- Let $A$ be an $m \times n$ matrix. Suppose that the matrix equation $AX = Y$ is consistent for any $Y$ that is an element of $R^m$ . (a) What is the range of $T_A$? Justify your answer. (b) Under the assumptions above, is it possible that m > n? Justify your answer.
For part (a), I said the range of $T_A$ is $R^m$ because the codomain $Y$ are all possible solutions to the linear transformation $T_A$ with vectors $X$, then $R^m$ contains both consistent and nonconsistent solutions to all linear transformations.
For part (b), I wrote $m > n$ is possible, but I'm having a hard time explaining so. It is fairly obvious because the matrix $A$ is the transformation $T: R^n \rightarrow R^m$. Using the formula $\mathrm{rank}(A) + \mathrm{null}(A) = n$, then $m \geq n$ . Is that a sufficient answer? Any help is welcome.
For part a, consider row reducing the matrix to reduced row echelon form. What is the only way for the matrix to be consistent? How many linearly independent rows does it have then? Columns? And what do the columns of $A$ span?
For part b, you have it exactly backwards. You can use your approach from the rank-nullity theorem from which we get $$n = \rm{rank}(A) + \rm{nullity}(A)$$ Since your matrix is surjective into $\mathbb{R}^m$ we have $\rm{rank}(A) = m$. Since $\rm{nullity}(A) \ge 0$ we must then have $$n - m = \rm{nullity}(A) \ge 0$$ Alternatively, for an approach closer in style to part a, again consider the reduced row echelon form of $A$. If we have more rows than columns what must we necessarily introduce in the reduced row echelon form? How does that violate part a?