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Question: Can anyone use some linear algebra to justify why $R^{-1} = R^{T}$ in this case. (Matrices are not equal to their transpositions in general.) In other words, what is special about matrix $R$?


R is special because its columns are orthonormal vectors. We know that the dot product of a vector with itself is the norm of that vector. We also know that the dot product of two orthogonal vectors is 0.

We want a matrix $R^{-1}$ such that when multiplied with R we get the identity matrix. Now we observe that if we transpose R, each row i in $R^{T}$ will be column i in $R$. This means when we multiply $R^{T}$ by $R$, the only non zero value in each column i of the resulting matrix will be the value at the jth row where j = i since all other values are the result of a dot product between orthogonal vectors. But when j = i we have the product of two identical vectors of unit length, so their dot product is 1. This gives us the diagonal matrix with 1s along the diagonal, i.e. the identity matrix.


@lucida, That is a great explanation!