Setting up the standard matrix for orthogonal projection
$begingroup$
For a subspace $V$ of $mathbb{R}^4$, you are given these three ordered bases:
$A= (mathbf{a}=(1,-2,-1,3), mathbf{b}=(1,3,-2,-4))$
$B= (mathbf{a}=(1,-2,-1,3), mathbf{c}=(2,1,-3,-1))$
$C= (mathbf{b}=(1,3,-2,4), mathbf{d}=(0,5,-1,-7))$
Let $T: mathbb{R}^4 rightarrow mathbb{R}^4$ be the orthogonal projection of $mathbb{R}^4$ onto $V$.
$a)$ Explain why $T$ is or is not invertible.
$b)$ Find the matrix $[T]$ which represents $T$ with respect to the standard coordinates.
I really need some guidance with this... Why isn't the matrix invertible? Is it because the matrix $T$ is not a square matrix?
For part $b)$, how do I do it? I tried to use the projection formula $[T]=A(A^TA)^{-1}A^T$ but the matrix isn't invertible...
How do I approach this? I'm very much stumped on this.
linear-algebra inverse transformation
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
$endgroup$
add a comment |
$begingroup$
For a subspace $V$ of $mathbb{R}^4$, you are given these three ordered bases:
$A= (mathbf{a}=(1,-2,-1,3), mathbf{b}=(1,3,-2,-4))$
$B= (mathbf{a}=(1,-2,-1,3), mathbf{c}=(2,1,-3,-1))$
$C= (mathbf{b}=(1,3,-2,4), mathbf{d}=(0,5,-1,-7))$
Let $T: mathbb{R}^4 rightarrow mathbb{R}^4$ be the orthogonal projection of $mathbb{R}^4$ onto $V$.
$a)$ Explain why $T$ is or is not invertible.
$b)$ Find the matrix $[T]$ which represents $T$ with respect to the standard coordinates.
I really need some guidance with this... Why isn't the matrix invertible? Is it because the matrix $T$ is not a square matrix?
For part $b)$, how do I do it? I tried to use the projection formula $[T]=A(A^TA)^{-1}A^T$ but the matrix isn't invertible...
How do I approach this? I'm very much stumped on this.
linear-algebra inverse transformation
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
$endgroup$
$begingroup$
$T$ has better be square since it represents a map from $mathbb R^4$ into itself. Think about the rank of $T$.
$endgroup$
– amd
Dec 23 '18 at 7:18
$begingroup$
Ohh yeah. Point made. Is it not invertible since the matrix is not onto? The transformation takes vectors in 4D space and spits out a vectors which span a 2D space which does span $mathbb{R}^4$. If that's nonsense, how does the rank work? Rank+nullity=dimension. So that would mean $4+nullity=4$ and thus $nullity=0$. Woudn't that mean the matrix IS invertible though?
$endgroup$
– Future Math person
Dec 23 '18 at 7:29
add a comment |
$begingroup$
For a subspace $V$ of $mathbb{R}^4$, you are given these three ordered bases:
$A= (mathbf{a}=(1,-2,-1,3), mathbf{b}=(1,3,-2,-4))$
$B= (mathbf{a}=(1,-2,-1,3), mathbf{c}=(2,1,-3,-1))$
$C= (mathbf{b}=(1,3,-2,4), mathbf{d}=(0,5,-1,-7))$
Let $T: mathbb{R}^4 rightarrow mathbb{R}^4$ be the orthogonal projection of $mathbb{R}^4$ onto $V$.
$a)$ Explain why $T$ is or is not invertible.
$b)$ Find the matrix $[T]$ which represents $T$ with respect to the standard coordinates.
I really need some guidance with this... Why isn't the matrix invertible? Is it because the matrix $T$ is not a square matrix?
For part $b)$, how do I do it? I tried to use the projection formula $[T]=A(A^TA)^{-1}A^T$ but the matrix isn't invertible...
How do I approach this? I'm very much stumped on this.
linear-algebra inverse transformation
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
$endgroup$
For a subspace $V$ of $mathbb{R}^4$, you are given these three ordered bases:
$A= (mathbf{a}=(1,-2,-1,3), mathbf{b}=(1,3,-2,-4))$
$B= (mathbf{a}=(1,-2,-1,3), mathbf{c}=(2,1,-3,-1))$
$C= (mathbf{b}=(1,3,-2,4), mathbf{d}=(0,5,-1,-7))$
Let $T: mathbb{R}^4 rightarrow mathbb{R}^4$ be the orthogonal projection of $mathbb{R}^4$ onto $V$.
$a)$ Explain why $T$ is or is not invertible.
$b)$ Find the matrix $[T]$ which represents $T$ with respect to the standard coordinates.
I really need some guidance with this... Why isn't the matrix invertible? Is it because the matrix $T$ is not a square matrix?
For part $b)$, how do I do it? I tried to use the projection formula $[T]=A(A^TA)^{-1}A^T$ but the matrix isn't invertible...
How do I approach this? I'm very much stumped on this.
linear-algebra inverse transformation
linear-algebra inverse transformation
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
edited Dec 23 '18 at 9:03
Future Math person
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
993818
993818
$begingroup$
$T$ has better be square since it represents a map from $mathbb R^4$ into itself. Think about the rank of $T$.
$endgroup$
– amd
Dec 23 '18 at 7:18
$begingroup$
Ohh yeah. Point made. Is it not invertible since the matrix is not onto? The transformation takes vectors in 4D space and spits out a vectors which span a 2D space which does span $mathbb{R}^4$. If that's nonsense, how does the rank work? Rank+nullity=dimension. So that would mean $4+nullity=4$ and thus $nullity=0$. Woudn't that mean the matrix IS invertible though?
$endgroup$
– Future Math person
Dec 23 '18 at 7:29
add a comment |
$begingroup$
$T$ has better be square since it represents a map from $mathbb R^4$ into itself. Think about the rank of $T$.
$endgroup$
– amd
Dec 23 '18 at 7:18
$begingroup$
Ohh yeah. Point made. Is it not invertible since the matrix is not onto? The transformation takes vectors in 4D space and spits out a vectors which span a 2D space which does span $mathbb{R}^4$. If that's nonsense, how does the rank work? Rank+nullity=dimension. So that would mean $4+nullity=4$ and thus $nullity=0$. Woudn't that mean the matrix IS invertible though?
$endgroup$
– Future Math person
Dec 23 '18 at 7:29
$begingroup$
$T$ has better be square since it represents a map from $mathbb R^4$ into itself. Think about the rank of $T$.
$endgroup$
– amd
Dec 23 '18 at 7:18
$begingroup$
$T$ has better be square since it represents a map from $mathbb R^4$ into itself. Think about the rank of $T$.
$endgroup$
– amd
Dec 23 '18 at 7:18
$begingroup$
Ohh yeah. Point made. Is it not invertible since the matrix is not onto? The transformation takes vectors in 4D space and spits out a vectors which span a 2D space which does span $mathbb{R}^4$. If that's nonsense, how does the rank work? Rank+nullity=dimension. So that would mean $4+nullity=4$ and thus $nullity=0$. Woudn't that mean the matrix IS invertible though?
$endgroup$
– Future Math person
Dec 23 '18 at 7:29
$begingroup$
Ohh yeah. Point made. Is it not invertible since the matrix is not onto? The transformation takes vectors in 4D space and spits out a vectors which span a 2D space which does span $mathbb{R}^4$. If that's nonsense, how does the rank work? Rank+nullity=dimension. So that would mean $4+nullity=4$ and thus $nullity=0$. Woudn't that mean the matrix IS invertible though?
$endgroup$
– Future Math person
Dec 23 '18 at 7:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall that a linear transformation from $A^mto B^n$, where $m,n$ are dimensions of $A,B$ respectively, has a matrix of order $ntimes m$. Thus, $T:Bbb R^4toBbb R^4$ is a linear map having a $4times4$ square matrix, $[T]$.
For showing that $T$ is not invertible, you can show that there is an element in $ker(T)$ other than $(0,0,0,0)$, that is, $dim(ker(T))>0$. For example, $mathbf u=(7,1,5,0)$ has $bf0$ projection on $V$. Alternatively, you can solve $(b)$ first and show that the matrix of $T$ isn't invertible, that is, $text{rank}([T])<4$.
For part $(b)$, let $mathbf v=(x,y,z,w)inBbb R^4$. The basis $B={mathbf a,mathbf c}$ of $V$ consists of orthogonal linearly independent vectors, so the orthogonal projection of $mathbf v=(x,y,z,w)$ on $V$ is the vector sum $$displaystylefrac{langlemathbf v,mathbf arangle}{langlemathbf a,mathbf arangle}cdotmathbf a+frac{langlemathbf v,mathbf crangle}{langlemathbf c,mathbf crangle}cdotmathbf c$$
Note that you are only able to do this because the basis vectors are orthogonal.
This gives the orthogonal projection of $mathbf v=(x,y,z,w)$ as
$displaystyle=Big[frac{x-2y-z+3w}{15}Big]cdot(1,-2,-1,3)+Big[frac{2x+y-3z-w}{15}Big]cdot(2,1,-3,-1)\displaystyle=Big(frac{5x-7z+w}{15},frac{5y-z-7w}{15},frac{-7x-y+10z}{15},frac{x-7y+10w}{15}Big)$
Thus, the matrix of $T,[T]=begin{bmatrix}1/3&0&-7/15&1/15\0&1/3&-1/15&-7/15\-7/15&-1/15&2/3&0\1/15&-7/15&0&2/3end{bmatrix}$.
Edit. As hinted by you in a comment, we can show that $T$ is not invertible by noting that the range $R(T)$ is two-dimensional, so nullity of $T>0$.
edited Dec 23 '18 at 9:43
Future Math person
993818
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
$endgroup$
1
$begingroup$
I just read your comment on the original post. You gave a fine reasoning for $T$ not being invertible: it is not onto. In other words, its range $R(T)$ is two-dimensional, so the rank of $[T]$, which is nothing but the dimension of the range of $T$, is $2$. This means the nullity of $T=2>0$, so $T$ is not invertible.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:41
1
$begingroup$
As for how I calculated $bf u$, recall that the cross-product of two vectors gives you a vector that is orthogonal to both the vectors. A vector that is orthogonal to another vector has $bf0$ projection along the latter. But this orthogonality property of the cross-product doesn't exist for vectors in $Bbb R^4$. So I took the cross-product of the first $3$ co-ordinates of the basis vectors of $A$, and made the fourth co-ordinate of $bf u$ zero.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:48
1
$begingroup$
As for why $bf u$ has $bf0$ projection on $V$, you know that any vector in $V$ can be written as $c_1mathbf a+c_2mathbf b$, so $langlemathbf u,c_1mathbf a+c_2mathbf brangle=c_1langlemathbf u,mathbf arangle+c_2langlemathbf u,mathbf brangle=0+0$
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:52
1
$begingroup$
Let's say we have any two basis vectors $bf v_1,v_2$ of $V$. Then the projection of $bf u$ along $bf v_1$ is $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}$ and similarly for its projection along $bf v_2$. But in general, the projection of $bf u$ on $V$ is not equal to $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}+displaystylefrac{langlemathbf u,mathbf{v_2}rangle}{langlemathbf{v_2},mathbf{v_2}rangle}cdotmathbf{v_2}$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:10
1
$begingroup$
For this to happen, $bf v_1,v_2$ must be orthogonal. The only basis containing orthogonal linearly independent vectors is $A$
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:11
|
show 11 more comments
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050130%2fsetting-up-the-standard-matrix-for-orthogonal-projection%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Recall that a linear transformation from $A^mto B^n$, where $m,n$ are dimensions of $A,B$ respectively, has a matrix of order $ntimes m$. Thus, $T:Bbb R^4toBbb R^4$ is a linear map having a $4times4$ square matrix, $[T]$.
For showing that $T$ is not invertible, you can show that there is an element in $ker(T)$ other than $(0,0,0,0)$, that is, $dim(ker(T))>0$. For example, $mathbf u=(7,1,5,0)$ has $bf0$ projection on $V$. Alternatively, you can solve $(b)$ first and show that the matrix of $T$ isn't invertible, that is, $text{rank}([T])<4$.
For part $(b)$, let $mathbf v=(x,y,z,w)inBbb R^4$. The basis $B={mathbf a,mathbf c}$ of $V$ consists of orthogonal linearly independent vectors, so the orthogonal projection of $mathbf v=(x,y,z,w)$ on $V$ is the vector sum $$displaystylefrac{langlemathbf v,mathbf arangle}{langlemathbf a,mathbf arangle}cdotmathbf a+frac{langlemathbf v,mathbf crangle}{langlemathbf c,mathbf crangle}cdotmathbf c$$
Note that you are only able to do this because the basis vectors are orthogonal.
This gives the orthogonal projection of $mathbf v=(x,y,z,w)$ as
$displaystyle=Big[frac{x-2y-z+3w}{15}Big]cdot(1,-2,-1,3)+Big[frac{2x+y-3z-w}{15}Big]cdot(2,1,-3,-1)\displaystyle=Big(frac{5x-7z+w}{15},frac{5y-z-7w}{15},frac{-7x-y+10z}{15},frac{x-7y+10w}{15}Big)$
Thus, the matrix of $T,[T]=begin{bmatrix}1/3&0&-7/15&1/15\0&1/3&-1/15&-7/15\-7/15&-1/15&2/3&0\1/15&-7/15&0&2/3end{bmatrix}$.
Edit. As hinted by you in a comment, we can show that $T$ is not invertible by noting that the range $R(T)$ is two-dimensional, so nullity of $T>0$.
edited Dec 23 '18 at 9:43
Future Math person
993818
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
$endgroup$
1
$begingroup$
I just read your comment on the original post. You gave a fine reasoning for $T$ not being invertible: it is not onto. In other words, its range $R(T)$ is two-dimensional, so the rank of $[T]$, which is nothing but the dimension of the range of $T$, is $2$. This means the nullity of $T=2>0$, so $T$ is not invertible.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:41
1
$begingroup$
As for how I calculated $bf u$, recall that the cross-product of two vectors gives you a vector that is orthogonal to both the vectors. A vector that is orthogonal to another vector has $bf0$ projection along the latter. But this orthogonality property of the cross-product doesn't exist for vectors in $Bbb R^4$. So I took the cross-product of the first $3$ co-ordinates of the basis vectors of $A$, and made the fourth co-ordinate of $bf u$ zero.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:48
1
$begingroup$
As for why $bf u$ has $bf0$ projection on $V$, you know that any vector in $V$ can be written as $c_1mathbf a+c_2mathbf b$, so $langlemathbf u,c_1mathbf a+c_2mathbf brangle=c_1langlemathbf u,mathbf arangle+c_2langlemathbf u,mathbf brangle=0+0$
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:52
1
$begingroup$
Let's say we have any two basis vectors $bf v_1,v_2$ of $V$. Then the projection of $bf u$ along $bf v_1$ is $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}$ and similarly for its projection along $bf v_2$. But in general, the projection of $bf u$ on $V$ is not equal to $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}+displaystylefrac{langlemathbf u,mathbf{v_2}rangle}{langlemathbf{v_2},mathbf{v_2}rangle}cdotmathbf{v_2}$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:10
1
$begingroup$
For this to happen, $bf v_1,v_2$ must be orthogonal. The only basis containing orthogonal linearly independent vectors is $A$
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:11
|
show 11 more comments
$begingroup$
Recall that a linear transformation from $A^mto B^n$, where $m,n$ are dimensions of $A,B$ respectively, has a matrix of order $ntimes m$. Thus, $T:Bbb R^4toBbb R^4$ is a linear map having a $4times4$ square matrix, $[T]$.
For showing that $T$ is not invertible, you can show that there is an element in $ker(T)$ other than $(0,0,0,0)$, that is, $dim(ker(T))>0$. For example, $mathbf u=(7,1,5,0)$ has $bf0$ projection on $V$. Alternatively, you can solve $(b)$ first and show that the matrix of $T$ isn't invertible, that is, $text{rank}([T])<4$.
For part $(b)$, let $mathbf v=(x,y,z,w)inBbb R^4$. The basis $B={mathbf a,mathbf c}$ of $V$ consists of orthogonal linearly independent vectors, so the orthogonal projection of $mathbf v=(x,y,z,w)$ on $V$ is the vector sum $$displaystylefrac{langlemathbf v,mathbf arangle}{langlemathbf a,mathbf arangle}cdotmathbf a+frac{langlemathbf v,mathbf crangle}{langlemathbf c,mathbf crangle}cdotmathbf c$$
Note that you are only able to do this because the basis vectors are orthogonal.
This gives the orthogonal projection of $mathbf v=(x,y,z,w)$ as
$displaystyle=Big[frac{x-2y-z+3w}{15}Big]cdot(1,-2,-1,3)+Big[frac{2x+y-3z-w}{15}Big]cdot(2,1,-3,-1)\displaystyle=Big(frac{5x-7z+w}{15},frac{5y-z-7w}{15},frac{-7x-y+10z}{15},frac{x-7y+10w}{15}Big)$
Thus, the matrix of $T,[T]=begin{bmatrix}1/3&0&-7/15&1/15\0&1/3&-1/15&-7/15\-7/15&-1/15&2/3&0\1/15&-7/15&0&2/3end{bmatrix}$.
Edit. As hinted by you in a comment, we can show that $T$ is not invertible by noting that the range $R(T)$ is two-dimensional, so nullity of $T>0$.
edited Dec 23 '18 at 9:43
Future Math person
993818
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
$endgroup$
1
$begingroup$
I just read your comment on the original post. You gave a fine reasoning for $T$ not being invertible: it is not onto. In other words, its range $R(T)$ is two-dimensional, so the rank of $[T]$, which is nothing but the dimension of the range of $T$, is $2$. This means the nullity of $T=2>0$, so $T$ is not invertible.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:41
1
$begingroup$
As for how I calculated $bf u$, recall that the cross-product of two vectors gives you a vector that is orthogonal to both the vectors. A vector that is orthogonal to another vector has $bf0$ projection along the latter. But this orthogonality property of the cross-product doesn't exist for vectors in $Bbb R^4$. So I took the cross-product of the first $3$ co-ordinates of the basis vectors of $A$, and made the fourth co-ordinate of $bf u$ zero.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:48
1
$begingroup$
As for why $bf u$ has $bf0$ projection on $V$, you know that any vector in $V$ can be written as $c_1mathbf a+c_2mathbf b$, so $langlemathbf u,c_1mathbf a+c_2mathbf brangle=c_1langlemathbf u,mathbf arangle+c_2langlemathbf u,mathbf brangle=0+0$
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:52
1
$begingroup$
Let's say we have any two basis vectors $bf v_1,v_2$ of $V$. Then the projection of $bf u$ along $bf v_1$ is $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}$ and similarly for its projection along $bf v_2$. But in general, the projection of $bf u$ on $V$ is not equal to $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}+displaystylefrac{langlemathbf u,mathbf{v_2}rangle}{langlemathbf{v_2},mathbf{v_2}rangle}cdotmathbf{v_2}$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:10
1
$begingroup$
For this to happen, $bf v_1,v_2$ must be orthogonal. The only basis containing orthogonal linearly independent vectors is $A$
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:11
|
show 11 more comments
$begingroup$
Recall that a linear transformation from $A^mto B^n$, where $m,n$ are dimensions of $A,B$ respectively, has a matrix of order $ntimes m$. Thus, $T:Bbb R^4toBbb R^4$ is a linear map having a $4times4$ square matrix, $[T]$.
For showing that $T$ is not invertible, you can show that there is an element in $ker(T)$ other than $(0,0,0,0)$, that is, $dim(ker(T))>0$. For example, $mathbf u=(7,1,5,0)$ has $bf0$ projection on $V$. Alternatively, you can solve $(b)$ first and show that the matrix of $T$ isn't invertible, that is, $text{rank}([T])<4$.
For part $(b)$, let $mathbf v=(x,y,z,w)inBbb R^4$. The basis $B={mathbf a,mathbf c}$ of $V$ consists of orthogonal linearly independent vectors, so the orthogonal projection of $mathbf v=(x,y,z,w)$ on $V$ is the vector sum $$displaystylefrac{langlemathbf v,mathbf arangle}{langlemathbf a,mathbf arangle}cdotmathbf a+frac{langlemathbf v,mathbf crangle}{langlemathbf c,mathbf crangle}cdotmathbf c$$
Note that you are only able to do this because the basis vectors are orthogonal.
This gives the orthogonal projection of $mathbf v=(x,y,z,w)$ as
$displaystyle=Big[frac{x-2y-z+3w}{15}Big]cdot(1,-2,-1,3)+Big[frac{2x+y-3z-w}{15}Big]cdot(2,1,-3,-1)\displaystyle=Big(frac{5x-7z+w}{15},frac{5y-z-7w}{15},frac{-7x-y+10z}{15},frac{x-7y+10w}{15}Big)$
Thus, the matrix of $T,[T]=begin{bmatrix}1/3&0&-7/15&1/15\0&1/3&-1/15&-7/15\-7/15&-1/15&2/3&0\1/15&-7/15&0&2/3end{bmatrix}$.
Edit. As hinted by you in a comment, we can show that $T$ is not invertible by noting that the range $R(T)$ is two-dimensional, so nullity of $T>0$.
edited Dec 23 '18 at 9:43
Future Math person
993818
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
$endgroup$
Recall that a linear transformation from $A^mto B^n$, where $m,n$ are dimensions of $A,B$ respectively, has a matrix of order $ntimes m$. Thus, $T:Bbb R^4toBbb R^4$ is a linear map having a $4times4$ square matrix, $[T]$.
For showing that $T$ is not invertible, you can show that there is an element in $ker(T)$ other than $(0,0,0,0)$, that is, $dim(ker(T))>0$. For example, $mathbf u=(7,1,5,0)$ has $bf0$ projection on $V$. Alternatively, you can solve $(b)$ first and show that the matrix of $T$ isn't invertible, that is, $text{rank}([T])<4$.
For part $(b)$, let $mathbf v=(x,y,z,w)inBbb R^4$. The basis $B={mathbf a,mathbf c}$ of $V$ consists of orthogonal linearly independent vectors, so the orthogonal projection of $mathbf v=(x,y,z,w)$ on $V$ is the vector sum $$displaystylefrac{langlemathbf v,mathbf arangle}{langlemathbf a,mathbf arangle}cdotmathbf a+frac{langlemathbf v,mathbf crangle}{langlemathbf c,mathbf crangle}cdotmathbf c$$
Note that you are only able to do this because the basis vectors are orthogonal.
This gives the orthogonal projection of $mathbf v=(x,y,z,w)$ as
$displaystyle=Big[frac{x-2y-z+3w}{15}Big]cdot(1,-2,-1,3)+Big[frac{2x+y-3z-w}{15}Big]cdot(2,1,-3,-1)\displaystyle=Big(frac{5x-7z+w}{15},frac{5y-z-7w}{15},frac{-7x-y+10z}{15},frac{x-7y+10w}{15}Big)$
Thus, the matrix of $T,[T]=begin{bmatrix}1/3&0&-7/15&1/15\0&1/3&-1/15&-7/15\-7/15&-1/15&2/3&0\1/15&-7/15&0&2/3end{bmatrix}$.
Edit. As hinted by you in a comment, we can show that $T$ is not invertible by noting that the range $R(T)$ is two-dimensional, so nullity of $T>0$.
edited Dec 23 '18 at 9:43
Future Math person
993818
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
edited Dec 23 '18 at 9:43
Future Math person
993818
edited Dec 23 '18 at 9:43
Future Math person
993818
edited Dec 23 '18 at 9:43
Future Math person
993818
993818
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,668918
5,668918
1
$begingroup$
I just read your comment on the original post. You gave a fine reasoning for $T$ not being invertible: it is not onto. In other words, its range $R(T)$ is two-dimensional, so the rank of $[T]$, which is nothing but the dimension of the range of $T$, is $2$. This means the nullity of $T=2>0$, so $T$ is not invertible.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:41
1
$begingroup$
As for how I calculated $bf u$, recall that the cross-product of two vectors gives you a vector that is orthogonal to both the vectors. A vector that is orthogonal to another vector has $bf0$ projection along the latter. But this orthogonality property of the cross-product doesn't exist for vectors in $Bbb R^4$. So I took the cross-product of the first $3$ co-ordinates of the basis vectors of $A$, and made the fourth co-ordinate of $bf u$ zero.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:48
1
$begingroup$
As for why $bf u$ has $bf0$ projection on $V$, you know that any vector in $V$ can be written as $c_1mathbf a+c_2mathbf b$, so $langlemathbf u,c_1mathbf a+c_2mathbf brangle=c_1langlemathbf u,mathbf arangle+c_2langlemathbf u,mathbf brangle=0+0$
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:52
1
$begingroup$
Let's say we have any two basis vectors $bf v_1,v_2$ of $V$. Then the projection of $bf u$ along $bf v_1$ is $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}$ and similarly for its projection along $bf v_2$. But in general, the projection of $bf u$ on $V$ is not equal to $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}+displaystylefrac{langlemathbf u,mathbf{v_2}rangle}{langlemathbf{v_2},mathbf{v_2}rangle}cdotmathbf{v_2}$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:10
1
$begingroup$
For this to happen, $bf v_1,v_2$ must be orthogonal. The only basis containing orthogonal linearly independent vectors is $A$
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:11
|
show 11 more comments
1
$begingroup$
I just read your comment on the original post. You gave a fine reasoning for $T$ not being invertible: it is not onto. In other words, its range $R(T)$ is two-dimensional, so the rank of $[T]$, which is nothing but the dimension of the range of $T$, is $2$. This means the nullity of $T=2>0$, so $T$ is not invertible.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:41
1
$begingroup$
As for how I calculated $bf u$, recall that the cross-product of two vectors gives you a vector that is orthogonal to both the vectors. A vector that is orthogonal to another vector has $bf0$ projection along the latter. But this orthogonality property of the cross-product doesn't exist for vectors in $Bbb R^4$. So I took the cross-product of the first $3$ co-ordinates of the basis vectors of $A$, and made the fourth co-ordinate of $bf u$ zero.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:48
1
$begingroup$
As for why $bf u$ has $bf0$ projection on $V$, you know that any vector in $V$ can be written as $c_1mathbf a+c_2mathbf b$, so $langlemathbf u,c_1mathbf a+c_2mathbf brangle=c_1langlemathbf u,mathbf arangle+c_2langlemathbf u,mathbf brangle=0+0$
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:52
1
$begingroup$
Let's say we have any two basis vectors $bf v_1,v_2$ of $V$. Then the projection of $bf u$ along $bf v_1$ is $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}$ and similarly for its projection along $bf v_2$. But in general, the projection of $bf u$ on $V$ is not equal to $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}+displaystylefrac{langlemathbf u,mathbf{v_2}rangle}{langlemathbf{v_2},mathbf{v_2}rangle}cdotmathbf{v_2}$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:10
1
$begingroup$
For this to happen, $bf v_1,v_2$ must be orthogonal. The only basis containing orthogonal linearly independent vectors is $A$
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:11
1
1
$begingroup$
I just read your comment on the original post. You gave a fine reasoning for $T$ not being invertible: it is not onto. In other words, its range $R(T)$ is two-dimensional, so the rank of $[T]$, which is nothing but the dimension of the range of $T$, is $2$. This means the nullity of $T=2>0$, so $T$ is not invertible.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:41
$begingroup$
I just read your comment on the original post. You gave a fine reasoning for $T$ not being invertible: it is not onto. In other words, its range $R(T)$ is two-dimensional, so the rank of $[T]$, which is nothing but the dimension of the range of $T$, is $2$. This means the nullity of $T=2>0$, so $T$ is not invertible.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:41
1
1
$begingroup$
As for how I calculated $bf u$, recall that the cross-product of two vectors gives you a vector that is orthogonal to both the vectors. A vector that is orthogonal to another vector has $bf0$ projection along the latter. But this orthogonality property of the cross-product doesn't exist for vectors in $Bbb R^4$. So I took the cross-product of the first $3$ co-ordinates of the basis vectors of $A$, and made the fourth co-ordinate of $bf u$ zero.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:48
$begingroup$
As for how I calculated $bf u$, recall that the cross-product of two vectors gives you a vector that is orthogonal to both the vectors. A vector that is orthogonal to another vector has $bf0$ projection along the latter. But this orthogonality property of the cross-product doesn't exist for vectors in $Bbb R^4$. So I took the cross-product of the first $3$ co-ordinates of the basis vectors of $A$, and made the fourth co-ordinate of $bf u$ zero.
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:48
1
1
$begingroup$
As for why $bf u$ has $bf0$ projection on $V$, you know that any vector in $V$ can be written as $c_1mathbf a+c_2mathbf b$, so $langlemathbf u,c_1mathbf a+c_2mathbf brangle=c_1langlemathbf u,mathbf arangle+c_2langlemathbf u,mathbf brangle=0+0$
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:52
$begingroup$
As for why $bf u$ has $bf0$ projection on $V$, you know that any vector in $V$ can be written as $c_1mathbf a+c_2mathbf b$, so $langlemathbf u,c_1mathbf a+c_2mathbf brangle=c_1langlemathbf u,mathbf arangle+c_2langlemathbf u,mathbf brangle=0+0$
$endgroup$
– Shubham Johri
Dec 23 '18 at 8:52
1
1
$begingroup$
Let's say we have any two basis vectors $bf v_1,v_2$ of $V$. Then the projection of $bf u$ along $bf v_1$ is $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}$ and similarly for its projection along $bf v_2$. But in general, the projection of $bf u$ on $V$ is not equal to $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}+displaystylefrac{langlemathbf u,mathbf{v_2}rangle}{langlemathbf{v_2},mathbf{v_2}rangle}cdotmathbf{v_2}$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:10
$begingroup$
Let's say we have any two basis vectors $bf v_1,v_2$ of $V$. Then the projection of $bf u$ along $bf v_1$ is $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}$ and similarly for its projection along $bf v_2$. But in general, the projection of $bf u$ on $V$ is not equal to $displaystylefrac{langlemathbf u,mathbf{v_1}rangle}{langlemathbf{v_1},mathbf{v_1}rangle}cdotmathbf{v_1}+displaystylefrac{langlemathbf u,mathbf{v_2}rangle}{langlemathbf{v_2},mathbf{v_2}rangle}cdotmathbf{v_2}$.
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:10
1
1
$begingroup$
For this to happen, $bf v_1,v_2$ must be orthogonal. The only basis containing orthogonal linearly independent vectors is $A$
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:11
$begingroup$
For this to happen, $bf v_1,v_2$ must be orthogonal. The only basis containing orthogonal linearly independent vectors is $A$
$endgroup$
– Shubham Johri
Dec 23 '18 at 9:11
|
show 11 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050130%2fsetting-up-the-standard-matrix-for-orthogonal-projection%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$T$ has better be square since it represents a map from $mathbb R^4$ into itself. Think about the rank of $T$.
$endgroup$
– amd
Dec 23 '18 at 7:18
$begingroup$
Ohh yeah. Point made. Is it not invertible since the matrix is not onto? The transformation takes vectors in 4D space and spits out a vectors which span a 2D space which does span $mathbb{R}^4$. If that's nonsense, how does the rank work? Rank+nullity=dimension. So that would mean $4+nullity=4$ and thus $nullity=0$. Woudn't that mean the matrix IS invertible though?
$endgroup$
– Future Math person
Dec 23 '18 at 7:29