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Suppose that $A_{mn}$ is a matrix over some field, and that $C, R$ is its column space and row space, without using the fact that $rank(C) = rank(R)$, can we show that, there exists a bijection between $C$ and $R$?

Not an ID
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  • HINT: $rank(A)=rank(A^T)$ – meta_warrior Nov 03 '13 at 01:38
  • @freak_warrior Are you suggesting that $dim C_A = dim R_{A^T}$ (by def.) and $ R_{A^T} = dim R_A$ (by hint), where $C_A$ means the column space of $A$, thus $dim R_A = dim C_A$, hence they are isomorphic? In fact I want to use this to show that column rank equals row rank, therefor we can define the rank to be either one. – Not an ID Nov 03 '13 at 02:30

2 Answers2

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Hint: For $x\in R$, consider $Ax\in C$. Show this map is a bijection. (It might be useful to note that $\Bbb R^n=R\oplus N$, where $N=\{x: Ax=0\}$.)

Ted Shifrin
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  • Though it's limited to real numbers, I think this is the best I can get. Thanks. – Not an ID Nov 03 '13 at 11:22
  • No, you don't need to use the orthogonality. But avoiding dimension everywhere makes it a bit harder. – Ted Shifrin Nov 03 '13 at 11:40
  • Do you mean that $F^n = R \oplus N$ holds for any filed $F$? ($\diamond$) I used orthogonality for proving the formula $\mathbb{R}^n = R \oplus N$. If ($\diamond$) is true, can you give me some hint to prove it? :-) (I have some college linear algebra, hope it's sufficient) – Not an ID Nov 03 '13 at 14:02
  • On further thought, you're right. Any time we can have $x^\top x=0$ for some $x\ne 0$, we have a nontrivial intersection. However, the usual proof that pivots in echelon form give you a recipe for bases for both $R$ and $C$ works over any field. Indeed, my original isomorphism works only over $\Bbb R$, as the matrix $A=\begin{bmatrix} 1&i\-i&1\end{bmatrix}$ illustrates. – Ted Shifrin Nov 03 '13 at 14:58
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    "Indeed, my original isomorphism only works over $\mathbb{R}$". There are more fields than $\mathbb{R}$ and $\mathbb{C}$! Your argument works over a field for which $x_1^2 + \ldots + x_n^2 = 0 \implies x_1 = \ldots = x_n = 0$, i.e., a formally real field. But I honestly find this answer a bit strange even over $\mathbb{R}$: you're not using the fact that row rank equals column rank, but you're using exactly the tools you need to prove this fact...and then you're essentially proving it. – Pete L. Clark Dec 05 '13 at 05:32
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The assertion that there is a linear bijection between $C$ and $R$ is equivalent to the assertion that they have the same dimension. So if your bijection is required to be a linear map, the answer must be "not really".

What if your bijection is not required to be linear? Well, if the scalar field $F$ is finite, then two $F$-vector spaces $V_1$ and $V_2$ are isomorphic if and only if there is a bijection between them...so again, "not really".

If your field is infinite and your vector spaces are finite-dimensional (as $R$ and $C$ are), then there is a bijection between them exactly when they are either both $0$ or neither $0$. So we get something here: could it be easier to show that the row space is zero if and only if the column space is zero than to show that they have the same dimension? Yes, it's much easier!

Pete L. Clark
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