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
979817
$endgroup$
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
979817
$endgroup$
$begingroup$
$T$ 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
979817
$endgroup$
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
979817
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
979817
edited Dec 23 '18 at 9:03
Future Math person
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
979817
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
979817
asked Dec 23 '18 at 7:16
Future Math personFuture Math person
979817
979817
$begingroup$
$T$ 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
979817
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
$endgroup$
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
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$begingroup$
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
979817
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
$endgroup$
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
979817
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
$endgroup$
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
979817
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
$endgroup$
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
979817
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
edited Dec 23 '18 at 9:43
Future Math person
979817
edited Dec 23 '18 at 9:43
Future Math person
979817
edited Dec 23 '18 at 9:43
Future Math person
979817
979817
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
answered Dec 23 '18 at 8:24
Shubham JohriShubham Johri
5,204718
5,204718
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
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$begingroup$
$T$ 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