Compute dim(Ker(T))? The Next CEO of Stack OverflowConstructing a linear mapProof of $dim ker...
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Compute dim(Ker(T))?
The Next CEO of Stack OverflowConstructing a linear mapProof of $dim ker (f circ g) le dim ker (f) + dim ker (g)$Show dim(Ker$(T)) = $dim(Ker$(QTP))$$renewcommand{Im}{text{Im}}$Prove that $dim(ker(T))=dim(ker(T^2))$ if $dim(Im(T))=dim(Im(T^2))$$dim(kervarphicapkerpsi)=n-2$ proofProve that if $T[V]=ker(T)$, then $n$ is even, where $dim(V)=n$Show the kernel is the zero-vector spaceFind $dim (Ker T + ImT) $ and $dim(Ker T cap ImT) $Showing the “boundaries” of linear mapping $mathbb{R}^{2} rightarrow mathbb{R}^{n}$Showing that image of a certain linear map is either trivial or a straight line
$begingroup$
I have the following question regarding Kernel and Image:
Part b of the question asks to find $dim(ker(T))$. I am not sure how to go about this part, as I haven't dealt with finding kernel $T$ where the input and output vector space is $M_{ntimes n} (mathbb{R})$. I understand that the dimension of the vector space is $n^2$, but I could not figure out how to find $ker(T)$ and its dimension. Could anyone help me with this problem?
linear-algebra
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
$endgroup$
add a comment |
$begingroup$
I have the following question regarding Kernel and Image:
Part b of the question asks to find $dim(ker(T))$. I am not sure how to go about this part, as I haven't dealt with finding kernel $T$ where the input and output vector space is $M_{ntimes n} (mathbb{R})$. I understand that the dimension of the vector space is $n^2$, but I could not figure out how to find $ker(T)$ and its dimension. Could anyone help me with this problem?
linear-algebra
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
$endgroup$
$begingroup$
The dimension of $M_{ntimes n}(Bbb R)$ is $n^2$, not $n^n$.
$endgroup$
– Shubham Johri
Mar 16 at 19:07
$begingroup$
When in doubt, go back to basic definitions. What is the definition of the kernel of a linear map?
$endgroup$
– amd
Mar 16 at 21:50
add a comment |
$begingroup$
I have the following question regarding Kernel and Image:
Part b of the question asks to find $dim(ker(T))$. I am not sure how to go about this part, as I haven't dealt with finding kernel $T$ where the input and output vector space is $M_{ntimes n} (mathbb{R})$. I understand that the dimension of the vector space is $n^2$, but I could not figure out how to find $ker(T)$ and its dimension. Could anyone help me with this problem?
linear-algebra
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
$endgroup$
I have the following question regarding Kernel and Image:
Part b of the question asks to find $dim(ker(T))$. I am not sure how to go about this part, as I haven't dealt with finding kernel $T$ where the input and output vector space is $M_{ntimes n} (mathbb{R})$. I understand that the dimension of the vector space is $n^2$, but I could not figure out how to find $ker(T)$ and its dimension. Could anyone help me with this problem?
linear-algebra
linear-algebra
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
edited Mar 16 at 19:34
Edeinielle
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
asked Mar 16 at 18:55
EdeinielleEdeinielle
12
12
$begingroup$
The dimension of $M_{ntimes n}(Bbb R)$ is $n^2$, not $n^n$.
$endgroup$
– Shubham Johri
Mar 16 at 19:07
$begingroup$
When in doubt, go back to basic definitions. What is the definition of the kernel of a linear map?
$endgroup$
– amd
Mar 16 at 21:50
add a comment |
$begingroup$
The dimension of $M_{ntimes n}(Bbb R)$ is $n^2$, not $n^n$.
$endgroup$
– Shubham Johri
Mar 16 at 19:07
$begingroup$
When in doubt, go back to basic definitions. What is the definition of the kernel of a linear map?
$endgroup$
– amd
Mar 16 at 21:50
$begingroup$
The dimension of $M_{ntimes n}(Bbb R)$ is $n^2$, not $n^n$.
$endgroup$
– Shubham Johri
Mar 16 at 19:07
$begingroup$
The dimension of $M_{ntimes n}(Bbb R)$ is $n^2$, not $n^n$.
$endgroup$
– Shubham Johri
Mar 16 at 19:07
$begingroup$
When in doubt, go back to basic definitions. What is the definition of the kernel of a linear map?
$endgroup$
– amd
Mar 16 at 21:50
$begingroup$
When in doubt, go back to basic definitions. What is the definition of the kernel of a linear map?
$endgroup$
– amd
Mar 16 at 21:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Kernel of a linear transformation $L$ is the set of all vectors $v$ such that $L(v)=0$. In your case, that would mean all $Ain M_{ntimes n}(Bbb R)|A+A^T=0$. That is, $A=-A^T$. Thus you need to find the dimension of the space of $ntimes n$ skew-symmetric matrices.
Note that in an $ntimes n$ skew-symmetric matrix, the diagonal contains $0$ while the entries above (or below) the diagonal also control the corresponding entry below (or above) the diagonal. So you have the freedom of choosing independently the entries above (or below) the diagonal, which are $(n^2-n)/2$ in number, which is its dimension.
For better insight, this is a basis of the kernel for $n=3$:
$$left{begin{bmatrix}0&1&0\-1&0&0\0&0&0end{bmatrix},begin{bmatrix}0&0&1\0&0&0\-1&0&0end{bmatrix},begin{bmatrix}0&0&0\0&0&1\0&-1&0end{bmatrix}right}$$
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
$endgroup$
add a comment |
$begingroup$
I guess that exercise 13 is about proving that every $ntimes n$ matrix $A$ is uniquely the sum of a symmetric and an antisymmetric matrix, precisely
$$
A=frac{1}{2}(A+A')+frac{1}{2}(A-A')
$$
Since every symmetric matrix $A$ satisfies $T(A)=A$ and $T(A)$ is symmetric for every $A$, you see that the image of $T$ is the subspace consisting of all symmetric matrices.
The kernel of $T$ consists of all matrices $A$ such that $A+A'=0$, that is, the antisymmetric matrices.
Now you have two ways of determining $dimker(T)$: either determine it directly or use the rank-nullity theorem after determining $dimoperatorname{im}(T)$, because
$$
n^2=dimoperatorname{im}(T)+dimker(T)
$$
If you call $x_{ij}$ the entries of an antisymmetric matrix, the equations are
$$
x_{ji}=-x_{ij} qquad 1le ile n,quad 1le jle n
$$
This implies that $x_{ii}=0$ for $1le ile n$, so you have to count the number of unordered pairs ${i,j}$, with $ine j$. These are
$$
binom{n}{2}=frac{n(n-1)}{2}
$$
which thus is the dimension of $dimker(T)$, because this gives the number of free variables for a matrix to belong to $ker(T)$.
answered Mar 16 at 21:22
egregegreg
185k1486206
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Kernel of a linear transformation $L$ is the set of all vectors $v$ such that $L(v)=0$. In your case, that would mean all $Ain M_{ntimes n}(Bbb R)|A+A^T=0$. That is, $A=-A^T$. Thus you need to find the dimension of the space of $ntimes n$ skew-symmetric matrices.
Note that in an $ntimes n$ skew-symmetric matrix, the diagonal contains $0$ while the entries above (or below) the diagonal also control the corresponding entry below (or above) the diagonal. So you have the freedom of choosing independently the entries above (or below) the diagonal, which are $(n^2-n)/2$ in number, which is its dimension.
For better insight, this is a basis of the kernel for $n=3$:
$$left{begin{bmatrix}0&1&0\-1&0&0\0&0&0end{bmatrix},begin{bmatrix}0&0&1\0&0&0\-1&0&0end{bmatrix},begin{bmatrix}0&0&0\0&0&1\0&-1&0end{bmatrix}right}$$
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
$endgroup$
add a comment |
$begingroup$
Kernel of a linear transformation $L$ is the set of all vectors $v$ such that $L(v)=0$. In your case, that would mean all $Ain M_{ntimes n}(Bbb R)|A+A^T=0$. That is, $A=-A^T$. Thus you need to find the dimension of the space of $ntimes n$ skew-symmetric matrices.
Note that in an $ntimes n$ skew-symmetric matrix, the diagonal contains $0$ while the entries above (or below) the diagonal also control the corresponding entry below (or above) the diagonal. So you have the freedom of choosing independently the entries above (or below) the diagonal, which are $(n^2-n)/2$ in number, which is its dimension.
For better insight, this is a basis of the kernel for $n=3$:
$$left{begin{bmatrix}0&1&0\-1&0&0\0&0&0end{bmatrix},begin{bmatrix}0&0&1\0&0&0\-1&0&0end{bmatrix},begin{bmatrix}0&0&0\0&0&1\0&-1&0end{bmatrix}right}$$
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
$endgroup$
add a comment |
$begingroup$
Kernel of a linear transformation $L$ is the set of all vectors $v$ such that $L(v)=0$. In your case, that would mean all $Ain M_{ntimes n}(Bbb R)|A+A^T=0$. That is, $A=-A^T$. Thus you need to find the dimension of the space of $ntimes n$ skew-symmetric matrices.
Note that in an $ntimes n$ skew-symmetric matrix, the diagonal contains $0$ while the entries above (or below) the diagonal also control the corresponding entry below (or above) the diagonal. So you have the freedom of choosing independently the entries above (or below) the diagonal, which are $(n^2-n)/2$ in number, which is its dimension.
For better insight, this is a basis of the kernel for $n=3$:
$$left{begin{bmatrix}0&1&0\-1&0&0\0&0&0end{bmatrix},begin{bmatrix}0&0&1\0&0&0\-1&0&0end{bmatrix},begin{bmatrix}0&0&0\0&0&1\0&-1&0end{bmatrix}right}$$
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
$endgroup$
Kernel of a linear transformation $L$ is the set of all vectors $v$ such that $L(v)=0$. In your case, that would mean all $Ain M_{ntimes n}(Bbb R)|A+A^T=0$. That is, $A=-A^T$. Thus you need to find the dimension of the space of $ntimes n$ skew-symmetric matrices.
Note that in an $ntimes n$ skew-symmetric matrix, the diagonal contains $0$ while the entries above (or below) the diagonal also control the corresponding entry below (or above) the diagonal. So you have the freedom of choosing independently the entries above (or below) the diagonal, which are $(n^2-n)/2$ in number, which is its dimension.
For better insight, this is a basis of the kernel for $n=3$:
$$left{begin{bmatrix}0&1&0\-1&0&0\0&0&0end{bmatrix},begin{bmatrix}0&0&1\0&0&0\-1&0&0end{bmatrix},begin{bmatrix}0&0&0\0&0&1\0&-1&0end{bmatrix}right}$$
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
answered Mar 16 at 19:15
Shubham JohriShubham Johri
5,477818
5,477818
add a comment |
add a comment |
$begingroup$
I guess that exercise 13 is about proving that every $ntimes n$ matrix $A$ is uniquely the sum of a symmetric and an antisymmetric matrix, precisely
$$
A=frac{1}{2}(A+A')+frac{1}{2}(A-A')
$$
Since every symmetric matrix $A$ satisfies $T(A)=A$ and $T(A)$ is symmetric for every $A$, you see that the image of $T$ is the subspace consisting of all symmetric matrices.
The kernel of $T$ consists of all matrices $A$ such that $A+A'=0$, that is, the antisymmetric matrices.
Now you have two ways of determining $dimker(T)$: either determine it directly or use the rank-nullity theorem after determining $dimoperatorname{im}(T)$, because
$$
n^2=dimoperatorname{im}(T)+dimker(T)
$$
If you call $x_{ij}$ the entries of an antisymmetric matrix, the equations are
$$
x_{ji}=-x_{ij} qquad 1le ile n,quad 1le jle n
$$
This implies that $x_{ii}=0$ for $1le ile n$, so you have to count the number of unordered pairs ${i,j}$, with $ine j$. These are
$$
binom{n}{2}=frac{n(n-1)}{2}
$$
which thus is the dimension of $dimker(T)$, because this gives the number of free variables for a matrix to belong to $ker(T)$.
answered Mar 16 at 21:22
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
I guess that exercise 13 is about proving that every $ntimes n$ matrix $A$ is uniquely the sum of a symmetric and an antisymmetric matrix, precisely
$$
A=frac{1}{2}(A+A')+frac{1}{2}(A-A')
$$
Since every symmetric matrix $A$ satisfies $T(A)=A$ and $T(A)$ is symmetric for every $A$, you see that the image of $T$ is the subspace consisting of all symmetric matrices.
The kernel of $T$ consists of all matrices $A$ such that $A+A'=0$, that is, the antisymmetric matrices.
Now you have two ways of determining $dimker(T)$: either determine it directly or use the rank-nullity theorem after determining $dimoperatorname{im}(T)$, because
$$
n^2=dimoperatorname{im}(T)+dimker(T)
$$
If you call $x_{ij}$ the entries of an antisymmetric matrix, the equations are
$$
x_{ji}=-x_{ij} qquad 1le ile n,quad 1le jle n
$$
This implies that $x_{ii}=0$ for $1le ile n$, so you have to count the number of unordered pairs ${i,j}$, with $ine j$. These are
$$
binom{n}{2}=frac{n(n-1)}{2}
$$
which thus is the dimension of $dimker(T)$, because this gives the number of free variables for a matrix to belong to $ker(T)$.
answered Mar 16 at 21:22
egregegreg
185k1486206
$endgroup$
add a comment |
$begingroup$
I guess that exercise 13 is about proving that every $ntimes n$ matrix $A$ is uniquely the sum of a symmetric and an antisymmetric matrix, precisely
$$
A=frac{1}{2}(A+A')+frac{1}{2}(A-A')
$$
Since every symmetric matrix $A$ satisfies $T(A)=A$ and $T(A)$ is symmetric for every $A$, you see that the image of $T$ is the subspace consisting of all symmetric matrices.
The kernel of $T$ consists of all matrices $A$ such that $A+A'=0$, that is, the antisymmetric matrices.
Now you have two ways of determining $dimker(T)$: either determine it directly or use the rank-nullity theorem after determining $dimoperatorname{im}(T)$, because
$$
n^2=dimoperatorname{im}(T)+dimker(T)
$$
If you call $x_{ij}$ the entries of an antisymmetric matrix, the equations are
$$
x_{ji}=-x_{ij} qquad 1le ile n,quad 1le jle n
$$
This implies that $x_{ii}=0$ for $1le ile n$, so you have to count the number of unordered pairs ${i,j}$, with $ine j$. These are
$$
binom{n}{2}=frac{n(n-1)}{2}
$$
which thus is the dimension of $dimker(T)$, because this gives the number of free variables for a matrix to belong to $ker(T)$.
answered Mar 16 at 21:22
egregegreg
185k1486206
$endgroup$
I guess that exercise 13 is about proving that every $ntimes n$ matrix $A$ is uniquely the sum of a symmetric and an antisymmetric matrix, precisely
$$
A=frac{1}{2}(A+A')+frac{1}{2}(A-A')
$$
Since every symmetric matrix $A$ satisfies $T(A)=A$ and $T(A)$ is symmetric for every $A$, you see that the image of $T$ is the subspace consisting of all symmetric matrices.
The kernel of $T$ consists of all matrices $A$ such that $A+A'=0$, that is, the antisymmetric matrices.
Now you have two ways of determining $dimker(T)$: either determine it directly or use the rank-nullity theorem after determining $dimoperatorname{im}(T)$, because
$$
n^2=dimoperatorname{im}(T)+dimker(T)
$$
If you call $x_{ij}$ the entries of an antisymmetric matrix, the equations are
$$
x_{ji}=-x_{ij} qquad 1le ile n,quad 1le jle n
$$
This implies that $x_{ii}=0$ for $1le ile n$, so you have to count the number of unordered pairs ${i,j}$, with $ine j$. These are
$$
binom{n}{2}=frac{n(n-1)}{2}
$$
which thus is the dimension of $dimker(T)$, because this gives the number of free variables for a matrix to belong to $ker(T)$.
answered Mar 16 at 21:22
egregegreg
185k1486206
answered Mar 16 at 21:22
egregegreg
185k1486206
answered Mar 16 at 21:22
egregegreg
185k1486206
answered Mar 16 at 21:22
egregegreg
185k1486206
185k1486206
add a comment |
add a comment |
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$begingroup$
The dimension of $M_{ntimes n}(Bbb R)$ is $n^2$, not $n^n$.
$endgroup$
– Shubham Johri
Mar 16 at 19:07
$begingroup$
When in doubt, go back to basic definitions. What is the definition of the kernel of a linear map?
$endgroup$
– amd
Mar 16 at 21:50