This is a somewhat long and laborious process. Here's an outline of it.

You have to start by finding the eigenvalues and eigenvectors of T*T and TT*. Take TT* first, because that is a 2×2 matrix. So, . The eigenvalues of this matrix are 1 and 5, with corresponding normalised eigenvectors and .

Next, . You don't have to calculate the eigenvectors again, because the theory tell you that these will be the same as those of TT*, together with enough zeros added to make up the right number. Therefore they are 1, 5 and 0. You still have to find corresponding normalised eigenvectors, which are , , .

You can now write down the matrices for the singular value decomposition of T. These are constructed as follows (apart from a slight modification that will come at the end): U is the matrix whose columns are the normalised eigenvectors of T*T, V is the matrix whose columns are the normalised eigenvectors of TT*, and Σ is a matrix having the same shape as T, with zeros everywhere except down the main diagonal, where the entries are thesingular valuesof T, namely the square roots of the eigenvalues. Thus , , .

There's one final catch. You should check your answer by writing down the adjoint V* of V and computing the product UΣV*, which should be equal to T. But when you do that, you find that . This looks very different from T. But in fact it's not so different. The process of finding U, Σ and V is only determined up to a scalar factor of modulus 1, which in this case is . If you multiply each entry of UΣV* by then you do indeed get T. So to get the final correct solution, you should multiply each entry of U by . With that modified U it is true that T=UΣV*.