- #1
DumpmeAdrenaline
- 78
- 2
Let $$ X \in R^{m*n} $$ where m=n with rank(X)<m then there is at-least one equation which can be written as a linear combination of other equations. Let $$ \beta \in R^{n} $$.
$$ X\beta=y $$
Suppose we have x<m independent equations (the equations are consistent) formed by taking the dot product of x row vectors of with the column vector. Each independent equation represents a geometrical object of dimension n-1 (n-1 degrees of freedom) . We have x geometrical objects in n dimensions and we are trying to find the intersection of all these geometrical objects that satisfies the RHS represented by y. The dimension of row space is x which corresponds to the # of independent equations. Can we say that we are reducing the problem to finding a vector x that is perpendicular to all the x row vectors that all lie on some geometric object n-1? If this is the case, why there are infinite solutions?
I understand why we have infinite solutions if we think of X in terms of column vectors. If y is in the column space, we can check this by comparing the rank of X and rank of X augmented with y. We have infinite possibilities for scalars of the independent column vectors in the sum of independent columns that would yield y.
$$ X\beta=y $$
Suppose we have x<m independent equations (the equations are consistent) formed by taking the dot product of x row vectors of with the column vector. Each independent equation represents a geometrical object of dimension n-1 (n-1 degrees of freedom) . We have x geometrical objects in n dimensions and we are trying to find the intersection of all these geometrical objects that satisfies the RHS represented by y. The dimension of row space is x which corresponds to the # of independent equations. Can we say that we are reducing the problem to finding a vector x that is perpendicular to all the x row vectors that all lie on some geometric object n-1? If this is the case, why there are infinite solutions?
I understand why we have infinite solutions if we think of X in terms of column vectors. If y is in the column space, we can check this by comparing the rank of X and rank of X augmented with y. We have infinite possibilities for scalars of the independent column vectors in the sum of independent columns that would yield y.