Find the matrix of $T$ with respect to standard basis of $V$.
$begingroup$
Let $T:Vto V$ be the rotation by an angle $theta$ counterclockwise in the plane passing through the origin perpendicular to $(1,2,3)$ where $V=Bbb R^3$
Find the matrix of $T$ with respect to standard basis of $V$.
I know that the equation of the plane passing through the origin perpendicular to $(1,2,3)$ is $x+2y+3z=0$ but I dont know how to find the matrix .
I really dont understand where does $T$ map the vector $(1,0,0)$ .
Can someone kindly help me?
Hints will suffice.I dont see any way out
linear-algebra geometry linear-transformations
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
$endgroup$
add a comment |
$begingroup$
Let $T:Vto V$ be the rotation by an angle $theta$ counterclockwise in the plane passing through the origin perpendicular to $(1,2,3)$ where $V=Bbb R^3$
Find the matrix of $T$ with respect to standard basis of $V$.
I know that the equation of the plane passing through the origin perpendicular to $(1,2,3)$ is $x+2y+3z=0$ but I dont know how to find the matrix .
I really dont understand where does $T$ map the vector $(1,0,0)$ .
Can someone kindly help me?
Hints will suffice.I dont see any way out
linear-algebra geometry linear-transformations
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
$endgroup$
add a comment |
$begingroup$
Let $T:Vto V$ be the rotation by an angle $theta$ counterclockwise in the plane passing through the origin perpendicular to $(1,2,3)$ where $V=Bbb R^3$
Find the matrix of $T$ with respect to standard basis of $V$.
I know that the equation of the plane passing through the origin perpendicular to $(1,2,3)$ is $x+2y+3z=0$ but I dont know how to find the matrix .
I really dont understand where does $T$ map the vector $(1,0,0)$ .
Can someone kindly help me?
Hints will suffice.I dont see any way out
linear-algebra geometry linear-transformations
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
$endgroup$
Let $T:Vto V$ be the rotation by an angle $theta$ counterclockwise in the plane passing through the origin perpendicular to $(1,2,3)$ where $V=Bbb R^3$
Find the matrix of $T$ with respect to standard basis of $V$.
I know that the equation of the plane passing through the origin perpendicular to $(1,2,3)$ is $x+2y+3z=0$ but I dont know how to find the matrix .
I really dont understand where does $T$ map the vector $(1,0,0)$ .
Can someone kindly help me?
Hints will suffice.I dont see any way out
linear-algebra geometry linear-transformations
linear-algebra geometry linear-transformations
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
asked Dec 6 '18 at 9:38
Join_PhDJoin_PhD
4018
4018
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Extend $v_1=frac{1}{sqrt{14}} (1,2,3)$ to an orthonormal basis $B=(v_1, v_2, v_3)$. Write down the matrix of $T$ with respect to this basis. Perform a change of basis to the standard basis $S=(e_1, e_2, e_3)$.
The vector $v_1$ I picked is a vector in the direction $(1,2,3)$ having length $|v_1|=1$. To extend this to an orthonormal basis, you can proceed in (at least) two ways:
- extend to any basis first, e.g. $v_1, e_1, e_2$ and apply Gram–Schmidt orhonormalization
- find one unit vector $v_2$ orthogonal to $v_1$ and then compute the cross product $v_3=v_1times v_2$.
Having done that, we can find the matrix of $T$ with respect to $B=(v_1, v_2, v_3)$.
Since $v_1$ is a vector along the axis of rotation, we have $T(v_1)=v_1$. The vectors $v_2,v_3$ form an orthonormal basis of the plane perpendicular to $v_1$, so they get rotated by $theta$ in this plane and we have
begin{align*}
T(v_2) &= cos(theta), v_2 + sin(theta), v_3, \
T(v_2) &= -sin(theta), v_2 + cos(theta), v_3.
end{align*}
Thus, the matrix of $T$ with respect to the basis $B$ is given by
$$M(T)_B = begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}.$$
The change of base matrix from $v_1,v_2,v_3$ to the standard basis is the matrix $C_{S,B}=(v_1 ,|, v_2 ,|, v_3)$ with columns $v_1, v_2, v_3$.
Hence, the matrix of $T$ with respect to the standard basis is
$$
M(T)_S = C_{S,B} ,M(T)_B ,C_{B,S} = C_{S,B} ,M(T)_B ,C_{S,B}^{-1}.
$$
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
$endgroup$
$begingroup$
But T is the rotation about this plane,I dont understand why should I follow the above procedure to find its matrix
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Can you please give some details using geometry
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Do you know how to write down a rotation about the axis $(1,0,0)$ by an angle $theta$?
$endgroup$
– Christoph
Dec 6 '18 at 10:09
$begingroup$
Yes it is $begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}$
$endgroup$
– Join_PhD
Dec 6 '18 at 10:10
$begingroup$
This is the matrix of $T$ with respect to $v_1, v_2, v_3$. Now perform a change of basis.
$endgroup$
– Christoph
Dec 6 '18 at 10:12
|
show 10 more comments
Your Answer
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1 Answer
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$begingroup$
Extend $v_1=frac{1}{sqrt{14}} (1,2,3)$ to an orthonormal basis $B=(v_1, v_2, v_3)$. Write down the matrix of $T$ with respect to this basis. Perform a change of basis to the standard basis $S=(e_1, e_2, e_3)$.
The vector $v_1$ I picked is a vector in the direction $(1,2,3)$ having length $|v_1|=1$. To extend this to an orthonormal basis, you can proceed in (at least) two ways:
- extend to any basis first, e.g. $v_1, e_1, e_2$ and apply Gram–Schmidt orhonormalization
- find one unit vector $v_2$ orthogonal to $v_1$ and then compute the cross product $v_3=v_1times v_2$.
Having done that, we can find the matrix of $T$ with respect to $B=(v_1, v_2, v_3)$.
Since $v_1$ is a vector along the axis of rotation, we have $T(v_1)=v_1$. The vectors $v_2,v_3$ form an orthonormal basis of the plane perpendicular to $v_1$, so they get rotated by $theta$ in this plane and we have
begin{align*}
T(v_2) &= cos(theta), v_2 + sin(theta), v_3, \
T(v_2) &= -sin(theta), v_2 + cos(theta), v_3.
end{align*}
Thus, the matrix of $T$ with respect to the basis $B$ is given by
$$M(T)_B = begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}.$$
The change of base matrix from $v_1,v_2,v_3$ to the standard basis is the matrix $C_{S,B}=(v_1 ,|, v_2 ,|, v_3)$ with columns $v_1, v_2, v_3$.
Hence, the matrix of $T$ with respect to the standard basis is
$$
M(T)_S = C_{S,B} ,M(T)_B ,C_{B,S} = C_{S,B} ,M(T)_B ,C_{S,B}^{-1}.
$$
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
$endgroup$
$begingroup$
But T is the rotation about this plane,I dont understand why should I follow the above procedure to find its matrix
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Can you please give some details using geometry
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Do you know how to write down a rotation about the axis $(1,0,0)$ by an angle $theta$?
$endgroup$
– Christoph
Dec 6 '18 at 10:09
$begingroup$
Yes it is $begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}$
$endgroup$
– Join_PhD
Dec 6 '18 at 10:10
$begingroup$
This is the matrix of $T$ with respect to $v_1, v_2, v_3$. Now perform a change of basis.
$endgroup$
– Christoph
Dec 6 '18 at 10:12
|
show 10 more comments
$begingroup$
Extend $v_1=frac{1}{sqrt{14}} (1,2,3)$ to an orthonormal basis $B=(v_1, v_2, v_3)$. Write down the matrix of $T$ with respect to this basis. Perform a change of basis to the standard basis $S=(e_1, e_2, e_3)$.
The vector $v_1$ I picked is a vector in the direction $(1,2,3)$ having length $|v_1|=1$. To extend this to an orthonormal basis, you can proceed in (at least) two ways:
- extend to any basis first, e.g. $v_1, e_1, e_2$ and apply Gram–Schmidt orhonormalization
- find one unit vector $v_2$ orthogonal to $v_1$ and then compute the cross product $v_3=v_1times v_2$.
Having done that, we can find the matrix of $T$ with respect to $B=(v_1, v_2, v_3)$.
Since $v_1$ is a vector along the axis of rotation, we have $T(v_1)=v_1$. The vectors $v_2,v_3$ form an orthonormal basis of the plane perpendicular to $v_1$, so they get rotated by $theta$ in this plane and we have
begin{align*}
T(v_2) &= cos(theta), v_2 + sin(theta), v_3, \
T(v_2) &= -sin(theta), v_2 + cos(theta), v_3.
end{align*}
Thus, the matrix of $T$ with respect to the basis $B$ is given by
$$M(T)_B = begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}.$$
The change of base matrix from $v_1,v_2,v_3$ to the standard basis is the matrix $C_{S,B}=(v_1 ,|, v_2 ,|, v_3)$ with columns $v_1, v_2, v_3$.
Hence, the matrix of $T$ with respect to the standard basis is
$$
M(T)_S = C_{S,B} ,M(T)_B ,C_{B,S} = C_{S,B} ,M(T)_B ,C_{S,B}^{-1}.
$$
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
$endgroup$
$begingroup$
But T is the rotation about this plane,I dont understand why should I follow the above procedure to find its matrix
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Can you please give some details using geometry
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Do you know how to write down a rotation about the axis $(1,0,0)$ by an angle $theta$?
$endgroup$
– Christoph
Dec 6 '18 at 10:09
$begingroup$
Yes it is $begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}$
$endgroup$
– Join_PhD
Dec 6 '18 at 10:10
$begingroup$
This is the matrix of $T$ with respect to $v_1, v_2, v_3$. Now perform a change of basis.
$endgroup$
– Christoph
Dec 6 '18 at 10:12
|
show 10 more comments
$begingroup$
Extend $v_1=frac{1}{sqrt{14}} (1,2,3)$ to an orthonormal basis $B=(v_1, v_2, v_3)$. Write down the matrix of $T$ with respect to this basis. Perform a change of basis to the standard basis $S=(e_1, e_2, e_3)$.
The vector $v_1$ I picked is a vector in the direction $(1,2,3)$ having length $|v_1|=1$. To extend this to an orthonormal basis, you can proceed in (at least) two ways:
- extend to any basis first, e.g. $v_1, e_1, e_2$ and apply Gram–Schmidt orhonormalization
- find one unit vector $v_2$ orthogonal to $v_1$ and then compute the cross product $v_3=v_1times v_2$.
Having done that, we can find the matrix of $T$ with respect to $B=(v_1, v_2, v_3)$.
Since $v_1$ is a vector along the axis of rotation, we have $T(v_1)=v_1$. The vectors $v_2,v_3$ form an orthonormal basis of the plane perpendicular to $v_1$, so they get rotated by $theta$ in this plane and we have
begin{align*}
T(v_2) &= cos(theta), v_2 + sin(theta), v_3, \
T(v_2) &= -sin(theta), v_2 + cos(theta), v_3.
end{align*}
Thus, the matrix of $T$ with respect to the basis $B$ is given by
$$M(T)_B = begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}.$$
The change of base matrix from $v_1,v_2,v_3$ to the standard basis is the matrix $C_{S,B}=(v_1 ,|, v_2 ,|, v_3)$ with columns $v_1, v_2, v_3$.
Hence, the matrix of $T$ with respect to the standard basis is
$$
M(T)_S = C_{S,B} ,M(T)_B ,C_{B,S} = C_{S,B} ,M(T)_B ,C_{S,B}^{-1}.
$$
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
$endgroup$
Extend $v_1=frac{1}{sqrt{14}} (1,2,3)$ to an orthonormal basis $B=(v_1, v_2, v_3)$. Write down the matrix of $T$ with respect to this basis. Perform a change of basis to the standard basis $S=(e_1, e_2, e_3)$.
The vector $v_1$ I picked is a vector in the direction $(1,2,3)$ having length $|v_1|=1$. To extend this to an orthonormal basis, you can proceed in (at least) two ways:
- extend to any basis first, e.g. $v_1, e_1, e_2$ and apply Gram–Schmidt orhonormalization
- find one unit vector $v_2$ orthogonal to $v_1$ and then compute the cross product $v_3=v_1times v_2$.
Having done that, we can find the matrix of $T$ with respect to $B=(v_1, v_2, v_3)$.
Since $v_1$ is a vector along the axis of rotation, we have $T(v_1)=v_1$. The vectors $v_2,v_3$ form an orthonormal basis of the plane perpendicular to $v_1$, so they get rotated by $theta$ in this plane and we have
begin{align*}
T(v_2) &= cos(theta), v_2 + sin(theta), v_3, \
T(v_2) &= -sin(theta), v_2 + cos(theta), v_3.
end{align*}
Thus, the matrix of $T$ with respect to the basis $B$ is given by
$$M(T)_B = begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}.$$
The change of base matrix from $v_1,v_2,v_3$ to the standard basis is the matrix $C_{S,B}=(v_1 ,|, v_2 ,|, v_3)$ with columns $v_1, v_2, v_3$.
Hence, the matrix of $T$ with respect to the standard basis is
$$
M(T)_S = C_{S,B} ,M(T)_B ,C_{B,S} = C_{S,B} ,M(T)_B ,C_{S,B}^{-1}.
$$
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
edited Dec 6 '18 at 10:35
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
answered Dec 6 '18 at 9:46
ChristophChristoph
12k1642
12k1642
$begingroup$
But T is the rotation about this plane,I dont understand why should I follow the above procedure to find its matrix
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Can you please give some details using geometry
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Do you know how to write down a rotation about the axis $(1,0,0)$ by an angle $theta$?
$endgroup$
– Christoph
Dec 6 '18 at 10:09
$begingroup$
Yes it is $begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}$
$endgroup$
– Join_PhD
Dec 6 '18 at 10:10
$begingroup$
This is the matrix of $T$ with respect to $v_1, v_2, v_3$. Now perform a change of basis.
$endgroup$
– Christoph
Dec 6 '18 at 10:12
|
show 10 more comments
$begingroup$
But T is the rotation about this plane,I dont understand why should I follow the above procedure to find its matrix
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Can you please give some details using geometry
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Do you know how to write down a rotation about the axis $(1,0,0)$ by an angle $theta$?
$endgroup$
– Christoph
Dec 6 '18 at 10:09
$begingroup$
Yes it is $begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}$
$endgroup$
– Join_PhD
Dec 6 '18 at 10:10
$begingroup$
This is the matrix of $T$ with respect to $v_1, v_2, v_3$. Now perform a change of basis.
$endgroup$
– Christoph
Dec 6 '18 at 10:12
$begingroup$
But T is the rotation about this plane,I dont understand why should I follow the above procedure to find its matrix
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
But T is the rotation about this plane,I dont understand why should I follow the above procedure to find its matrix
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Can you please give some details using geometry
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Can you please give some details using geometry
$endgroup$
– Join_PhD
Dec 6 '18 at 10:08
$begingroup$
Do you know how to write down a rotation about the axis $(1,0,0)$ by an angle $theta$?
$endgroup$
– Christoph
Dec 6 '18 at 10:09
$begingroup$
Do you know how to write down a rotation about the axis $(1,0,0)$ by an angle $theta$?
$endgroup$
– Christoph
Dec 6 '18 at 10:09
$begingroup$
Yes it is $begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}$
$endgroup$
– Join_PhD
Dec 6 '18 at 10:10
$begingroup$
Yes it is $begin{bmatrix} 1 & 0&0\0 &cos theta&-sin theta\0& sintheta &costhetaend{bmatrix}$
$endgroup$
– Join_PhD
Dec 6 '18 at 10:10
$begingroup$
This is the matrix of $T$ with respect to $v_1, v_2, v_3$. Now perform a change of basis.
$endgroup$
– Christoph
Dec 6 '18 at 10:12
$begingroup$
This is the matrix of $T$ with respect to $v_1, v_2, v_3$. Now perform a change of basis.
$endgroup$
– Christoph
Dec 6 '18 at 10:12
|
show 10 more comments
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