Recent content by liengen

I see your point. Nothing else is stated though. If the rows or columns are linearly dependent, I still think it's obvious, since then we just have either the same equations (rows) or the unknowns equal each other (columns) and the pseudoinverse just kills these. We still get the right...

Homework Statement Let's say I have a non invertible system of linear equations: Ax=b Then the pseudoinverse gives a approximate solution: x'=A^+*b (1) Given the property: A*A^+*A = A, prove that x' is a vector which lies in the image of A and minimises the error = ||b-Ax'|| (2) show...

Ah, I see. By using the identity sin(theta)=(+-)sqrt(1-cos^2(theta)): A = |v3|*|v4|*sin(theta) A = |v3|*|v4|*(+-)sqrt(1-cos^2(theta)) A^2 = |v3|^2*|v4|^2*(1-cos^2(theta)) = |v3|^2*|v4|^2 - dot(v3,v4)^2 Again, thanks a lot man :-D

Ah, sorry, I was a little to fast earlier. When I expanded the dot product and got the actual result, I thought I was finished. So, the rotated unit v3 vector is, v3rot = [-b a]/sqrt(b^2+a^2) Now we can do the scalar projection from v4 onto this, and we get the height of the parallelogram: h...

I see ^^. Or actually not. Dot product between rotated vector and the basis: [-b a].[c d] = -bc + ad = ad - bc (which is the area) Generally speaking, let's say we have a parallelogram spanned by two vectors, say v1 and v2 and the rotated version of for example v1 is v3. Then, why do the...

Yes, I've tried that, but I'm definitely doing something wrong along the way, since my result is wrong... Let's say the parallelogram is spanned by v3 and v4 and my base is v3. According to my calculations, the height is: sqrt(|v4|^2 - dot(v3,v4)^2) That is, v4 is the hypotenuse and...

Homework Statement So, I'm in a 2D Euclidean space V. The two vectors v1 and v2 forms the orthonormal basis set for V. Consider the transformation T(v1) = a*v1 + b*v2, T(v2) = c*v1 + d*v2 Show, using scalar products (not vector products), that the area scale factor of the transformation T is...