By Prabhat Choudhary

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**Example text**

Rotating an object and projecting it is equivalent to looking at it from di erent points. However, this method does not give a very realistic picture, because it does not take into account the perspective, the fact that the objects that are further away look smaller. To get a more realistic picture one needs to use the so-called perspective projection. To: Qefine a perspective projection one needs to pick a point the centre of projection or the 24 Basic Notions focal point) and a plane to project onto.

En} there. , it is just the matrix B whose kth column is the vector (column) vk . And in the other direction [l]BS = ([1] SB )-1 = 15 1. For example, consider a basis un B={(1), in }R2 , and let S denote the standard basis there. Then [1]SB = Ui) =: B and 1]-1 S[1]BS= [ SB 1(-I2 -I2) = B-1 ="3 (we know how to compute inverses, and it is also easy to check that the above matrix is indeed the inverse of B) An example: going through the standard basis. In the space of polynomials of degree at most 1 we have bases .

Vm E ~m is complete in ~n ifand only if the equation +x2v2 +··· +xmvm=b has a solution for any right side b E ~ n . By statement 2 above, it happens if and only XlvI if there is a pivot in every column in echelon form of the matrix. And finally, the system VI' v2' ••. , vm E ~m is a basis in ~n ifand only if the equation +x2v2 +··· +xmvm=b has unique solution for any right side b XlVI ~n. By statement 3 this happens if and only ifthere is a pivot in every column and in every row of echelon form of A.