[NeoKDarkmatter]

Matrices A and B as follows: All are 3 x 3 matrices A=[first row 1/root2 -1/root2 0 second row 1/root2 1/root2 0 third row 0 0 1 ] B=[first row 1/root2 0 -1/root2 second row 0 1 0 third row 1/root2 0 1/root2] A is a partitioned matrix I) Describe the homogeneous linear transformation T(X) = AX as a mapping points of (x1, x2) plane and corresponding mapping of x3. ii) Can a similar description be given to the transformation S(X) = BX? iii) Find matrix product c = BA? So I basically work out the determinant of matrix A which works out to be 1 (correct?)

[jcaron2]

If you write out the matrix A, it's very obvious that it's partitioned, as you said. It consists of a 2x2 square matrix in the upper left, 2^(-1/2)*[1 -1; 1 1], and a 1x1 identity matrix in the lower right, [1]. Everything else is zero. Hence, as you said, that means the matrix is performing a transformation on the x1 and x2 dimensions (it's a 45 degree rotation in the x1-x2 plane), while it leaves the x3 dimension unaltered.

In the case of the B matrix, it's a little less obvious that it's partitioned, but you could simply rearrange the order of the dimensions to give it the same form as A. Instead of the vector X being [x1; x2; x3] (in that order), rearrange it so that it's [x1; x3; x2]. Likewise, S(X) gets rearranged from [S1; S2; S3] to [S1; S3; S2]. Those reordering operations require you to also swap second and third rows, as well as the second and third columns, of the matrix B. That makes it a partitioned matrix of the same form as matrix A (in fact, it makes it exactly equal to matrix A).

Thus, the matrix B does the same transformation as A, except that it rotates in the x1-x3 plane while leaving the x2 dimension unchanged. So B is actually a 90-degree-rotated version of A.

If that's a little difficult to understand, a good way to see what's going on is to reconstruct the system of equations in the old fashioned form like you would have seen back in Algebra II:

T1(X) = 2^(-1/2)*(x1 - x2)
T2(X) = 2^(-1/2)*(x1 + x2)
T3(X) = x3

Notice that the third equation is linearly independent from the other two (it doesn't contain any of the variables from the other two equations, and those two equations don't contain any of its variables). When written in matrix form, this scenario results in a partitioned matrix.

If you write out the equations for S(X)=BX, you'll see that it's the second equation, S2(x)=x2, which is independent. But if you were to simply write the equations in a different order (swapping the second and third), the resulting system looks just like the one for T(X)=AX.