
Matrix of Linear Trans. in basis i, j, k
Let $\displaystyle \phi: R^3 \rightarrow R^3$ be the projection of the threedimensional space onto the plane $\displaystyle x + y + z = 0$. Find the matrix of this linear transformation in the standard basis i, j, k.
[Hint: Find the coordinates of the normal vector N to the plane. Then use the fact that the projection of a basis vector m (m = i, j, or k) onto the plane is m  a where a is the projection of m onto the direction of N.]

Re: Matrix of Linear Trans. in basis i, j, k
Hint 2: A normal vector to the plane ax + by + cz = d is (a, b, c).
Hint 3: The projection of m onto the direction of N is (m, N) / N (here (m, N) is the scalar, or dot, product).

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
emakarov Hint 2: A normal vector to the plane ax + by + cz = d is (a, b, c).
Hint 3: The projection of m onto the direction of N is (m, N) / N (here (m, N) is the scalar, or dot, product).
So from your hints I would say that the coordinates of the normal vector N to the plane is (1,1,1) right? So N = (1 1 1)? I am not sure I understand what m is?

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
vidomagru So from your hints I would say that the coordinates of the normal vector N to the plane is (1,1,1) right? So N = (1 1 1)?
Yes.
Quote:
Originally Posted by
vidomagru I am not sure I understand what m is?
Have you read the hint in your own post?

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
emakarov Yes.
Have you read the hint in your own post?
I really do not understand what you are getting at here.
m = i, j, or k for i = (1,0,0), j = (0,1,0), k = (0,0,1) and N = (1,1,1), but you N is not a n x n matrix so how would you take the determinant?

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
vidomagru I really do not understand what you are getting at here.
Even though you are saying this, you answered one of your own questions (about what m is), so you are making progress.
Quote:
Originally Posted by
vidomagru N is not a n x n matrix so how would you take the determinant?
Vertical bars denote many things in mathematics. Here, N is the length of vector N.
You may wish to review the material about projections and scalar (dot) product.

Re: Matrix of Linear Trans. in basis i, j, k
I am sorry, hint 3 from post #2 should be as follows.
Hint 3: The projection of m onto the direction of N is ((m, N) / N^{2}) N, where (m, N) is the scalar, or dot, product.
Previously I said that the projection was (m, N) / N, but this is the scalar projection, not a vector. To get the vector projection onto the direction of N, the scalar projection must be multiplied by the unit vector in the direction of N, which is N / N. This article in Wikipedia describes projections well.

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
emakarov I am sorry, hint 3 from post #2 should be as follows.
Hint 3: The projection of
m onto the direction of
N is ((
m,
N) / 
N
^{2})
N, where (
m,
N) is the scalar, or dot, product.
Previously I said that the projection was (
m,
N) / 
N, but this is the scalar projection, not a vector. To get the vector projection onto the direction of
N, the scalar projection must be multiplied by the unit vector in the direction of
N, which is
N / 
N.
This article in Wikipedia describes projections well.
Ok, yes in the research I was just doing I had seen that it should be ((m, N) / N^{2}) N. So here is what I have done:
$\displaystyle a = proj_N(m) =$ (m, N) / N^{2}) N = m, N) / 3) N.
Now for m = i = (1,0,0) we have $\displaystyle a_1 = \dfrac{1}{3}$.
Now for m = j = (0,1,0) we have $\displaystyle a_2 = \dfrac{1}{3}$.
Now for m = k = (0,0,1) we have $\displaystyle a_3 = \dfrac{1}{3}$.
so $\displaystyle a = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$. From here we should do m  a. Do I do (i, j, k)  $\displaystyle (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$ to give me a 3x3 matrix?

Re: Matrix of Linear Trans. in basis i, j, k
I'll drop the bold font.
Quote:
Originally Posted by
vidomagru $\displaystyle a = proj_N(m) =$ (m, N) / N^{2}) N = m, N) / 3) N.
I agree, but what is the type of the left and righthand sides? It's a vector, not a number. Getting a vector was the reason the scalar projection (m, N) / N was multiplied by the unit vector N / N.
Quote:
Originally Posted by
vidomagru Now for m = i = (1,0,0) we have $\displaystyle a_1 = \dfrac{1}{3}$.
But here I understand that a₁ is a number. Which formula did you use to get a number?
Quote:
Originally Posted by
vidomagru so $\displaystyle a = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$.
You used three vectors i, j and k to get one? Where did you get such formula?
((i, N) / N²) N is the projection of i onto N (the result is a vector), and similarly for j and k. You subtract that projection from the corresponding original vector (i, j or k), and you get the projection of that vector onto the plane.

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
emakarov I'll drop the bold font.
I agree, but what is the type of the left and righthand sides? It's a vector, not a number. Getting a vector was the reason the scalar projection (m, N) / N was multiplied by the unit vector N / N.
But here I understand that a₁ is a number. Which formula did you use to get a number?
You used three vectors i, j and k to get one? Where did you get such formula?
((i, N) / N²) N is the projection of i onto N (the result is a vector), and similarly for j and k. You subtract that projection from the corresponding original vector (i, j or k), and you get the projection of that vector onto the plane.
From ((i, N) / N²) N one has ((1,0,0)$\displaystyle \cdot$(1,1,1)/3)N = $\displaystyle \dfrac{1}{3}$N = $\displaystyle (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$. So for m = i one has i  a = $\displaystyle (\dfrac{2}{3},\dfrac{1}{3},\dfrac{1}{3})$? Then for m = j one has j  a = $\displaystyle (\dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3})$? And finally for m = k one has k  a = $\displaystyle (\dfrac{1}{3},\dfrac{1}{3},\dfrac{2}{3})$? How does that translate into the matrix of linear transformation?

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
vidomagru From ((i, N) / N²) N one has ((1,0,0)$\displaystyle \cdot$(1,1,1)/3)N = $\displaystyle \dfrac{1}{3}$N = $\displaystyle (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$. So for m = i one has i  a = $\displaystyle (\dfrac{2}{3},\dfrac{1}{3},\dfrac{1}{3})$? Then for m = j one has j  a = $\displaystyle (\dfrac{1}{3},\dfrac{2}{3},\dfrac{1}{3})$? And finally for m = k one has k  a = $\displaystyle (\dfrac{1}{3},\dfrac{1}{3},\dfrac{2}{3})$?
Correct.
Quote:
Originally Posted by
vidomagru How does that translate into the matrix of linear transformation?
The matrix consists of the coordinates of images of basis vectors written as columns.

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
emakarov Correct.
The matrix consists of the coordinates of images of basis vectors written as columns.
So just $\displaystyle \begin{bmatrix} \dfrac{2}{3}&\dfrac{1}{3}&\dfrac{1}{3}\\ \\ \dfrac{1}{3}&\dfrac{2}{3}&\dfrac{1}{3}\\ \\ \dfrac{1}{3}&\dfrac{1}{3}&\dfrac{2}{3} \end{bmatrix}$?

Re: Matrix of Linear Trans. in basis i, j, k
I $\displaystyle \{u_1, u_2, ..., u_m\}$ is a basis for U and $\displaystyle \{v_1, v_2, ..., v_n\}$ is a basis for V, A is a linear transformation from U to V,
then the "matrix representation" of A in those two bases requires calculating [tex]A(u_i)[tex] and writing as a linear combination of the $\displaystyle v_i$ vectors. The coefficents of $\displaystyle v_i$ in $\displaystyle A(u_j)$ form the $\displaystyle i^{th}$ column of matrix.
So what are (1, 0, 0), (0, 1, 0), and (0, 0, 1) projected to?

Re: Matrix of Linear Trans. in basis i, j, k
Quote:
Originally Posted by
vidomagru So just $\displaystyle \begin{bmatrix} \dfrac{2}{3}&\dfrac{1}{3}&\dfrac{1}{3}\\ \\ \dfrac{1}{3}&\dfrac{2}{3}&\dfrac{1}{3}\\ \\ \dfrac{1}{3}&\dfrac{1}{3}&\dfrac{2}{3} \end{bmatrix}$?
Yes.