# Yet another Linear Algebra question!

Discussion in 'Physics & Math' started by oxymoron, Oct 10, 2003.

1. ### oxymoronRegistered Senior Member

Messages:
454
I am sure everyone is sick of me filling up the forum with my problems. Well guess what!? Here is another, but this time slightly less irritating...

Okay I just need some checking. That's all. Unless there is a problem of course

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__Question__
For some angle (theta) which is confined between -pi/2 < (theta) < pi/2 we let T be the orthogonal projection of R^2 onto the line L through the origin which makes an angle (theta) with the x-axis.

(i) What is the standard matrix for T?

(ai) [T] = [T(e_1 | T(e_2) where e_1 and e_2 are the standard basis vectors in R^2. ie.
e_1 = (1, 0)
e_2 = (0, 1)

Now we consider the case where 0 <= (theta) < pi/2 (because it is the same for -pi/2 <= (theta) < 0 only reversed).

||T(e_1|| = cos(theta) ... Using trigonometry
T(e_1) =
| . . . cos^2(theta) . . .|
| sin(theta)cos(theta) |

Likewise for e_2 the standard matrix is...

| sin(theta)cos(theta) |
| . . . cos^2(theta) . . .|

ii) Does the same matrix work when theta = pi/2?

aii) When theta = pi/2 the matrix becomes...

| 0 0 |
| 0 1 |

And the projection becomes (0, 1) on the plane. Which works.

iii) Find bases for Image(T) and Kernel(T)

aiii) e_1 and e_2 span R^2 so T(e_1) and T(e_2) span Im(T). ie...
(let q = theta)

Im(T) = [ a * {cos^2(q), sin(q)cos(q)}^-1 + b * {sin(q)cos(q), sin^2(q)}^-1 ] such that a and b are scalars.

Hence {cos^2(q), sin(q)cos(q)}^-1 and {sin(q)cos(q), sin^2(q)}^-1 are a basis for Im(T).

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For the basis of Kernel(T) it is too hard to write so I will just say my answer and hope it is right.

The basis vector for Ker(T) = {-tan(q), 1}^-1

iv) Show that the matrix in (i) has determinant zero.

aiv) det(T) = 0

But why???

Thankyou in advance for anyone who can check this for me.

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3. ### HallsofIvyRegistered Senior Member

Messages:
307
You don't have different "standard matrices" for e_1 and e_2. The standard matrix is the matrix whose columns are the vectors e_1 and e_2 are transformed into. I think what you mean is that the standard matrix is [cos^2([theta]) sin(theta)cos(theta) ]
[sin(theta) cos(theta) sin^2(theta) ]
except that I notice you have two "cos^2 " terms.

Actually the matrix you gave would give [0 0]
[0 0]
You need that "sin^2(theta)" to get 1.

The image of the linear transformation is, of course, the given straight line. A "basis" for that is any vector in that direction. In particular, {(cos(theta), sin(theta))} works fine.

The kernal is the line (through the origin) perpendicular to the given line. Since (cos(theta), sin(theta)) is in the direction of the line, (sin(theta),-cos(theta)) is orthogonal to that and gives a basis for the kernel.

Of course the determinant is 0. The linear transformation maps all of R^2 into a line- many vectors are mapped into the same result and so the linear transformation does not have an inverse. If the determinant were not 0, the matrix would have an inverse.

(and, of course, the determinant is cos^(theta)sin^2(theta)-
(sin(theta)cos(theta))^2= 0.)