Mixing
Matrix of an orthogonal projector represented in the canonical basis
$begingroup$
Find the matrix (represented in the canonical base ${(1,0,0),(0,1,0),(0,0,1)}$) of the projector $P:mathbb{C^3}rightarrowmathbb{C^3}$ onto the subspace $$M=[{f_1,f_2}]=[{(0,0,1),(frac{2}{sqrt{5}},frac{-1}{sqrt{5}},0)}]$$
where the field is $mathbb{C}$ so the dimension is $3$.
My attempt:
The basis for $M^{perp}$ is ${f_3}={(frac{-1}{sqrt{5}},frac{2}{sqrt{5}},0)}$
Obviously, $(f)$ is an orthonormal basis for $mathbb{C^3}$.
Now I represent the canonical vectors $e_1,e_2$ and $e_3$ using this basis, but since I need to project them onto $M$ I only use the parts from vectors $f_1$ and $f_2$, or more specifically, the projection of a vector $x$ will be $$(x,f_1)f_1 + (x,f_2)f_2$$ where $(a,b)$ denotes the standard scalar product.
So, e.g. $$P(e_1)=(frac{-2}{sqrt{5}})cdot f_2$$
Now, what I'm interested in is which basis I'm working in? Let us denote $A(c,d)$ where $A$ is a linear operator on $mathbb{C}^3$ $c$ is a basis for the codomain and $d$ is a basis for the domain.
The matrix would be, after computing the projections of all canonical vectors:
$$begin{bmatrix}
0 & 0 & 1\
-2/5 & 1/5 & 0 \
0 & 0 & 0 \
end{bmatrix}$$
meaning the projection of $e_2$ onto $M$ is $frac{1}{5}f_2$ and for $e_3$ it's just $f_1$
Now, what I've got is $P(f,e)$, right?
I still need to compute $P(e,e)=I(e,f)*P(f,e)$ to get my final solution?
vector-spaces projection-matrices
asked Feb 2 at 18:50
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723520
$endgroup$
add a comment |
$begingroup$
Find the matrix (represented in the canonical base ${(1,0,0),(0,1,0),(0,0,1)}$) of the projector $P:mathbb{C^3}rightarrowmathbb{C^3}$ onto the subspace $$M=[{f_1,f_2}]=[{(0,0,1),(frac{2}{sqrt{5}},frac{-1}{sqrt{5}},0)}]$$
where the field is $mathbb{C}$ so the dimension is $3$.
My attempt:
The basis for $M^{perp}$ is ${f_3}={(frac{-1}{sqrt{5}},frac{2}{sqrt{5}},0)}$
Obviously, $(f)$ is an orthonormal basis for $mathbb{C^3}$.
Now I represent the canonical vectors $e_1,e_2$ and $e_3$ using this basis, but since I need to project them onto $M$ I only use the parts from vectors $f_1$ and $f_2$, or more specifically, the projection of a vector $x$ will be $$(x,f_1)f_1 + (x,f_2)f_2$$ where $(a,b)$ denotes the standard scalar product.
So, e.g. $$P(e_1)=(frac{-2}{sqrt{5}})cdot f_2$$
Now, what I'm interested in is which basis I'm working in? Let us denote $A(c,d)$ where $A$ is a linear operator on $mathbb{C}^3$ $c$ is a basis for the codomain and $d$ is a basis for the domain.
The matrix would be, after computing the projections of all canonical vectors:
$$begin{bmatrix}
0 & 0 & 1\
-2/5 & 1/5 & 0 \
0 & 0 & 0 \
end{bmatrix}$$
meaning the projection of $e_2$ onto $M$ is $frac{1}{5}f_2$ and for $e_3$ it's just $f_1$
Now, what I've got is $P(f,e)$, right?
I still need to compute $P(e,e)=I(e,f)*P(f,e)$ to get my final solution?
vector-spaces projection-matrices
asked Feb 2 at 18:50
CollapseCollapse
723520
$endgroup$
$begingroup$
How did you go from $left(frac{-2}{sqrt5}right)f_2$ to $(0,-2/5,0)^T$? The coordinates of $P(e_1)$ relative to the basis you’re calling $f$ are $(0,-2/sqrt5,0)^T$.
$endgroup$
– amd
Feb 2 at 19:13
$begingroup$
Incidentally, the simplest way to produce the required matrix, given that you’ve already found a basis for $M^perp$, (which, BTW, has a sign error), is to work out the matrix of the orthogonal projection onto that space—easy, because there’s only one vector to deal with—and then subtract that from the identity.
$endgroup$
– amd
Feb 2 at 23:23
add a comment |
$begingroup$
Find the matrix (represented in the canonical base ${(1,0,0),(0,1,0),(0,0,1)}$) of the projector $P:mathbb{C^3}rightarrowmathbb{C^3}$ onto the subspace $$M=[{f_1,f_2}]=[{(0,0,1),(frac{2}{sqrt{5}},frac{-1}{sqrt{5}},0)}]$$
where the field is $mathbb{C}$ so the dimension is $3$.
My attempt:
The basis for $M^{perp}$ is ${f_3}={(frac{-1}{sqrt{5}},frac{2}{sqrt{5}},0)}$
Obviously, $(f)$ is an orthonormal basis for $mathbb{C^3}$.
Now I represent the canonical vectors $e_1,e_2$ and $e_3$ using this basis, but since I need to project them onto $M$ I only use the parts from vectors $f_1$ and $f_2$, or more specifically, the projection of a vector $x$ will be $$(x,f_1)f_1 + (x,f_2)f_2$$ where $(a,b)$ denotes the standard scalar product.
So, e.g. $$P(e_1)=(frac{-2}{sqrt{5}})cdot f_2$$
Now, what I'm interested in is which basis I'm working in? Let us denote $A(c,d)$ where $A$ is a linear operator on $mathbb{C}^3$ $c$ is a basis for the codomain and $d$ is a basis for the domain.
The matrix would be, after computing the projections of all canonical vectors:
$$begin{bmatrix}
0 & 0 & 1\
-2/5 & 1/5 & 0 \
0 & 0 & 0 \
end{bmatrix}$$
meaning the projection of $e_2$ onto $M$ is $frac{1}{5}f_2$ and for $e_3$ it's just $f_1$
Now, what I've got is $P(f,e)$, right?
I still need to compute $P(e,e)=I(e,f)*P(f,e)$ to get my final solution?
vector-spaces projection-matrices
asked Feb 2 at 18:50
CollapseCollapse
723520
$endgroup$
Find the matrix (represented in the canonical base ${(1,0,0),(0,1,0),(0,0,1)}$) of the projector $P:mathbb{C^3}rightarrowmathbb{C^3}$ onto the subspace $$M=[{f_1,f_2}]=[{(0,0,1),(frac{2}{sqrt{5}},frac{-1}{sqrt{5}},0)}]$$
where the field is $mathbb{C}$ so the dimension is $3$.
My attempt:
The basis for $M^{perp}$ is ${f_3}={(frac{-1}{sqrt{5}},frac{2}{sqrt{5}},0)}$
Obviously, $(f)$ is an orthonormal basis for $mathbb{C^3}$.
Now I represent the canonical vectors $e_1,e_2$ and $e_3$ using this basis, but since I need to project them onto $M$ I only use the parts from vectors $f_1$ and $f_2$, or more specifically, the projection of a vector $x$ will be $$(x,f_1)f_1 + (x,f_2)f_2$$ where $(a,b)$ denotes the standard scalar product.
So, e.g. $$P(e_1)=(frac{-2}{sqrt{5}})cdot f_2$$
Now, what I'm interested in is which basis I'm working in? Let us denote $A(c,d)$ where $A$ is a linear operator on $mathbb{C}^3$ $c$ is a basis for the codomain and $d$ is a basis for the domain.
The matrix would be, after computing the projections of all canonical vectors:
$$begin{bmatrix}
0 & 0 & 1\
-2/5 & 1/5 & 0 \
0 & 0 & 0 \
end{bmatrix}$$
meaning the projection of $e_2$ onto $M$ is $frac{1}{5}f_2$ and for $e_3$ it's just $f_1$
Now, what I've got is $P(f,e)$, right?
I still need to compute $P(e,e)=I(e,f)*P(f,e)$ to get my final solution?
vector-spaces projection-matrices
vector-spaces projection-matrices
asked Feb 2 at 18:50
CollapseCollapse
723520
asked Feb 2 at 18:50
CollapseCollapse
723520
asked Feb 2 at 18:50
CollapseCollapse
723520
asked Feb 2 at 18:50
CollapseCollapse
723520
asked Feb 2 at 18:50
CollapseCollapse
723520
723520
$begingroup$
How did you go from $left(frac{-2}{sqrt5}right)f_2$ to $(0,-2/5,0)^T$? The coordinates of $P(e_1)$ relative to the basis you’re calling $f$ are $(0,-2/sqrt5,0)^T$.
$endgroup$
– amd
Feb 2 at 19:13
$begingroup$
Incidentally, the simplest way to produce the required matrix, given that you’ve already found a basis for $M^perp$, (which, BTW, has a sign error), is to work out the matrix of the orthogonal projection onto that space—easy, because there’s only one vector to deal with—and then subtract that from the identity.
$endgroup$
– amd
Feb 2 at 23:23
add a comment |
$begingroup$
How did you go from $left(frac{-2}{sqrt5}right)f_2$ to $(0,-2/5,0)^T$? The coordinates of $P(e_1)$ relative to the basis you’re calling $f$ are $(0,-2/sqrt5,0)^T$.
$endgroup$
– amd
Feb 2 at 19:13
$begingroup$
Incidentally, the simplest way to produce the required matrix, given that you’ve already found a basis for $M^perp$, (which, BTW, has a sign error), is to work out the matrix of the orthogonal projection onto that space—easy, because there’s only one vector to deal with—and then subtract that from the identity.
$endgroup$
– amd
Feb 2 at 23:23
$begingroup$
How did you go from $left(frac{-2}{sqrt5}right)f_2$ to $(0,-2/5,0)^T$? The coordinates of $P(e_1)$ relative to the basis you’re calling $f$ are $(0,-2/sqrt5,0)^T$.
$endgroup$
– amd
Feb 2 at 19:13
$begingroup$
How did you go from $left(frac{-2}{sqrt5}right)f_2$ to $(0,-2/5,0)^T$? The coordinates of $P(e_1)$ relative to the basis you’re calling $f$ are $(0,-2/sqrt5,0)^T$.
$endgroup$
– amd
Feb 2 at 19:13
$begingroup$
Incidentally, the simplest way to produce the required matrix, given that you’ve already found a basis for $M^perp$, (which, BTW, has a sign error), is to work out the matrix of the orthogonal projection onto that space—easy, because there’s only one vector to deal with—and then subtract that from the identity.
$endgroup$
– amd
Feb 2 at 23:23
$begingroup$
Incidentally, the simplest way to produce the required matrix, given that you’ve already found a basis for $M^perp$, (which, BTW, has a sign error), is to work out the matrix of the orthogonal projection onto that space—easy, because there’s only one vector to deal with—and then subtract that from the identity.
$endgroup$
– amd
Feb 2 at 23:23
add a comment |
2 Answers
2
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$begingroup$
Since $f_1$ and $f_2$ form an orthonormal basis, then, as you wrote,$$P(x)=langle x,f_1rangle f_1+langle x,f_2rangle f_2.$$So,$$P(e_1)=frac2{sqrt5}f_2=left(frac45,-frac25,0right),$$$$P(e_2)=-frac1{sqrt5}f_2=left(-frac25,frac15,0right),$$and$$P(e_3)=f_1=(0,0,1).$$Therefore, the matrix of $P$ with respect to the canonical basis is$$begin{bmatrix}frac45&-frac25&0\-frac25&frac15&0\0&0&1end{bmatrix}.$$
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
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$begingroup$
You started out reasonably, but then made some critical errors in constructing $P(f,e)$. You’d already stated earlier that $P(e_1)=-{2oversqrt5}f_2$ (which contains a sign error, by the way), i.e., that its coordinates in the $f$-basis are $left(0,-2/sqrt5,0right)^T$, not $(0,-2/5,0)^T$. Similarly, $P(e_2)=-{1oversqrt5}f_2$, so the second column of the matrix should be $left(0,-1/sqrt5,0right)^T$, not $(0,1/5,0)^T$. Using the correct matrix for $P(f,e)$ you should then get $$P(e,e) = begin{bmatrix}0 & {2oversqrt5} & {1oversqrt5} \ 0 & -{1oversqrt5} & {2oversqrt5} \ 1&0&0end{bmatrix} begin{bmatrix}0&0&1 \ {2oversqrt5} & -{1oversqrt5} & 0 \ 0&0&0 end{bmatrix} = begin{bmatrix}frac45 & -frac25 & 0 \ -frac25 & frac45 & 0 \ 0&0&1end{bmatrix}.$$ Note, by the way, that you’ve also made a sign error computing a basis for $M^perp$.
answered Feb 2 at 19:36
amdamd
31.8k21053
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2 Answers
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$begingroup$
Since $f_1$ and $f_2$ form an orthonormal basis, then, as you wrote,$$P(x)=langle x,f_1rangle f_1+langle x,f_2rangle f_2.$$So,$$P(e_1)=frac2{sqrt5}f_2=left(frac45,-frac25,0right),$$$$P(e_2)=-frac1{sqrt5}f_2=left(-frac25,frac15,0right),$$and$$P(e_3)=f_1=(0,0,1).$$Therefore, the matrix of $P$ with respect to the canonical basis is$$begin{bmatrix}frac45&-frac25&0\-frac25&frac15&0\0&0&1end{bmatrix}.$$
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
$begingroup$
Since $f_1$ and $f_2$ form an orthonormal basis, then, as you wrote,$$P(x)=langle x,f_1rangle f_1+langle x,f_2rangle f_2.$$So,$$P(e_1)=frac2{sqrt5}f_2=left(frac45,-frac25,0right),$$$$P(e_2)=-frac1{sqrt5}f_2=left(-frac25,frac15,0right),$$and$$P(e_3)=f_1=(0,0,1).$$Therefore, the matrix of $P$ with respect to the canonical basis is$$begin{bmatrix}frac45&-frac25&0\-frac25&frac15&0\0&0&1end{bmatrix}.$$
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
add a comment |
$begingroup$
Since $f_1$ and $f_2$ form an orthonormal basis, then, as you wrote,$$P(x)=langle x,f_1rangle f_1+langle x,f_2rangle f_2.$$So,$$P(e_1)=frac2{sqrt5}f_2=left(frac45,-frac25,0right),$$$$P(e_2)=-frac1{sqrt5}f_2=left(-frac25,frac15,0right),$$and$$P(e_3)=f_1=(0,0,1).$$Therefore, the matrix of $P$ with respect to the canonical basis is$$begin{bmatrix}frac45&-frac25&0\-frac25&frac15&0\0&0&1end{bmatrix}.$$
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
$endgroup$
Since $f_1$ and $f_2$ form an orthonormal basis, then, as you wrote,$$P(x)=langle x,f_1rangle f_1+langle x,f_2rangle f_2.$$So,$$P(e_1)=frac2{sqrt5}f_2=left(frac45,-frac25,0right),$$$$P(e_2)=-frac1{sqrt5}f_2=left(-frac25,frac15,0right),$$and$$P(e_3)=f_1=(0,0,1).$$Therefore, the matrix of $P$ with respect to the canonical basis is$$begin{bmatrix}frac45&-frac25&0\-frac25&frac15&0\0&0&1end{bmatrix}.$$
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
answered Feb 2 at 19:00
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
add a comment |
add a comment |
$begingroup$
You started out reasonably, but then made some critical errors in constructing $P(f,e)$. You’d already stated earlier that $P(e_1)=-{2oversqrt5}f_2$ (which contains a sign error, by the way), i.e., that its coordinates in the $f$-basis are $left(0,-2/sqrt5,0right)^T$, not $(0,-2/5,0)^T$. Similarly, $P(e_2)=-{1oversqrt5}f_2$, so the second column of the matrix should be $left(0,-1/sqrt5,0right)^T$, not $(0,1/5,0)^T$. Using the correct matrix for $P(f,e)$ you should then get $$P(e,e) = begin{bmatrix}0 & {2oversqrt5} & {1oversqrt5} \ 0 & -{1oversqrt5} & {2oversqrt5} \ 1&0&0end{bmatrix} begin{bmatrix}0&0&1 \ {2oversqrt5} & -{1oversqrt5} & 0 \ 0&0&0 end{bmatrix} = begin{bmatrix}frac45 & -frac25 & 0 \ -frac25 & frac45 & 0 \ 0&0&1end{bmatrix}.$$ Note, by the way, that you’ve also made a sign error computing a basis for $M^perp$.
answered Feb 2 at 19:36
amdamd
31.8k21053
$endgroup$
add a comment |
$begingroup$
You started out reasonably, but then made some critical errors in constructing $P(f,e)$. You’d already stated earlier that $P(e_1)=-{2oversqrt5}f_2$ (which contains a sign error, by the way), i.e., that its coordinates in the $f$-basis are $left(0,-2/sqrt5,0right)^T$, not $(0,-2/5,0)^T$. Similarly, $P(e_2)=-{1oversqrt5}f_2$, so the second column of the matrix should be $left(0,-1/sqrt5,0right)^T$, not $(0,1/5,0)^T$. Using the correct matrix for $P(f,e)$ you should then get $$P(e,e) = begin{bmatrix}0 & {2oversqrt5} & {1oversqrt5} \ 0 & -{1oversqrt5} & {2oversqrt5} \ 1&0&0end{bmatrix} begin{bmatrix}0&0&1 \ {2oversqrt5} & -{1oversqrt5} & 0 \ 0&0&0 end{bmatrix} = begin{bmatrix}frac45 & -frac25 & 0 \ -frac25 & frac45 & 0 \ 0&0&1end{bmatrix}.$$ Note, by the way, that you’ve also made a sign error computing a basis for $M^perp$.
answered Feb 2 at 19:36
amdamd
31.8k21053
$endgroup$
add a comment |
$begingroup$
You started out reasonably, but then made some critical errors in constructing $P(f,e)$. You’d already stated earlier that $P(e_1)=-{2oversqrt5}f_2$ (which contains a sign error, by the way), i.e., that its coordinates in the $f$-basis are $left(0,-2/sqrt5,0right)^T$, not $(0,-2/5,0)^T$. Similarly, $P(e_2)=-{1oversqrt5}f_2$, so the second column of the matrix should be $left(0,-1/sqrt5,0right)^T$, not $(0,1/5,0)^T$. Using the correct matrix for $P(f,e)$ you should then get $$P(e,e) = begin{bmatrix}0 & {2oversqrt5} & {1oversqrt5} \ 0 & -{1oversqrt5} & {2oversqrt5} \ 1&0&0end{bmatrix} begin{bmatrix}0&0&1 \ {2oversqrt5} & -{1oversqrt5} & 0 \ 0&0&0 end{bmatrix} = begin{bmatrix}frac45 & -frac25 & 0 \ -frac25 & frac45 & 0 \ 0&0&1end{bmatrix}.$$ Note, by the way, that you’ve also made a sign error computing a basis for $M^perp$.
answered Feb 2 at 19:36
amdamd
31.8k21053
$endgroup$
You started out reasonably, but then made some critical errors in constructing $P(f,e)$. You’d already stated earlier that $P(e_1)=-{2oversqrt5}f_2$ (which contains a sign error, by the way), i.e., that its coordinates in the $f$-basis are $left(0,-2/sqrt5,0right)^T$, not $(0,-2/5,0)^T$. Similarly, $P(e_2)=-{1oversqrt5}f_2$, so the second column of the matrix should be $left(0,-1/sqrt5,0right)^T$, not $(0,1/5,0)^T$. Using the correct matrix for $P(f,e)$ you should then get $$P(e,e) = begin{bmatrix}0 & {2oversqrt5} & {1oversqrt5} \ 0 & -{1oversqrt5} & {2oversqrt5} \ 1&0&0end{bmatrix} begin{bmatrix}0&0&1 \ {2oversqrt5} & -{1oversqrt5} & 0 \ 0&0&0 end{bmatrix} = begin{bmatrix}frac45 & -frac25 & 0 \ -frac25 & frac45 & 0 \ 0&0&1end{bmatrix}.$$ Note, by the way, that you’ve also made a sign error computing a basis for $M^perp$.
answered Feb 2 at 19:36
amdamd
31.8k21053
edited Feb 2 at 23:25
answered Feb 2 at 19:36
amdamd
31.8k21053
answered Feb 2 at 19:36
amdamd
31.8k21053
answered Feb 2 at 19:36
amdamd
31.8k21053
31.8k21053
add a comment |
add a comment |
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$begingroup$
How did you go from $left(frac{-2}{sqrt5}right)f_2$ to $(0,-2/5,0)^T$? The coordinates of $P(e_1)$ relative to the basis you’re calling $f$ are $(0,-2/sqrt5,0)^T$.
$endgroup$
– amd
Feb 2 at 19:13
$begingroup$
Incidentally, the simplest way to produce the required matrix, given that you’ve already found a basis for $M^perp$, (which, BTW, has a sign error), is to work out the matrix of the orthogonal projection onto that space—easy, because there’s only one vector to deal with—and then subtract that from the identity.
$endgroup$
– amd
Feb 2 at 23:23