope. Take $n=2.$ Consider the $2 times 2$ matrix $$A=beginpmatrix 1 & 0 \ 0 & 0 endpmatrix.$$ Is $A$ invertible?
$endgroup$
– Dbchatto67
Mar 28 at 15:03
$begingroup$
Ah ok, I see now.
$endgroup$
– Dr.Stone
Mar 28 at 15:05
add a comment |
$begingroup$
But doesn't the fact that A is n x n matrix implies that it is invertible?
$endgroup$
– Dr.Stone
Mar 28 at 15:01
1
$begingroup$
Nope. Take $n=2.$ Consider the $2 times 2$ matrix $$A=beginpmatrix 1 & 0 \ 0 & 0 endpmatrix.$$ Is $A$ invertible?
$endgroup$
– Dbchatto67
Mar 28 at 15:03
$begingroup$
Ah ok, I see now.
$endgroup$
– Dr.Stone
Mar 28 at 15:05
$begingroup$
But doesn't the fact that A is n x n matrix implies that it is invertible?
$endgroup$
– Dr.Stone
Mar 28 at 15:01
$begingroup$
But doesn't the fact that A is n x n matrix implies that it is invertible?
$endgroup$
– Dr.Stone
Mar 28 at 15:01
1
1
$begingroup$
Nope. Take $n=2.$ Consider the $2 times 2$ matrix $$A=beginpmatrix 1 & 0 \ 0 & 0 endpmatrix.$$ Is $A$ invertible?
$endgroup$
– Dbchatto67
Mar 28 at 15:03
$begingroup$
Nope. Take $n=2.$ Consider the $2 times 2$ matrix $$A=beginpmatrix 1 & 0 \ 0 & 0 endpmatrix.$$ Is $A$ invertible?
$endgroup$
– Dbchatto67
Mar 28 at 15:03
$begingroup$
Ah ok, I see now.
$endgroup$
– Dr.Stone
Mar 28 at 15:05
$begingroup$
Ah ok, I see now.
$endgroup$
– Dr.Stone
Mar 28 at 15:05
add a comment |
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
$0 leq dim(ker(A)) leq n$, depending on how many linearly independent columns there are.
$endgroup$
– Hyperion
Mar 28 at 15:06