Mixing
Prove that the orthogonal projection operator is idempotent
$begingroup$
Let ${u_{1},u_{2},cdots,u_{n}}$ e an orthonormal basis for a subspace $U$ in an inner product space $X$.
Define the orthogonal projection of $X$ onto $U$, $P:X to U$, to be $Px = sum_{i=1}^{n}langle x, u_{i} rangle u_{i}$, where $langle cdot, u_{i} rangle$ represents the inner product.
I need to prove that $P = P^{2}$; i.e., that $P$ is idempotent. I have already proven that $P$ is linear, and am therefore free to use it.
So far, I set up what I am trying to show as follows:
$P^{2}x = sum_{i=1}^{n} langle Px, u_{i}rangle u_{i} =sum_{i=1}^{n}langle sum_{i=1}^{n}langle x, u_{i} rangle u_{i},u_{i}rangle u_{i}$
Then, I thought that perhaps expanding out the inner sum might be helpful, and then somewhere along the line I might be able to use linearity to get $sum_{i=1}^{n}langle x, u_{i}rangle u_{i}$ eventually on the RHS.
This is about as far as I got playing around with the sums:
$sum_{i=1}^{n}langle langle x, u_{1}rangle u_{1}+langle x, u_{2}rangle u_{2} + cdots + langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} = sum_{i=1}^{n}left(langle langle x, u_{1} rangle u_{1}, u_{i} rangle + langle langle x, u_{2}rangle u_{2}, u_{i} rangle + cdots + langle langle x, u_{n} rangle u_{n}, u_{i} rangle right)u_{i} = sum_{i=1}^{n}left[left(langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i}right) + left(langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i}right) + cdots + left(langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} right)right] = sum_{i=1}^{n} langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i} + sum_{i=1}^{n}langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i} + cdots + sum_{i=1}^{n}langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i}$
But, it's still not looking anywhere closer to where I need to be.
Could somebody please help me finish this?
Thank you.
inner-product-space linear-transformations orthogonality orthonormal
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
$endgroup$
add a comment |
$begingroup$
Let ${u_{1},u_{2},cdots,u_{n}}$ e an orthonormal basis for a subspace $U$ in an inner product space $X$.
Define the orthogonal projection of $X$ onto $U$, $P:X to U$, to be $Px = sum_{i=1}^{n}langle x, u_{i} rangle u_{i}$, where $langle cdot, u_{i} rangle$ represents the inner product.
I need to prove that $P = P^{2}$; i.e., that $P$ is idempotent. I have already proven that $P$ is linear, and am therefore free to use it.
So far, I set up what I am trying to show as follows:
$P^{2}x = sum_{i=1}^{n} langle Px, u_{i}rangle u_{i} =sum_{i=1}^{n}langle sum_{i=1}^{n}langle x, u_{i} rangle u_{i},u_{i}rangle u_{i}$
Then, I thought that perhaps expanding out the inner sum might be helpful, and then somewhere along the line I might be able to use linearity to get $sum_{i=1}^{n}langle x, u_{i}rangle u_{i}$ eventually on the RHS.
This is about as far as I got playing around with the sums:
$sum_{i=1}^{n}langle langle x, u_{1}rangle u_{1}+langle x, u_{2}rangle u_{2} + cdots + langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} = sum_{i=1}^{n}left(langle langle x, u_{1} rangle u_{1}, u_{i} rangle + langle langle x, u_{2}rangle u_{2}, u_{i} rangle + cdots + langle langle x, u_{n} rangle u_{n}, u_{i} rangle right)u_{i} = sum_{i=1}^{n}left[left(langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i}right) + left(langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i}right) + cdots + left(langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} right)right] = sum_{i=1}^{n} langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i} + sum_{i=1}^{n}langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i} + cdots + sum_{i=1}^{n}langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i}$
But, it's still not looking anywhere closer to where I need to be.
Could somebody please help me finish this?
Thank you.
inner-product-space linear-transformations orthogonality orthonormal
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
$endgroup$
add a comment |
$begingroup$
Let ${u_{1},u_{2},cdots,u_{n}}$ e an orthonormal basis for a subspace $U$ in an inner product space $X$.
Define the orthogonal projection of $X$ onto $U$, $P:X to U$, to be $Px = sum_{i=1}^{n}langle x, u_{i} rangle u_{i}$, where $langle cdot, u_{i} rangle$ represents the inner product.
I need to prove that $P = P^{2}$; i.e., that $P$ is idempotent. I have already proven that $P$ is linear, and am therefore free to use it.
So far, I set up what I am trying to show as follows:
$P^{2}x = sum_{i=1}^{n} langle Px, u_{i}rangle u_{i} =sum_{i=1}^{n}langle sum_{i=1}^{n}langle x, u_{i} rangle u_{i},u_{i}rangle u_{i}$
Then, I thought that perhaps expanding out the inner sum might be helpful, and then somewhere along the line I might be able to use linearity to get $sum_{i=1}^{n}langle x, u_{i}rangle u_{i}$ eventually on the RHS.
This is about as far as I got playing around with the sums:
$sum_{i=1}^{n}langle langle x, u_{1}rangle u_{1}+langle x, u_{2}rangle u_{2} + cdots + langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} = sum_{i=1}^{n}left(langle langle x, u_{1} rangle u_{1}, u_{i} rangle + langle langle x, u_{2}rangle u_{2}, u_{i} rangle + cdots + langle langle x, u_{n} rangle u_{n}, u_{i} rangle right)u_{i} = sum_{i=1}^{n}left[left(langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i}right) + left(langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i}right) + cdots + left(langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} right)right] = sum_{i=1}^{n} langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i} + sum_{i=1}^{n}langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i} + cdots + sum_{i=1}^{n}langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i}$
But, it's still not looking anywhere closer to where I need to be.
Could somebody please help me finish this?
Thank you.
inner-product-space linear-transformations orthogonality orthonormal
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
$endgroup$
Let ${u_{1},u_{2},cdots,u_{n}}$ e an orthonormal basis for a subspace $U$ in an inner product space $X$.
Define the orthogonal projection of $X$ onto $U$, $P:X to U$, to be $Px = sum_{i=1}^{n}langle x, u_{i} rangle u_{i}$, where $langle cdot, u_{i} rangle$ represents the inner product.
I need to prove that $P = P^{2}$; i.e., that $P$ is idempotent. I have already proven that $P$ is linear, and am therefore free to use it.
So far, I set up what I am trying to show as follows:
$P^{2}x = sum_{i=1}^{n} langle Px, u_{i}rangle u_{i} =sum_{i=1}^{n}langle sum_{i=1}^{n}langle x, u_{i} rangle u_{i},u_{i}rangle u_{i}$
Then, I thought that perhaps expanding out the inner sum might be helpful, and then somewhere along the line I might be able to use linearity to get $sum_{i=1}^{n}langle x, u_{i}rangle u_{i}$ eventually on the RHS.
This is about as far as I got playing around with the sums:
$sum_{i=1}^{n}langle langle x, u_{1}rangle u_{1}+langle x, u_{2}rangle u_{2} + cdots + langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} = sum_{i=1}^{n}left(langle langle x, u_{1} rangle u_{1}, u_{i} rangle + langle langle x, u_{2}rangle u_{2}, u_{i} rangle + cdots + langle langle x, u_{n} rangle u_{n}, u_{i} rangle right)u_{i} = sum_{i=1}^{n}left[left(langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i}right) + left(langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i}right) + cdots + left(langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i} right)right] = sum_{i=1}^{n} langle langle x, u_{1} rangle u_{1}, u_{i}rangle u_{i} + sum_{i=1}^{n}langle langle x, u_{2} rangle u_{2}, u_{i} rangle u_{i} + cdots + sum_{i=1}^{n}langle langle x, u_{n} rangle u_{n}, u_{i} rangle u_{i}$
But, it's still not looking anywhere closer to where I need to be.
Could somebody please help me finish this?
Thank you.
inner-product-space linear-transformations orthogonality orthonormal
inner-product-space linear-transformations orthogonality orthonormal
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
asked Dec 14 '15 at 5:50
ALannisterALannister
1,75231451
1,75231451
add a comment |
add a comment |
4 Answers
4
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oldest
votes
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It is simpler to start with $P(u_i)=sum_{j=1}^nlangle u_i,u_j rangle u_j$ and the only non vanishing term corresponds to $j=i$. So $P(u_i)=u_i$. Then we have by linearity
$$begin{align}P(P(x))&=sum_{i=1}^nlangle x,u_i rangle P(u_i)\&=sum_{i=1}^nlangle x,u_i rangle u_i\&=P(x)end{align}$$
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
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add a comment |
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Just a hint: You should make use of the orthonormal condition of the basis, i.e. $ langle u_i,u_jrangle=delta_{ij}$.
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
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add a comment |
$begingroup$
Applying $P$ to both sides of the equation $$ P(x) = sumlimits_{i=1}^n langle x, u_i rangle u_i$$ and using linearity, you get $$P(P(x)) = sumlimits_{i=1}^n langle x,u_i rangle P(u_i)$$ Now, what is $P(u_i)$?
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
$endgroup$
$begingroup$
$P(u_{i}) = sum_{i=1}^{n} langle u_{i}, u_{i} rangle u_{i}$. How does that help me?
$endgroup$
– ALannister
Dec 14 '15 at 5:59
$begingroup$
That's not quite right..
$endgroup$
– D_S
Dec 14 '15 at 6:00
$begingroup$
For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
$endgroup$
– D_S
Dec 14 '15 at 6:01
$begingroup$
enlighten me. Obviously, I don't know what I'm doing.
$endgroup$
– ALannister
Dec 14 '15 at 6:01
$begingroup$
Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = langle u_j, u_1 rangle u_1 + langle u_j, u_2 rangle u_2 + cdots + langle u_j, u_n rangle u_n$$
$endgroup$
– D_S
Dec 14 '15 at 6:02
add a comment |
$begingroup$
An approach might be: let's show that $P|_U=id_U$.
This is sufficient, because, since $operatorname{im}Psubseteq U$, $$P^2=Pcirc P=P|_Ucirc P=id_Ucirc P=P$$
Indeed, since ${u_1,cdots,u_n}$ is a basis, you only need to show that $P(u_i)=u_i$. But $$P(u_i)=sum_{j=1}^nlangle u_i,u_jrangle u_j=langle u_i,u_irangle u_i=u_i$$ $square$
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is simpler to start with $P(u_i)=sum_{j=1}^nlangle u_i,u_j rangle u_j$ and the only non vanishing term corresponds to $j=i$. So $P(u_i)=u_i$. Then we have by linearity
$$begin{align}P(P(x))&=sum_{i=1}^nlangle x,u_i rangle P(u_i)\&=sum_{i=1}^nlangle x,u_i rangle u_i\&=P(x)end{align}$$
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
$endgroup$
add a comment |
$begingroup$
It is simpler to start with $P(u_i)=sum_{j=1}^nlangle u_i,u_j rangle u_j$ and the only non vanishing term corresponds to $j=i$. So $P(u_i)=u_i$. Then we have by linearity
$$begin{align}P(P(x))&=sum_{i=1}^nlangle x,u_i rangle P(u_i)\&=sum_{i=1}^nlangle x,u_i rangle u_i\&=P(x)end{align}$$
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
$endgroup$
add a comment |
$begingroup$
It is simpler to start with $P(u_i)=sum_{j=1}^nlangle u_i,u_j rangle u_j$ and the only non vanishing term corresponds to $j=i$. So $P(u_i)=u_i$. Then we have by linearity
$$begin{align}P(P(x))&=sum_{i=1}^nlangle x,u_i rangle P(u_i)\&=sum_{i=1}^nlangle x,u_i rangle u_i\&=P(x)end{align}$$
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
$endgroup$
It is simpler to start with $P(u_i)=sum_{j=1}^nlangle u_i,u_j rangle u_j$ and the only non vanishing term corresponds to $j=i$. So $P(u_i)=u_i$. Then we have by linearity
$$begin{align}P(P(x))&=sum_{i=1}^nlangle x,u_i rangle P(u_i)\&=sum_{i=1}^nlangle x,u_i rangle u_i\&=P(x)end{align}$$
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
answered Dec 14 '15 at 5:59
marwalixmarwalix
13.3k12338
13.3k12338
add a comment |
add a comment |
$begingroup$
Just a hint: You should make use of the orthonormal condition of the basis, i.e. $ langle u_i,u_jrangle=delta_{ij}$.
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
$endgroup$
add a comment |
$begingroup$
Just a hint: You should make use of the orthonormal condition of the basis, i.e. $ langle u_i,u_jrangle=delta_{ij}$.
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
$endgroup$
add a comment |
$begingroup$
Just a hint: You should make use of the orthonormal condition of the basis, i.e. $ langle u_i,u_jrangle=delta_{ij}$.
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
$endgroup$
Just a hint: You should make use of the orthonormal condition of the basis, i.e. $ langle u_i,u_jrangle=delta_{ij}$.
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
edited Dec 14 '15 at 5:58
user228113
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
answered Dec 14 '15 at 5:57
Jie MinJie Min
1663
1663
add a comment |
add a comment |
$begingroup$
Applying $P$ to both sides of the equation $$ P(x) = sumlimits_{i=1}^n langle x, u_i rangle u_i$$ and using linearity, you get $$P(P(x)) = sumlimits_{i=1}^n langle x,u_i rangle P(u_i)$$ Now, what is $P(u_i)$?
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
$endgroup$
$begingroup$
$P(u_{i}) = sum_{i=1}^{n} langle u_{i}, u_{i} rangle u_{i}$. How does that help me?
$endgroup$
– ALannister
Dec 14 '15 at 5:59
$begingroup$
That's not quite right..
$endgroup$
– D_S
Dec 14 '15 at 6:00
$begingroup$
For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
$endgroup$
– D_S
Dec 14 '15 at 6:01
$begingroup$
enlighten me. Obviously, I don't know what I'm doing.
$endgroup$
– ALannister
Dec 14 '15 at 6:01
$begingroup$
Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = langle u_j, u_1 rangle u_1 + langle u_j, u_2 rangle u_2 + cdots + langle u_j, u_n rangle u_n$$
$endgroup$
– D_S
Dec 14 '15 at 6:02
add a comment |
$begingroup$
Applying $P$ to both sides of the equation $$ P(x) = sumlimits_{i=1}^n langle x, u_i rangle u_i$$ and using linearity, you get $$P(P(x)) = sumlimits_{i=1}^n langle x,u_i rangle P(u_i)$$ Now, what is $P(u_i)$?
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
$endgroup$
$begingroup$
$P(u_{i}) = sum_{i=1}^{n} langle u_{i}, u_{i} rangle u_{i}$. How does that help me?
$endgroup$
– ALannister
Dec 14 '15 at 5:59
$begingroup$
That's not quite right..
$endgroup$
– D_S
Dec 14 '15 at 6:00
$begingroup$
For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
$endgroup$
– D_S
Dec 14 '15 at 6:01
$begingroup$
enlighten me. Obviously, I don't know what I'm doing.
$endgroup$
– ALannister
Dec 14 '15 at 6:01
$begingroup$
Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = langle u_j, u_1 rangle u_1 + langle u_j, u_2 rangle u_2 + cdots + langle u_j, u_n rangle u_n$$
$endgroup$
– D_S
Dec 14 '15 at 6:02
add a comment |
$begingroup$
Applying $P$ to both sides of the equation $$ P(x) = sumlimits_{i=1}^n langle x, u_i rangle u_i$$ and using linearity, you get $$P(P(x)) = sumlimits_{i=1}^n langle x,u_i rangle P(u_i)$$ Now, what is $P(u_i)$?
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
$endgroup$
Applying $P$ to both sides of the equation $$ P(x) = sumlimits_{i=1}^n langle x, u_i rangle u_i$$ and using linearity, you get $$P(P(x)) = sumlimits_{i=1}^n langle x,u_i rangle P(u_i)$$ Now, what is $P(u_i)$?
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
answered Dec 14 '15 at 5:57
D_SD_S
13.4k51551
13.4k51551
$begingroup$
$P(u_{i}) = sum_{i=1}^{n} langle u_{i}, u_{i} rangle u_{i}$. How does that help me?
$endgroup$
– ALannister
Dec 14 '15 at 5:59
$begingroup$
That's not quite right..
$endgroup$
– D_S
Dec 14 '15 at 6:00
$begingroup$
For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
$endgroup$
– D_S
Dec 14 '15 at 6:01
$begingroup$
enlighten me. Obviously, I don't know what I'm doing.
$endgroup$
– ALannister
Dec 14 '15 at 6:01
$begingroup$
Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = langle u_j, u_1 rangle u_1 + langle u_j, u_2 rangle u_2 + cdots + langle u_j, u_n rangle u_n$$
$endgroup$
– D_S
Dec 14 '15 at 6:02
add a comment |
$begingroup$
$P(u_{i}) = sum_{i=1}^{n} langle u_{i}, u_{i} rangle u_{i}$. How does that help me?
$endgroup$
– ALannister
Dec 14 '15 at 5:59
$begingroup$
That's not quite right..
$endgroup$
– D_S
Dec 14 '15 at 6:00
$begingroup$
For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
$endgroup$
– D_S
Dec 14 '15 at 6:01
$begingroup$
enlighten me. Obviously, I don't know what I'm doing.
$endgroup$
– ALannister
Dec 14 '15 at 6:01
$begingroup$
Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = langle u_j, u_1 rangle u_1 + langle u_j, u_2 rangle u_2 + cdots + langle u_j, u_n rangle u_n$$
$endgroup$
– D_S
Dec 14 '15 at 6:02
$begingroup$
$P(u_{i}) = sum_{i=1}^{n} langle u_{i}, u_{i} rangle u_{i}$. How does that help me?
$endgroup$
– ALannister
Dec 14 '15 at 5:59
$begingroup$
$P(u_{i}) = sum_{i=1}^{n} langle u_{i}, u_{i} rangle u_{i}$. How does that help me?
$endgroup$
– ALannister
Dec 14 '15 at 5:59
$begingroup$
That's not quite right..
$endgroup$
– D_S
Dec 14 '15 at 6:00
$begingroup$
That's not quite right..
$endgroup$
– D_S
Dec 14 '15 at 6:00
$begingroup$
For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
$endgroup$
– D_S
Dec 14 '15 at 6:01
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For any fixed number $j$ between $1$ and $n$, what is $P(u_j)$? This is what you should focus on.
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– D_S
Dec 14 '15 at 6:01
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enlighten me. Obviously, I don't know what I'm doing.
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– ALannister
Dec 14 '15 at 6:01
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enlighten me. Obviously, I don't know what I'm doing.
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– ALannister
Dec 14 '15 at 6:01
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Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = langle u_j, u_1 rangle u_1 + langle u_j, u_2 rangle u_2 + cdots + langle u_j, u_n rangle u_n$$
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– D_S
Dec 14 '15 at 6:02
$begingroup$
Maybe it's clearer if you write it without the sigma notation. $$P(u_j) = langle u_j, u_1 rangle u_1 + langle u_j, u_2 rangle u_2 + cdots + langle u_j, u_n rangle u_n$$
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– D_S
Dec 14 '15 at 6:02
add a comment |
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An approach might be: let's show that $P|_U=id_U$.
This is sufficient, because, since $operatorname{im}Psubseteq U$, $$P^2=Pcirc P=P|_Ucirc P=id_Ucirc P=P$$
Indeed, since ${u_1,cdots,u_n}$ is a basis, you only need to show that $P(u_i)=u_i$. But $$P(u_i)=sum_{j=1}^nlangle u_i,u_jrangle u_j=langle u_i,u_irangle u_i=u_i$$ $square$
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add a comment |
$begingroup$
An approach might be: let's show that $P|_U=id_U$.
This is sufficient, because, since $operatorname{im}Psubseteq U$, $$P^2=Pcirc P=P|_Ucirc P=id_Ucirc P=P$$
Indeed, since ${u_1,cdots,u_n}$ is a basis, you only need to show that $P(u_i)=u_i$. But $$P(u_i)=sum_{j=1}^nlangle u_i,u_jrangle u_j=langle u_i,u_irangle u_i=u_i$$ $square$
$endgroup$
add a comment |
$begingroup$
An approach might be: let's show that $P|_U=id_U$.
This is sufficient, because, since $operatorname{im}Psubseteq U$, $$P^2=Pcirc P=P|_Ucirc P=id_Ucirc P=P$$
Indeed, since ${u_1,cdots,u_n}$ is a basis, you only need to show that $P(u_i)=u_i$. But $$P(u_i)=sum_{j=1}^nlangle u_i,u_jrangle u_j=langle u_i,u_irangle u_i=u_i$$ $square$
$endgroup$
An approach might be: let's show that $P|_U=id_U$.
This is sufficient, because, since $operatorname{im}Psubseteq U$, $$P^2=Pcirc P=P|_Ucirc P=id_Ucirc P=P$$
Indeed, since ${u_1,cdots,u_n}$ is a basis, you only need to show that $P(u_i)=u_i$. But $$P(u_i)=sum_{j=1}^nlangle u_i,u_jrangle u_j=langle u_i,u_irangle u_i=u_i$$ $square$
answered Dec 14 '15 at 6:03
user228113
add a comment |
add a comment |
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