standard matrix of a orthogonal projection linear transformation
$begingroup$
I am try to solve part (b) of question 1 in:
I'm not exactly sure how I should approach part (b). My attempt at the question so far is to plug in the elementary basis vectors $e_1, e_2, e_3$ into the span of vector given to see what the linear transformation does to $e_1, e_2, e_3$ and form the standard matrix from there which hasn't worked so far. Any hints?
linear-algebra
edited Jan 18 at 22:10
idriskameni
749321
asked Jan 18 at 21:42
lohboyslohboys
10319
$endgroup$
|
show 3 more comments
$begingroup$
I am try to solve part (b) of question 1 in:
I'm not exactly sure how I should approach part (b). My attempt at the question so far is to plug in the elementary basis vectors $e_1, e_2, e_3$ into the span of vector given to see what the linear transformation does to $e_1, e_2, e_3$ and form the standard matrix from there which hasn't worked so far. Any hints?
linear-algebra
edited Jan 18 at 22:10
idriskameni
749321
asked Jan 18 at 21:42
lohboyslohboys
10319
$endgroup$
$begingroup$
Try to think about how you would (orthogonally) project a vector $x$ onto the span of $e_1$ and $e_2$. Once you've done this, see if you can write it as a matrix operation.
$endgroup$
– tch
Jan 18 at 21:44
$begingroup$
do you mean project $e1$ and $e2$ with the span vectors? i.e. ((1,1,0)($e1$))($e1$) and so forth with $e2$ ?
$endgroup$
– lohboys
Jan 18 at 21:48
$begingroup$
Basically. I assume by $((1,1,0)(e1))$ you mean the inner product?
$endgroup$
– tch
Jan 18 at 22:08
$begingroup$
yes, inner product. So by computing what I suggested, I should be able to figure out the standard matrix for the linear transformation?
$endgroup$
– lohboys
Jan 18 at 22:11
$begingroup$
Sorry, instead of $e_1$ and $e_2$ i should have said $v_1$ and $v_2$, your orthonormal basis vectors for the span of $V$.
$endgroup$
– tch
Jan 18 at 22:19
|
show 3 more comments
$begingroup$
I am try to solve part (b) of question 1 in:
I'm not exactly sure how I should approach part (b). My attempt at the question so far is to plug in the elementary basis vectors $e_1, e_2, e_3$ into the span of vector given to see what the linear transformation does to $e_1, e_2, e_3$ and form the standard matrix from there which hasn't worked so far. Any hints?
linear-algebra
edited Jan 18 at 22:10
idriskameni
749321
asked Jan 18 at 21:42
lohboyslohboys
10319
$endgroup$
I am try to solve part (b) of question 1 in:
I'm not exactly sure how I should approach part (b). My attempt at the question so far is to plug in the elementary basis vectors $e_1, e_2, e_3$ into the span of vector given to see what the linear transformation does to $e_1, e_2, e_3$ and form the standard matrix from there which hasn't worked so far. Any hints?
linear-algebra
linear-algebra
edited Jan 18 at 22:10
idriskameni
749321
asked Jan 18 at 21:42
lohboyslohboys
10319
edited Jan 18 at 22:10
idriskameni
749321
asked Jan 18 at 21:42
lohboyslohboys
10319
edited Jan 18 at 22:10
idriskameni
749321
edited Jan 18 at 22:10
idriskameni
749321
edited Jan 18 at 22:10
idriskameni
749321
749321
asked Jan 18 at 21:42
lohboyslohboys
10319
asked Jan 18 at 21:42
lohboyslohboys
10319
asked Jan 18 at 21:42
lohboyslohboys
10319
10319
$begingroup$
Try to think about how you would (orthogonally) project a vector $x$ onto the span of $e_1$ and $e_2$. Once you've done this, see if you can write it as a matrix operation.
$endgroup$
– tch
Jan 18 at 21:44
$begingroup$
do you mean project $e1$ and $e2$ with the span vectors? i.e. ((1,1,0)($e1$))($e1$) and so forth with $e2$ ?
$endgroup$
– lohboys
Jan 18 at 21:48
$begingroup$
Basically. I assume by $((1,1,0)(e1))$ you mean the inner product?
$endgroup$
– tch
Jan 18 at 22:08
$begingroup$
yes, inner product. So by computing what I suggested, I should be able to figure out the standard matrix for the linear transformation?
$endgroup$
– lohboys
Jan 18 at 22:11
$begingroup$
Sorry, instead of $e_1$ and $e_2$ i should have said $v_1$ and $v_2$, your orthonormal basis vectors for the span of $V$.
$endgroup$
– tch
Jan 18 at 22:19
|
show 3 more comments
$begingroup$
Try to think about how you would (orthogonally) project a vector $x$ onto the span of $e_1$ and $e_2$. Once you've done this, see if you can write it as a matrix operation.
$endgroup$
– tch
Jan 18 at 21:44
$begingroup$
do you mean project $e1$ and $e2$ with the span vectors? i.e. ((1,1,0)($e1$))($e1$) and so forth with $e2$ ?
$endgroup$
– lohboys
Jan 18 at 21:48
$begingroup$
Basically. I assume by $((1,1,0)(e1))$ you mean the inner product?
$endgroup$
– tch
Jan 18 at 22:08
$begingroup$
yes, inner product. So by computing what I suggested, I should be able to figure out the standard matrix for the linear transformation?
$endgroup$
– lohboys
Jan 18 at 22:11
$begingroup$
Sorry, instead of $e_1$ and $e_2$ i should have said $v_1$ and $v_2$, your orthonormal basis vectors for the span of $V$.
$endgroup$
– tch
Jan 18 at 22:19
$begingroup$
Try to think about how you would (orthogonally) project a vector $x$ onto the span of $e_1$ and $e_2$. Once you've done this, see if you can write it as a matrix operation.
$endgroup$
– tch
Jan 18 at 21:44
$begingroup$
Try to think about how you would (orthogonally) project a vector $x$ onto the span of $e_1$ and $e_2$. Once you've done this, see if you can write it as a matrix operation.
$endgroup$
– tch
Jan 18 at 21:44
$begingroup$
do you mean project $e1$ and $e2$ with the span vectors? i.e. ((1,1,0)($e1$))($e1$) and so forth with $e2$ ?
$endgroup$
– lohboys
Jan 18 at 21:48
$begingroup$
do you mean project $e1$ and $e2$ with the span vectors? i.e. ((1,1,0)($e1$))($e1$) and so forth with $e2$ ?
$endgroup$
– lohboys
Jan 18 at 21:48
$begingroup$
Basically. I assume by $((1,1,0)(e1))$ you mean the inner product?
$endgroup$
– tch
Jan 18 at 22:08
$begingroup$
Basically. I assume by $((1,1,0)(e1))$ you mean the inner product?
$endgroup$
– tch
Jan 18 at 22:08
$begingroup$
yes, inner product. So by computing what I suggested, I should be able to figure out the standard matrix for the linear transformation?
$endgroup$
– lohboys
Jan 18 at 22:11
$begingroup$
yes, inner product. So by computing what I suggested, I should be able to figure out the standard matrix for the linear transformation?
$endgroup$
– lohboys
Jan 18 at 22:11
$begingroup$
Sorry, instead of $e_1$ and $e_2$ i should have said $v_1$ and $v_2$, your orthonormal basis vectors for the span of $V$.
$endgroup$
– tch
Jan 18 at 22:19
$begingroup$
Sorry, instead of $e_1$ and $e_2$ i should have said $v_1$ and $v_2$, your orthonormal basis vectors for the span of $V$.
$endgroup$
– tch
Jan 18 at 22:19
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Here is a way. We note that for $yin mathbb{R}^3$ the projected vector $Ay$ is such that $y-Ay$ is orthogonal to $V$. In particular if we form a matrix $B$ whose columns are $(1,1,0)^T$ and $(0,0,1)^T$ respectively, we get that
$$
B'(y-Ay)=0.
$$
But $Ay=Bc$ for some $c$ whence
$$
B'(y-Bc)=0iff B'y=B'Bc
$$
Since $B$ is full-rank, it follows that $B'B$ is invertible whence
$$
c=(B'B)^{-1}B'yimplies Ay=Bc=B(B'B)^{-1}B'y.
$$
So $A=B(B'B)^{-1}B'$.
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
$endgroup$
$begingroup$
this is quite a complex explanation for myself, what exactly is $B'$ in this instance?
$endgroup$
– lohboys
Jan 18 at 22:30
1
$begingroup$
The transpose of the matrix.
$endgroup$
– Foobaz John
Jan 18 at 22:48
add a comment |
$begingroup$
To expand on my comments since they were getting long.
Suppose we have a set of orthonormal vectors $v_1$ and $v_2$. If we want to compute the orthogonal projection of $x$ onto the span of $v_1$ and $v_2$ we would first project $x$ onto $v_1$:
$$
y = P_1x = (v_1^Tx) v_1
$$
and then onto $v_2$:
$$
P_2x = (v_2^Tx)v_2
$$
Therefore,
$$
Px = (v_1^Tx)v_1 + (v_2^Tx)v_2
$$
We can write this as the sum of rank-1 outer products:
$$
Px = v_1v_1^Tx + v_2v_2^Tx = (v_1v_1^T+v_2v_2^T)x
$$
Therefore,
$$
P = VV^T = [v_1 v_2]
begin{bmatrix}
v_1^T \ v_2^T
end{bmatrix}
$$
If $v_1$ and $v_2$ are not orthonormal you can first orthonormalize them (using Gram-Schmidt) and then do this. Alternatively, you can use the formula Foobaz John gave.
answered Jan 19 at 0:18
tchtch
833310
$endgroup$
$begingroup$
if I'm interpreting this correctly, $P$ gives the standard matrix of this linear transformation yes?
$endgroup$
– lohboys
Jan 19 at 2:46
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a way. We note that for $yin mathbb{R}^3$ the projected vector $Ay$ is such that $y-Ay$ is orthogonal to $V$. In particular if we form a matrix $B$ whose columns are $(1,1,0)^T$ and $(0,0,1)^T$ respectively, we get that
$$
B'(y-Ay)=0.
$$
But $Ay=Bc$ for some $c$ whence
$$
B'(y-Bc)=0iff B'y=B'Bc
$$
Since $B$ is full-rank, it follows that $B'B$ is invertible whence
$$
c=(B'B)^{-1}B'yimplies Ay=Bc=B(B'B)^{-1}B'y.
$$
So $A=B(B'B)^{-1}B'$.
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
$endgroup$
$begingroup$
this is quite a complex explanation for myself, what exactly is $B'$ in this instance?
$endgroup$
– lohboys
Jan 18 at 22:30
1
$begingroup$
The transpose of the matrix.
$endgroup$
– Foobaz John
Jan 18 at 22:48
add a comment |
$begingroup$
Here is a way. We note that for $yin mathbb{R}^3$ the projected vector $Ay$ is such that $y-Ay$ is orthogonal to $V$. In particular if we form a matrix $B$ whose columns are $(1,1,0)^T$ and $(0,0,1)^T$ respectively, we get that
$$
B'(y-Ay)=0.
$$
But $Ay=Bc$ for some $c$ whence
$$
B'(y-Bc)=0iff B'y=B'Bc
$$
Since $B$ is full-rank, it follows that $B'B$ is invertible whence
$$
c=(B'B)^{-1}B'yimplies Ay=Bc=B(B'B)^{-1}B'y.
$$
So $A=B(B'B)^{-1}B'$.
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
$endgroup$
$begingroup$
this is quite a complex explanation for myself, what exactly is $B'$ in this instance?
$endgroup$
– lohboys
Jan 18 at 22:30
1
$begingroup$
The transpose of the matrix.
$endgroup$
– Foobaz John
Jan 18 at 22:48
add a comment |
$begingroup$
Here is a way. We note that for $yin mathbb{R}^3$ the projected vector $Ay$ is such that $y-Ay$ is orthogonal to $V$. In particular if we form a matrix $B$ whose columns are $(1,1,0)^T$ and $(0,0,1)^T$ respectively, we get that
$$
B'(y-Ay)=0.
$$
But $Ay=Bc$ for some $c$ whence
$$
B'(y-Bc)=0iff B'y=B'Bc
$$
Since $B$ is full-rank, it follows that $B'B$ is invertible whence
$$
c=(B'B)^{-1}B'yimplies Ay=Bc=B(B'B)^{-1}B'y.
$$
So $A=B(B'B)^{-1}B'$.
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
$endgroup$
Here is a way. We note that for $yin mathbb{R}^3$ the projected vector $Ay$ is such that $y-Ay$ is orthogonal to $V$. In particular if we form a matrix $B$ whose columns are $(1,1,0)^T$ and $(0,0,1)^T$ respectively, we get that
$$
B'(y-Ay)=0.
$$
But $Ay=Bc$ for some $c$ whence
$$
B'(y-Bc)=0iff B'y=B'Bc
$$
Since $B$ is full-rank, it follows that $B'B$ is invertible whence
$$
c=(B'B)^{-1}B'yimplies Ay=Bc=B(B'B)^{-1}B'y.
$$
So $A=B(B'B)^{-1}B'$.
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
answered Jan 18 at 22:15
Foobaz JohnFoobaz John
23k41552
23k41552
$begingroup$
this is quite a complex explanation for myself, what exactly is $B'$ in this instance?
$endgroup$
– lohboys
Jan 18 at 22:30
1
$begingroup$
The transpose of the matrix.
$endgroup$
– Foobaz John
Jan 18 at 22:48
add a comment |
$begingroup$
this is quite a complex explanation for myself, what exactly is $B'$ in this instance?
$endgroup$
– lohboys
Jan 18 at 22:30
1
$begingroup$
The transpose of the matrix.
$endgroup$
– Foobaz John
Jan 18 at 22:48
$begingroup$
this is quite a complex explanation for myself, what exactly is $B'$ in this instance?
$endgroup$
– lohboys
Jan 18 at 22:30
$begingroup$
this is quite a complex explanation for myself, what exactly is $B'$ in this instance?
$endgroup$
– lohboys
Jan 18 at 22:30
1
1
$begingroup$
The transpose of the matrix.
$endgroup$
– Foobaz John
Jan 18 at 22:48
$begingroup$
The transpose of the matrix.
$endgroup$
– Foobaz John
Jan 18 at 22:48
add a comment |
$begingroup$
To expand on my comments since they were getting long.
Suppose we have a set of orthonormal vectors $v_1$ and $v_2$. If we want to compute the orthogonal projection of $x$ onto the span of $v_1$ and $v_2$ we would first project $x$ onto $v_1$:
$$
y = P_1x = (v_1^Tx) v_1
$$
and then onto $v_2$:
$$
P_2x = (v_2^Tx)v_2
$$
Therefore,
$$
Px = (v_1^Tx)v_1 + (v_2^Tx)v_2
$$
We can write this as the sum of rank-1 outer products:
$$
Px = v_1v_1^Tx + v_2v_2^Tx = (v_1v_1^T+v_2v_2^T)x
$$
Therefore,
$$
P = VV^T = [v_1 v_2]
begin{bmatrix}
v_1^T \ v_2^T
end{bmatrix}
$$
If $v_1$ and $v_2$ are not orthonormal you can first orthonormalize them (using Gram-Schmidt) and then do this. Alternatively, you can use the formula Foobaz John gave.
answered Jan 19 at 0:18
tchtch
833310
$endgroup$
$begingroup$
if I'm interpreting this correctly, $P$ gives the standard matrix of this linear transformation yes?
$endgroup$
– lohboys
Jan 19 at 2:46
add a comment |
$begingroup$
To expand on my comments since they were getting long.
Suppose we have a set of orthonormal vectors $v_1$ and $v_2$. If we want to compute the orthogonal projection of $x$ onto the span of $v_1$ and $v_2$ we would first project $x$ onto $v_1$:
$$
y = P_1x = (v_1^Tx) v_1
$$
and then onto $v_2$:
$$
P_2x = (v_2^Tx)v_2
$$
Therefore,
$$
Px = (v_1^Tx)v_1 + (v_2^Tx)v_2
$$
We can write this as the sum of rank-1 outer products:
$$
Px = v_1v_1^Tx + v_2v_2^Tx = (v_1v_1^T+v_2v_2^T)x
$$
Therefore,
$$
P = VV^T = [v_1 v_2]
begin{bmatrix}
v_1^T \ v_2^T
end{bmatrix}
$$
If $v_1$ and $v_2$ are not orthonormal you can first orthonormalize them (using Gram-Schmidt) and then do this. Alternatively, you can use the formula Foobaz John gave.
answered Jan 19 at 0:18
tchtch
833310
$endgroup$
$begingroup$
if I'm interpreting this correctly, $P$ gives the standard matrix of this linear transformation yes?
$endgroup$
– lohboys
Jan 19 at 2:46
add a comment |
$begingroup$
To expand on my comments since they were getting long.
Suppose we have a set of orthonormal vectors $v_1$ and $v_2$. If we want to compute the orthogonal projection of $x$ onto the span of $v_1$ and $v_2$ we would first project $x$ onto $v_1$:
$$
y = P_1x = (v_1^Tx) v_1
$$
and then onto $v_2$:
$$
P_2x = (v_2^Tx)v_2
$$
Therefore,
$$
Px = (v_1^Tx)v_1 + (v_2^Tx)v_2
$$
We can write this as the sum of rank-1 outer products:
$$
Px = v_1v_1^Tx + v_2v_2^Tx = (v_1v_1^T+v_2v_2^T)x
$$
Therefore,
$$
P = VV^T = [v_1 v_2]
begin{bmatrix}
v_1^T \ v_2^T
end{bmatrix}
$$
If $v_1$ and $v_2$ are not orthonormal you can first orthonormalize them (using Gram-Schmidt) and then do this. Alternatively, you can use the formula Foobaz John gave.
answered Jan 19 at 0:18
tchtch
833310
$endgroup$
To expand on my comments since they were getting long.
Suppose we have a set of orthonormal vectors $v_1$ and $v_2$. If we want to compute the orthogonal projection of $x$ onto the span of $v_1$ and $v_2$ we would first project $x$ onto $v_1$:
$$
y = P_1x = (v_1^Tx) v_1
$$
and then onto $v_2$:
$$
P_2x = (v_2^Tx)v_2
$$
Therefore,
$$
Px = (v_1^Tx)v_1 + (v_2^Tx)v_2
$$
We can write this as the sum of rank-1 outer products:
$$
Px = v_1v_1^Tx + v_2v_2^Tx = (v_1v_1^T+v_2v_2^T)x
$$
Therefore,
$$
P = VV^T = [v_1 v_2]
begin{bmatrix}
v_1^T \ v_2^T
end{bmatrix}
$$
If $v_1$ and $v_2$ are not orthonormal you can first orthonormalize them (using Gram-Schmidt) and then do this. Alternatively, you can use the formula Foobaz John gave.
answered Jan 19 at 0:18
tchtch
833310
answered Jan 19 at 0:18
tchtch
833310
answered Jan 19 at 0:18
tchtch
833310
answered Jan 19 at 0:18
tchtch
833310
833310
$begingroup$
if I'm interpreting this correctly, $P$ gives the standard matrix of this linear transformation yes?
$endgroup$
– lohboys
Jan 19 at 2:46
add a comment |
$begingroup$
if I'm interpreting this correctly, $P$ gives the standard matrix of this linear transformation yes?
$endgroup$
– lohboys
Jan 19 at 2:46
$begingroup$
if I'm interpreting this correctly, $P$ gives the standard matrix of this linear transformation yes?
$endgroup$
– lohboys
Jan 19 at 2:46
$begingroup$
if I'm interpreting this correctly, $P$ gives the standard matrix of this linear transformation yes?
$endgroup$
– lohboys
Jan 19 at 2:46
add a comment |
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$begingroup$
Try to think about how you would (orthogonally) project a vector $x$ onto the span of $e_1$ and $e_2$. Once you've done this, see if you can write it as a matrix operation.
$endgroup$
– tch
Jan 18 at 21:44
$begingroup$
do you mean project $e1$ and $e2$ with the span vectors? i.e. ((1,1,0)($e1$))($e1$) and so forth with $e2$ ?
$endgroup$
– lohboys
Jan 18 at 21:48
$begingroup$
Basically. I assume by $((1,1,0)(e1))$ you mean the inner product?
$endgroup$
– tch
Jan 18 at 22:08
$begingroup$
yes, inner product. So by computing what I suggested, I should be able to figure out the standard matrix for the linear transformation?
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– lohboys
Jan 18 at 22:11
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Sorry, instead of $e_1$ and $e_2$ i should have said $v_1$ and $v_2$, your orthonormal basis vectors for the span of $V$.
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– tch
Jan 18 at 22:19