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  1. 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.

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Regarding your answer for part (b): The matrix $A$ does indeed define a linear transformation $T_A:\mathbb{R}^n\to\mathbb{R}^m$, but this by itself doesn't tell you anything about whether $m>n$. For example, the $m\times n$ zero matrix defines the zero transformation $0:\mathbb{R}^n\to\mathbb{R}^m$ for any choice of $m$ and $n$. –  Brad Nov 6 '12 at 16:44
    
Yes, sorry I forgot to add a couple things when i posted, i edited it to include what I forgot! –  tamefoxes Nov 6 '12 at 16:47
    
How do you conclude $m\geq n$ from that formula? –  Brad Nov 6 '12 at 16:50
    
I was using the logic that they sum to m, but since that is not so. I am back to where I started –  tamefoxes Nov 6 '12 at 16:52
    
Your answer for part (a) is correct, but your explanation is jumbled. What do you mean by "the codomain $Y$ are all possible solutions to the linear transformation $T_A$ with vectors $X$? What is a "nonconsistent solution"? (For that matter, linear transformations don't have "solutions", but rather equations involving linear transformations have solutions.) –  Brad Nov 6 '12 at 16:54

1 Answer 1

up vote 2 down vote accepted

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?

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The matrix is consistent if there are 3 linearly independent roew and columns. The columns of A spans R^m. –  tamefoxes Nov 7 '12 at 5:47

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