Dimension and solution to matrix-vector product

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.

DumpmeAdrenaline said:

Let $$ X \in R^{m*n} $$

Your notation is unusual, so difficult to follow.
Many textbooks would write something like this:
Let ##A \in M^{n \times n}## where the coefficients of A are real.

DumpmeAdrenaline said:

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 $$ \beta \in R^{n} $$.
$$ X\beta=y $$

Or more clearly,
Let ##\vec x \in \mathbb R^n## and let ##\vec y = A\vec x##.

DumpmeAdrenaline said:

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 don't see how this reduces the problem. The matrix equation (in my notation) ##\vec y = A\vec x## represents n - 1 equations in n unknowns. Each equation represents a hyperplane in n-dimensional space. If the n - 1 equations are linearly independent, then the solution space is a line in n-dimensional space, this there are an infinite number of solutions.

DumpmeAdrenaline said:

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.

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Mark44 said:

I don't see how this reduces the problem. The matrix equation (in my notation) ##\vec y = A\vec x## represents n - 1 equations in n unknowns. Each equation represents a hyperplane in n-dimensional space. If the n - 1 equations are linearly independent, then the solution space is a line in n-dimensional space, this there are an infinite number of solutions.

Unfortunately this is the notation used by the author of the textbook I am studying from. How do we have n-1 system of equations when we take the dot product of n rows of matrix A with n entries of the column vector x. Can we think of it this way:

Given a 3*3 matrix A, where 2 row vectors of the matrix A are linearly independent then the 2 row vectors span a plane in 3 D and the 3rd vector lies in that plane.

In a similar way, suppose we have have x<n linearly independent vectors. The x independent row vectors span some some geometrical object in n-dimensional space where all row vectors lie in. We are trying to find a column vector which when dotted to the n row vectors yields the entries in the column vector y.

DumpmeAdrenaline said:

Given a 3*3 matrix A, where 2 row vectors of the matrix A are linearly independent then the 2 row vectors span a plane in 3 D and the 3rd vector lies in that plane.

Again, I don't see how this reduces the problem. With a 3 x 3 matrix, if all three rows are linearly independent, then there is a unique solution. If two of the rows are linearly independent, and the third row is a linear combination of the other two rows, then the solution is all points on the line of intersection of the two planes represented by the independent rows. If two of the rows are multiples of one of the rows, then the solution is all points on the plane represented by one of the rows.

In what sense does this reduce the problem?

I thought it reduces the problem in the sense that all the independent vectors that span the row space will lie on a geometric object of dimension x.