How to rotate an image by applying rotational matrix

I guess rotating 3D image is same as rotating 2D image,right? for example, I have a 240x 240 2D image

    R = [0 -1;-1 0]; % rotate 90 degree
    Image = imread('pout.tif') % built-in image 
    Image = Image (1:240,1:240);
    Image = reshape(Image , (240*240)/2, 2) * R;
    Processed_image = reshape (Image,240,240);

However, It did not work.

Anyone help me?

Answer 1

[Matt J]

You can use imrotate to rotate a 2D image, or a 3D stack of 2D images slice-by-slice.

Multiplying the rotation matrix by the pixel values would not be appropriate. The matrix needs to be multiplied with pixel coordinates. Then you have to interpolate the pixel values at the rotated coordinates, as imrotate or imtransform do for you internally.

[tabw]

But in 3D images, I have to rotate angle phi and theta. Rotating a 2D image can only adjust one of these two angle.

only angle theta between x-y axis works. angle phi between y-z axis does not works.

I don't quite understand how imtransform works.

T = marketform('projective',A)

A is that where i should put the rotational matrix?

[Matt J]

You want 'affine' instead of 'projective'. The top 3x3 matrix in A will contain the rotation matrix.

[tabw]

Matt, I got a problem using marketform My Rotation Matrix is

R_t =

    0.3037   -0.9527   -0.0070

    0.9528    0.3037         0

    0.0021   -0.0067    1.0000

But it said The final column of A must consist of zeroes, except for a one in the last row.

How to tackle this problem??

It can't only be 0 /1 in the last column. Otherwise, how to rotate the image with certain angle?

 rotation matrix  from http://en.wikipedia.org/wiki/Rotation_matrix

[Matt J]

As it says in the documentation, A must be either 4x3 or 4x4. What is the dimension of your A matrix?