Subspace $X in Mat_{2 times 2}(mathbb{R}):tr(YX)=0$, where $Y =begin{bmatrix} 2 & 5 \ 3 & 4...
$begingroup$
I want to prove that set of $X in Mat_{2 times 2}(mathbb{R}):tr(YX)=0$, where $Y =begin{bmatrix}
2 & 5 \
3 & 4
end{bmatrix}$
is subspace of space $Mat_{2 times 2}(mathbb{R})$. But how can i find $textbf{1}:textbf{1} A = Atextbf{1} = A$, where A is an element of our subspace. And i also have problems with finding basis of this subspace. How can i do it?
linear-algebra matrices
edited Dec 25 '18 at 21:59
Namaste
1
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
$endgroup$
closed as off-topic by Namaste, Paul Frost, KReiser, Eevee Trainer, Saad Dec 26 '18 at 3:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Paul Frost, KReiser, Eevee Trainer, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I want to prove that set of $X in Mat_{2 times 2}(mathbb{R}):tr(YX)=0$, where $Y =begin{bmatrix}
2 & 5 \
3 & 4
end{bmatrix}$
is subspace of space $Mat_{2 times 2}(mathbb{R})$. But how can i find $textbf{1}:textbf{1} A = Atextbf{1} = A$, where A is an element of our subspace. And i also have problems with finding basis of this subspace. How can i do it?
linear-algebra matrices
edited Dec 25 '18 at 21:59
Namaste
1
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
$endgroup$
closed as off-topic by Namaste, Paul Frost, KReiser, Eevee Trainer, Saad Dec 26 '18 at 3:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Paul Frost, KReiser, Eevee Trainer, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Probably the most straightforward approach is to assume that the matrix $X$ consists of some real numbers $x_{ij}$, where $1 leq i, j leq 2$, calculate $YX$ explicitly and then check, under which conditions the trace of $YX$ equals $0$.
$endgroup$
– hyperkahler
Dec 25 '18 at 21:28
2
$begingroup$
The vector space axioms don't require a multiplicative identity, just an additive identity. The zero matrix is indeed in your subspace.
$endgroup$
– J. Pistachio
Dec 25 '18 at 21:33
add a comment |
$begingroup$
I want to prove that set of $X in Mat_{2 times 2}(mathbb{R}):tr(YX)=0$, where $Y =begin{bmatrix}
2 & 5 \
3 & 4
end{bmatrix}$
is subspace of space $Mat_{2 times 2}(mathbb{R})$. But how can i find $textbf{1}:textbf{1} A = Atextbf{1} = A$, where A is an element of our subspace. And i also have problems with finding basis of this subspace. How can i do it?
linear-algebra matrices
edited Dec 25 '18 at 21:59
Namaste
1
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
$endgroup$
I want to prove that set of $X in Mat_{2 times 2}(mathbb{R}):tr(YX)=0$, where $Y =begin{bmatrix}
2 & 5 \
3 & 4
end{bmatrix}$
is subspace of space $Mat_{2 times 2}(mathbb{R})$. But how can i find $textbf{1}:textbf{1} A = Atextbf{1} = A$, where A is an element of our subspace. And i also have problems with finding basis of this subspace. How can i do it?
linear-algebra matrices
linear-algebra matrices
edited Dec 25 '18 at 21:59
Namaste
1
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
edited Dec 25 '18 at 21:59
Namaste
1
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
edited Dec 25 '18 at 21:59
Namaste
1
edited Dec 25 '18 at 21:59
Namaste
1
edited Dec 25 '18 at 21:59
Namaste
1
1
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
asked Dec 25 '18 at 21:22
envy gruntenvy grunt
969
969
closed as off-topic by Namaste, Paul Frost, KReiser, Eevee Trainer, Saad Dec 26 '18 at 3:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Paul Frost, KReiser, Eevee Trainer, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Paul Frost, KReiser, Eevee Trainer, Saad Dec 26 '18 at 3:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Paul Frost, KReiser, Eevee Trainer, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Probably the most straightforward approach is to assume that the matrix $X$ consists of some real numbers $x_{ij}$, where $1 leq i, j leq 2$, calculate $YX$ explicitly and then check, under which conditions the trace of $YX$ equals $0$.
$endgroup$
– hyperkahler
Dec 25 '18 at 21:28
2
$begingroup$
The vector space axioms don't require a multiplicative identity, just an additive identity. The zero matrix is indeed in your subspace.
$endgroup$
– J. Pistachio
Dec 25 '18 at 21:33
add a comment |
1
$begingroup$
Probably the most straightforward approach is to assume that the matrix $X$ consists of some real numbers $x_{ij}$, where $1 leq i, j leq 2$, calculate $YX$ explicitly and then check, under which conditions the trace of $YX$ equals $0$.
$endgroup$
– hyperkahler
Dec 25 '18 at 21:28
2
$begingroup$
The vector space axioms don't require a multiplicative identity, just an additive identity. The zero matrix is indeed in your subspace.
$endgroup$
– J. Pistachio
Dec 25 '18 at 21:33
1
1
$begingroup$
Probably the most straightforward approach is to assume that the matrix $X$ consists of some real numbers $x_{ij}$, where $1 leq i, j leq 2$, calculate $YX$ explicitly and then check, under which conditions the trace of $YX$ equals $0$.
$endgroup$
– hyperkahler
Dec 25 '18 at 21:28
$begingroup$
Probably the most straightforward approach is to assume that the matrix $X$ consists of some real numbers $x_{ij}$, where $1 leq i, j leq 2$, calculate $YX$ explicitly and then check, under which conditions the trace of $YX$ equals $0$.
$endgroup$
– hyperkahler
Dec 25 '18 at 21:28
2
2
$begingroup$
The vector space axioms don't require a multiplicative identity, just an additive identity. The zero matrix is indeed in your subspace.
$endgroup$
– J. Pistachio
Dec 25 '18 at 21:33
$begingroup$
The vector space axioms don't require a multiplicative identity, just an additive identity. The zero matrix is indeed in your subspace.
$endgroup$
– J. Pistachio
Dec 25 '18 at 21:33
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You want to remember the a subspace is closed under addition and scalar multiplication, and contains the zero matrix Z.
So suppose A,B $in$ W. That is tr(YA)=tr(YB)=0. Now consider C = A+B. We have
$$YC = Y(A+B) = YA + YB$$
So tr(YC) = tr(YA+YB) = tr(YA)+tr(YB) = 0. $checkmark$
Now consider $k$A for any scalar k. We have Y$k$A = k(YA).
So tr($k$YA) = $k$tr(YA) = 0.$checkmark$
Obviously, tr(YZ) = tr(Z) = 0.$checkmark$
Thus W is a subspace. I don't think we need to know exactly what Y is in order to show that W is a subspace.
To find a basis, consider A = $left[begin{array}&a &b\c&dend{array}right]$. Then YA = $left[begin{array}&2a+5c &2b+5d\3a+4c&3b+4dend{array}right]$.
Suppose A $in$ W. Then $$tr(YA) = 2a+5c+3b+4d = 0$$ So once we pick 3 of the entires for A, the 4th will be determined. So the dimension is 3, and a possible basis would be $left[begin{array}&1 &0\0&frac{-1}{2}end{array}right]$,$left[begin{array}&0 &1\0&frac{-3}{4}end{array}right]$,$left[begin{array}&0 &0\1&frac{-5}{4}end{array}right]$.
Or if you scale $left[begin{array}&2 &0\0&-1end{array}right]$,$left[begin{array}&0 &4\0&-3end{array}right]$,$left[begin{array}&0 &0\4&-5end{array}right]$.
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
$endgroup$
$begingroup$
thank you, but what about the basis of this subspace, how can i find it?
$endgroup$
– envy grunt
Dec 25 '18 at 21:40
1
$begingroup$
i'll edit my answer shortly.
$endgroup$
– Joel Pereira
Dec 26 '18 at 1:11
add a comment |
$begingroup$
The application $varphi : X mapsto tr(YX)$ is linear as it is the composition of $varphi_1 : X mapsto YX$ and the trace applications that are both linear maps.
The set $mathcal S$ you’re looking for is the kernel of $varphi$ and is therefore a subspace of $Mat_{2 times 2}(mathbb R)$
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
The map $Xmapstooperatorname{Tr}(YX)$ is linear because it is a composition of linear maps: $Xmapsto YX$ and the trace. Your set is the kernel, so it is a subspace.
In order to find a basis of the kernel, you need to compute; if
$$
X=begin{bmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{bmatrix}
$$
then
$$
YX=begin{bmatrix}
2x_{11}+5x_{21} & 2x_{12}+5x_{22} \
3x_{11}+4x_{21} & 3x_{12}+4x_{22}
end{bmatrix}
$$
so the trace is zero when
$$
2x_{11}+3x_{12}+5x_{21}+4x_{22}=0
$$
and finding a basis should be easy.
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You want to remember the a subspace is closed under addition and scalar multiplication, and contains the zero matrix Z.
So suppose A,B $in$ W. That is tr(YA)=tr(YB)=0. Now consider C = A+B. We have
$$YC = Y(A+B) = YA + YB$$
So tr(YC) = tr(YA+YB) = tr(YA)+tr(YB) = 0. $checkmark$
Now consider $k$A for any scalar k. We have Y$k$A = k(YA).
So tr($k$YA) = $k$tr(YA) = 0.$checkmark$
Obviously, tr(YZ) = tr(Z) = 0.$checkmark$
Thus W is a subspace. I don't think we need to know exactly what Y is in order to show that W is a subspace.
To find a basis, consider A = $left[begin{array}&a &b\c&dend{array}right]$. Then YA = $left[begin{array}&2a+5c &2b+5d\3a+4c&3b+4dend{array}right]$.
Suppose A $in$ W. Then $$tr(YA) = 2a+5c+3b+4d = 0$$ So once we pick 3 of the entires for A, the 4th will be determined. So the dimension is 3, and a possible basis would be $left[begin{array}&1 &0\0&frac{-1}{2}end{array}right]$,$left[begin{array}&0 &1\0&frac{-3}{4}end{array}right]$,$left[begin{array}&0 &0\1&frac{-5}{4}end{array}right]$.
Or if you scale $left[begin{array}&2 &0\0&-1end{array}right]$,$left[begin{array}&0 &4\0&-3end{array}right]$,$left[begin{array}&0 &0\4&-5end{array}right]$.
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
$endgroup$
$begingroup$
thank you, but what about the basis of this subspace, how can i find it?
$endgroup$
– envy grunt
Dec 25 '18 at 21:40
1
$begingroup$
i'll edit my answer shortly.
$endgroup$
– Joel Pereira
Dec 26 '18 at 1:11
add a comment |
$begingroup$
You want to remember the a subspace is closed under addition and scalar multiplication, and contains the zero matrix Z.
So suppose A,B $in$ W. That is tr(YA)=tr(YB)=0. Now consider C = A+B. We have
$$YC = Y(A+B) = YA + YB$$
So tr(YC) = tr(YA+YB) = tr(YA)+tr(YB) = 0. $checkmark$
Now consider $k$A for any scalar k. We have Y$k$A = k(YA).
So tr($k$YA) = $k$tr(YA) = 0.$checkmark$
Obviously, tr(YZ) = tr(Z) = 0.$checkmark$
Thus W is a subspace. I don't think we need to know exactly what Y is in order to show that W is a subspace.
To find a basis, consider A = $left[begin{array}&a &b\c&dend{array}right]$. Then YA = $left[begin{array}&2a+5c &2b+5d\3a+4c&3b+4dend{array}right]$.
Suppose A $in$ W. Then $$tr(YA) = 2a+5c+3b+4d = 0$$ So once we pick 3 of the entires for A, the 4th will be determined. So the dimension is 3, and a possible basis would be $left[begin{array}&1 &0\0&frac{-1}{2}end{array}right]$,$left[begin{array}&0 &1\0&frac{-3}{4}end{array}right]$,$left[begin{array}&0 &0\1&frac{-5}{4}end{array}right]$.
Or if you scale $left[begin{array}&2 &0\0&-1end{array}right]$,$left[begin{array}&0 &4\0&-3end{array}right]$,$left[begin{array}&0 &0\4&-5end{array}right]$.
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
$endgroup$
$begingroup$
thank you, but what about the basis of this subspace, how can i find it?
$endgroup$
– envy grunt
Dec 25 '18 at 21:40
1
$begingroup$
i'll edit my answer shortly.
$endgroup$
– Joel Pereira
Dec 26 '18 at 1:11
add a comment |
$begingroup$
You want to remember the a subspace is closed under addition and scalar multiplication, and contains the zero matrix Z.
So suppose A,B $in$ W. That is tr(YA)=tr(YB)=0. Now consider C = A+B. We have
$$YC = Y(A+B) = YA + YB$$
So tr(YC) = tr(YA+YB) = tr(YA)+tr(YB) = 0. $checkmark$
Now consider $k$A for any scalar k. We have Y$k$A = k(YA).
So tr($k$YA) = $k$tr(YA) = 0.$checkmark$
Obviously, tr(YZ) = tr(Z) = 0.$checkmark$
Thus W is a subspace. I don't think we need to know exactly what Y is in order to show that W is a subspace.
To find a basis, consider A = $left[begin{array}&a &b\c&dend{array}right]$. Then YA = $left[begin{array}&2a+5c &2b+5d\3a+4c&3b+4dend{array}right]$.
Suppose A $in$ W. Then $$tr(YA) = 2a+5c+3b+4d = 0$$ So once we pick 3 of the entires for A, the 4th will be determined. So the dimension is 3, and a possible basis would be $left[begin{array}&1 &0\0&frac{-1}{2}end{array}right]$,$left[begin{array}&0 &1\0&frac{-3}{4}end{array}right]$,$left[begin{array}&0 &0\1&frac{-5}{4}end{array}right]$.
Or if you scale $left[begin{array}&2 &0\0&-1end{array}right]$,$left[begin{array}&0 &4\0&-3end{array}right]$,$left[begin{array}&0 &0\4&-5end{array}right]$.
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
$endgroup$
You want to remember the a subspace is closed under addition and scalar multiplication, and contains the zero matrix Z.
So suppose A,B $in$ W. That is tr(YA)=tr(YB)=0. Now consider C = A+B. We have
$$YC = Y(A+B) = YA + YB$$
So tr(YC) = tr(YA+YB) = tr(YA)+tr(YB) = 0. $checkmark$
Now consider $k$A for any scalar k. We have Y$k$A = k(YA).
So tr($k$YA) = $k$tr(YA) = 0.$checkmark$
Obviously, tr(YZ) = tr(Z) = 0.$checkmark$
Thus W is a subspace. I don't think we need to know exactly what Y is in order to show that W is a subspace.
To find a basis, consider A = $left[begin{array}&a &b\c&dend{array}right]$. Then YA = $left[begin{array}&2a+5c &2b+5d\3a+4c&3b+4dend{array}right]$.
Suppose A $in$ W. Then $$tr(YA) = 2a+5c+3b+4d = 0$$ So once we pick 3 of the entires for A, the 4th will be determined. So the dimension is 3, and a possible basis would be $left[begin{array}&1 &0\0&frac{-1}{2}end{array}right]$,$left[begin{array}&0 &1\0&frac{-3}{4}end{array}right]$,$left[begin{array}&0 &0\1&frac{-5}{4}end{array}right]$.
Or if you scale $left[begin{array}&2 &0\0&-1end{array}right]$,$left[begin{array}&0 &4\0&-3end{array}right]$,$left[begin{array}&0 &0\4&-5end{array}right]$.
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
edited Dec 26 '18 at 1:50
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
answered Dec 25 '18 at 21:31
Joel PereiraJoel Pereira
83719
83719
$begingroup$
thank you, but what about the basis of this subspace, how can i find it?
$endgroup$
– envy grunt
Dec 25 '18 at 21:40
1
$begingroup$
i'll edit my answer shortly.
$endgroup$
– Joel Pereira
Dec 26 '18 at 1:11
add a comment |
$begingroup$
thank you, but what about the basis of this subspace, how can i find it?
$endgroup$
– envy grunt
Dec 25 '18 at 21:40
1
$begingroup$
i'll edit my answer shortly.
$endgroup$
– Joel Pereira
Dec 26 '18 at 1:11
$begingroup$
thank you, but what about the basis of this subspace, how can i find it?
$endgroup$
– envy grunt
Dec 25 '18 at 21:40
$begingroup$
thank you, but what about the basis of this subspace, how can i find it?
$endgroup$
– envy grunt
Dec 25 '18 at 21:40
1
1
$begingroup$
i'll edit my answer shortly.
$endgroup$
– Joel Pereira
Dec 26 '18 at 1:11
$begingroup$
i'll edit my answer shortly.
$endgroup$
– Joel Pereira
Dec 26 '18 at 1:11
add a comment |
$begingroup$
The application $varphi : X mapsto tr(YX)$ is linear as it is the composition of $varphi_1 : X mapsto YX$ and the trace applications that are both linear maps.
The set $mathcal S$ you’re looking for is the kernel of $varphi$ and is therefore a subspace of $Mat_{2 times 2}(mathbb R)$
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
The application $varphi : X mapsto tr(YX)$ is linear as it is the composition of $varphi_1 : X mapsto YX$ and the trace applications that are both linear maps.
The set $mathcal S$ you’re looking for is the kernel of $varphi$ and is therefore a subspace of $Mat_{2 times 2}(mathbb R)$
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
add a comment |
$begingroup$
The application $varphi : X mapsto tr(YX)$ is linear as it is the composition of $varphi_1 : X mapsto YX$ and the trace applications that are both linear maps.
The set $mathcal S$ you’re looking for is the kernel of $varphi$ and is therefore a subspace of $Mat_{2 times 2}(mathbb R)$
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
$endgroup$
The application $varphi : X mapsto tr(YX)$ is linear as it is the composition of $varphi_1 : X mapsto YX$ and the trace applications that are both linear maps.
The set $mathcal S$ you’re looking for is the kernel of $varphi$ and is therefore a subspace of $Mat_{2 times 2}(mathbb R)$
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
edited Dec 25 '18 at 21:38
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
answered Dec 25 '18 at 21:31
mathcounterexamples.netmathcounterexamples.net
26.9k22158
26.9k22158
add a comment |
add a comment |
$begingroup$
The map $Xmapstooperatorname{Tr}(YX)$ is linear because it is a composition of linear maps: $Xmapsto YX$ and the trace. Your set is the kernel, so it is a subspace.
In order to find a basis of the kernel, you need to compute; if
$$
X=begin{bmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{bmatrix}
$$
then
$$
YX=begin{bmatrix}
2x_{11}+5x_{21} & 2x_{12}+5x_{22} \
3x_{11}+4x_{21} & 3x_{12}+4x_{22}
end{bmatrix}
$$
so the trace is zero when
$$
2x_{11}+3x_{12}+5x_{21}+4x_{22}=0
$$
and finding a basis should be easy.
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
The map $Xmapstooperatorname{Tr}(YX)$ is linear because it is a composition of linear maps: $Xmapsto YX$ and the trace. Your set is the kernel, so it is a subspace.
In order to find a basis of the kernel, you need to compute; if
$$
X=begin{bmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{bmatrix}
$$
then
$$
YX=begin{bmatrix}
2x_{11}+5x_{21} & 2x_{12}+5x_{22} \
3x_{11}+4x_{21} & 3x_{12}+4x_{22}
end{bmatrix}
$$
so the trace is zero when
$$
2x_{11}+3x_{12}+5x_{21}+4x_{22}=0
$$
and finding a basis should be easy.
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
$endgroup$
add a comment |
$begingroup$
The map $Xmapstooperatorname{Tr}(YX)$ is linear because it is a composition of linear maps: $Xmapsto YX$ and the trace. Your set is the kernel, so it is a subspace.
In order to find a basis of the kernel, you need to compute; if
$$
X=begin{bmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{bmatrix}
$$
then
$$
YX=begin{bmatrix}
2x_{11}+5x_{21} & 2x_{12}+5x_{22} \
3x_{11}+4x_{21} & 3x_{12}+4x_{22}
end{bmatrix}
$$
so the trace is zero when
$$
2x_{11}+3x_{12}+5x_{21}+4x_{22}=0
$$
and finding a basis should be easy.
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
$endgroup$
The map $Xmapstooperatorname{Tr}(YX)$ is linear because it is a composition of linear maps: $Xmapsto YX$ and the trace. Your set is the kernel, so it is a subspace.
In order to find a basis of the kernel, you need to compute; if
$$
X=begin{bmatrix} x_{11} & x_{12} \ x_{21} & x_{22} end{bmatrix}
$$
then
$$
YX=begin{bmatrix}
2x_{11}+5x_{21} & 2x_{12}+5x_{22} \
3x_{11}+4x_{21} & 3x_{12}+4x_{22}
end{bmatrix}
$$
so the trace is zero when
$$
2x_{11}+3x_{12}+5x_{21}+4x_{22}=0
$$
and finding a basis should be easy.
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
answered Dec 25 '18 at 21:45
egregegreg
185k1486208
185k1486208
add a comment |
add a comment |
1
$begingroup$
Probably the most straightforward approach is to assume that the matrix $X$ consists of some real numbers $x_{ij}$, where $1 leq i, j leq 2$, calculate $YX$ explicitly and then check, under which conditions the trace of $YX$ equals $0$.
$endgroup$
– hyperkahler
Dec 25 '18 at 21:28
2
$begingroup$
The vector space axioms don't require a multiplicative identity, just an additive identity. The zero matrix is indeed in your subspace.
$endgroup$
– J. Pistachio
Dec 25 '18 at 21:33