How to find Kernel with [ A | 0 ] if dim Ker > 1? The 2019 Stack Overflow Developer Survey Results Are InKernels and reduced row echelon form - explanationFind basis of im, ker and dim im, dim ker verificationKernel and image of linear operatorKernel and composition of linear transforms. I don't want to multiply matrices, should be a theorem.Consider nonzero $A_2times 2$ such that $A^2 = vec0$. Prove or disprove that dim$($ker$(A)) = 2$For linear transformation $f: Vlongrightarrow W$, $dim R(f) + dim ker(f) = dim V$.Relationship between # dimensions in image and kernel of linear transformation called A and # dimensions in basis of image and basis of kernel of A$dim(kervarphicapkerpsi)=n-2$ proofShow that $V=operatornamekernel(f)oplusoperatornameim(f)$Proof for Dim(V) = Dim(Kernel(T)) + Dim(Image(T)): Is this enough to prove it?Compute dim(Ker(T))?
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How to find Kernel with [ A | 0 ] if dim Ker > 1?
The 2019 Stack Overflow Developer Survey Results Are InKernels and reduced row echelon form - explanationFind basis of im, ker and dim im, dim ker verificationKernel and image of linear operatorKernel and composition of linear transforms. I don't want to multiply matrices, should be a theorem.Consider nonzero $A_2times 2$ such that $A^2 = vec0$. Prove or disprove that dim$($ker$(A)) = 2$For linear transformation $f: Vlongrightarrow W$, $dim R(f) + dim ker(f) = dim V$.Relationship between # dimensions in image and kernel of linear transformation called A and # dimensions in basis of image and basis of kernel of A$dim(kervarphicapkerpsi)=n-2$ proofShow that $V=operatornamekernel(f)oplusoperatornameim(f)$Proof for Dim(V) = Dim(Kernel(T)) + Dim(Image(T)): Is this enough to prove it?Compute dim(Ker(T))?
$begingroup$
I've been looking around for a method to find the basis of the kernel. What I've seen to be the simple method is:
- Do REF to the matrix A.
- Take the non-pivot columns we will call them k.
- Put them in [ k | 0 ] then solve.
Now the thing is that I have trouble in figuring out how this method would work with a dimension greater than 1. How do you get 2 vectors from this method if dim ker A = 2?
I now that by doing [k | 0 ] you will get either 0 or a general solution with free variables. You then need to set the free variable that we will call r to an arbitrary value to get the vector in your kernel basis.
So, when you have dim ker A = 2 and you get a general solution with a free variable r. Do you arbitrarily set r to any value? One time to get a first particular solution that represents your first vector of your basis and a second time to get a second particular solution for you second vector of your basis.
This way you get your two vectors needed to form a base of dimension 2. This is the problem, I am not sure if this is correct.
linear-algebra
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
$endgroup$
add a comment |
$begingroup$
I've been looking around for a method to find the basis of the kernel. What I've seen to be the simple method is:
- Do REF to the matrix A.
- Take the non-pivot columns we will call them k.
- Put them in [ k | 0 ] then solve.
Now the thing is that I have trouble in figuring out how this method would work with a dimension greater than 1. How do you get 2 vectors from this method if dim ker A = 2?
I now that by doing [k | 0 ] you will get either 0 or a general solution with free variables. You then need to set the free variable that we will call r to an arbitrary value to get the vector in your kernel basis.
So, when you have dim ker A = 2 and you get a general solution with a free variable r. Do you arbitrarily set r to any value? One time to get a first particular solution that represents your first vector of your basis and a second time to get a second particular solution for you second vector of your basis.
This way you get your two vectors needed to form a base of dimension 2. This is the problem, I am not sure if this is correct.
linear-algebra
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
$endgroup$
$begingroup$
See math.stackexchange.com/a/1521354/265466.
$endgroup$
– amd
Mar 30 at 18:39
$begingroup$
Its not that clear to me.
$endgroup$
– Dr.Stone
Apr 5 at 1:59
add a comment |
$begingroup$
I've been looking around for a method to find the basis of the kernel. What I've seen to be the simple method is:
- Do REF to the matrix A.
- Take the non-pivot columns we will call them k.
- Put them in [ k | 0 ] then solve.
Now the thing is that I have trouble in figuring out how this method would work with a dimension greater than 1. How do you get 2 vectors from this method if dim ker A = 2?
I now that by doing [k | 0 ] you will get either 0 or a general solution with free variables. You then need to set the free variable that we will call r to an arbitrary value to get the vector in your kernel basis.
So, when you have dim ker A = 2 and you get a general solution with a free variable r. Do you arbitrarily set r to any value? One time to get a first particular solution that represents your first vector of your basis and a second time to get a second particular solution for you second vector of your basis.
This way you get your two vectors needed to form a base of dimension 2. This is the problem, I am not sure if this is correct.
linear-algebra
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
$endgroup$
I've been looking around for a method to find the basis of the kernel. What I've seen to be the simple method is:
- Do REF to the matrix A.
- Take the non-pivot columns we will call them k.
- Put them in [ k | 0 ] then solve.
Now the thing is that I have trouble in figuring out how this method would work with a dimension greater than 1. How do you get 2 vectors from this method if dim ker A = 2?
I now that by doing [k | 0 ] you will get either 0 or a general solution with free variables. You then need to set the free variable that we will call r to an arbitrary value to get the vector in your kernel basis.
So, when you have dim ker A = 2 and you get a general solution with a free variable r. Do you arbitrarily set r to any value? One time to get a first particular solution that represents your first vector of your basis and a second time to get a second particular solution for you second vector of your basis.
This way you get your two vectors needed to form a base of dimension 2. This is the problem, I am not sure if this is correct.
linear-algebra
linear-algebra
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
edited Mar 30 at 16:29
Dr.Stone
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
asked Mar 30 at 15:09
Dr.StoneDr.Stone
375
375
$begingroup$
See math.stackexchange.com/a/1521354/265466.
$endgroup$
– amd
Mar 30 at 18:39
$begingroup$
Its not that clear to me.
$endgroup$
– Dr.Stone
Apr 5 at 1:59
add a comment |
$begingroup$
See math.stackexchange.com/a/1521354/265466.
$endgroup$
– amd
Mar 30 at 18:39
$begingroup$
Its not that clear to me.
$endgroup$
– Dr.Stone
Apr 5 at 1:59
$begingroup$
See math.stackexchange.com/a/1521354/265466.
$endgroup$
– amd
Mar 30 at 18:39
$begingroup$
See math.stackexchange.com/a/1521354/265466.
$endgroup$
– amd
Mar 30 at 18:39
$begingroup$
Its not that clear to me.
$endgroup$
– Dr.Stone
Apr 5 at 1:59
$begingroup$
Its not that clear to me.
$endgroup$
– Dr.Stone
Apr 5 at 1:59
add a comment |
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$begingroup$
See math.stackexchange.com/a/1521354/265466.
$endgroup$
– amd
Mar 30 at 18:39
$begingroup$
Its not that clear to me.
$endgroup$
– Dr.Stone
Apr 5 at 1:59