How does the SVD quarantine all of a matrix's difficulty inside one diagonal Σ, turning the pseudo-inverse into just inverted singular values?
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Building from the singular values
Start from , then form by inverting through , leaving the rest zero, and transposing the shape from by to by .
The products and
Multiply both ways, record the diagonal ones in sizes by and by , and name which projection each product performs.
The formula
Derive from using and , and write the finished formula down.
Minimality of the pseudo-inverse
Record that entries placed in the zero blocks of leave the products unchanged and only enlarge the matrix, so is the smallest one achieving them.