# Matrix of Linear Trans. in basis i, j, k

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• Oct 20th 2013, 11:27 AM
vidomagru
Matrix of Linear Trans. in basis i, j, k
Let $\phi: R^3 \rightarrow R^3$ be the projection of the three-dimensional space onto the plane $x + y + z = 0$. 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.]
• Oct 20th 2013, 12:12 PM
emakarov
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).
• Oct 21st 2013, 05:48 PM
vidomagru
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?
• Oct 22nd 2013, 12:33 AM
emakarov
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?
• Oct 22nd 2013, 02:39 PM
vidomagru
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?
• Oct 22nd 2013, 02:47 PM
emakarov
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.
• Oct 22nd 2013, 03:03 PM
emakarov
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.
• Oct 22nd 2013, 03:37 PM
vidomagru
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:
$a = proj_N(m) =$ (m, N) / |N|2) N = m, N) / 3) N.
Now for m = i = (1,0,0) we have $a_1 = \dfrac{1}{3}$.

Now for m = j = (0,1,0) we have $a_2 = \dfrac{1}{3}$.

Now for m = k = (0,0,1) we have $a_3 = \dfrac{1}{3}$.

so $a = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$. From here we should do m - a. Do I do (i, j, k) - $(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$ to give me a 3x3 matrix?
• Oct 22nd 2013, 04:07 PM
emakarov
Re: Matrix of Linear Trans. in basis i, j, k
I'll drop the bold font.

Quote:

Originally Posted by vidomagru
$a = proj_N(m) =$ (m, N) / |N|2) N = m, N) / 3) N.

I agree, but what is the type of the left- and right-hand 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 $a_1 = \dfrac{1}{3}$.

But here I understand that a₁ is a number. Which formula did you use to get a number?

Quote:

Originally Posted by vidomagru
so $a = (\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$.

You used three vectors i, j and k to get one? Where did you get such formula?

((i, N) / |N|²) 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.
• Oct 22nd 2013, 04:59 PM
vidomagru
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 right-hand 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) / |N|²) 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) / |N|²) N one has ((1,0,0) $\cdot$(1,1,1)/3)N = $\dfrac{1}{3}$N = $(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$. So for m = i one has i - a = $(\dfrac{2}{3},\dfrac{-1}{3},\dfrac{-1}{3})$? Then for m = j one has j - a = $(\dfrac{-1}{3},\dfrac{2}{3},\dfrac{-1}{3})$? And finally for m = k one has k - a = $(\dfrac{-1}{3},\dfrac{-1}{3},\dfrac{2}{3})$? How does that translate into the matrix of linear transformation?
• Oct 22nd 2013, 05:04 PM
emakarov
Re: Matrix of Linear Trans. in basis i, j, k
Quote:

Originally Posted by vidomagru
From ((i, N) / |N|²) N one has ((1,0,0) $\cdot$(1,1,1)/3)N = $\dfrac{1}{3}$N = $(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3})$. So for m = i one has i - a = $(\dfrac{2}{3},\dfrac{-1}{3},\dfrac{-1}{3})$? Then for m = j one has j - a = $(\dfrac{-1}{3},\dfrac{2}{3},\dfrac{-1}{3})$? And finally for m = k one has k - a = $(\dfrac{-1}{3},\dfrac{-1}{3},\dfrac{2}{3})$?

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.
• Oct 22nd 2013, 05:44 PM
vidomagru
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 $\begin{bmatrix} \dfrac{2}{3}&\dfrac{-1}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{2}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{-1}{3}&\dfrac{2}{3} \end{bmatrix}$?
• Oct 22nd 2013, 05:51 PM
HallsofIvy
Re: Matrix of Linear Trans. in basis i, j, k
I $\{u_1, u_2, ..., u_m\}$ is a basis for U and $\{v_1, v_2, ..., v_n\}$ 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 $v_i$ vectors. The coefficents of $v_i$ in $A(u_j)$ form the $i^{th}$ column of matrix.

So what are (1, 0, 0), (0, 1, 0), and (0, 0, 1) projected to?
• Oct 22nd 2013, 05:52 PM
emakarov
Re: Matrix of Linear Trans. in basis i, j, k
Quote:

Originally Posted by vidomagru
So just $\begin{bmatrix} \dfrac{2}{3}&\dfrac{-1}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{2}{3}&\dfrac{-1}{3}\\ \\ \dfrac{-1}{3}&\dfrac{-1}{3}&\dfrac{2}{3} \end{bmatrix}$?

Yes.