Is the rank of a matrix the same of its transpose? If yes, how can I prove it?

Question

I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view:

"The rank of a matrix A is the number of non-zero rows in the reduced row-echelon form of A".

The lecturer then explained that if the matrix $A$ has size $m * n$, then $rank(A) \leq m$ and $rank(A) \leq n$.

The way I had been taught about rank was that it was the smallest of

• the number of rows bringing new information
• the number of columns bringing new information.

I don't see how that would change if we transposed the matrix, so I said in the lecture:

"then the rank of a matrix is the same of its transpose, right?"

And the lecturer said:

"oh, not so fast! Hang on, I have to think about it".

As the class has about 100 students and the lecturer was just substituting for the "normal" lecturer, he was probably a bit nervous, so he just went on with the lecture.

I have tested "my theory" with one matrix and it works, but even if I tried with 100 matrices and it worked, I wouldn't have proven that it always works because there might be a case where it doesn't.

So my question is first whether I am right, that is, whether the rank of a matrix is the same as the rank of its transpose, and second, if that is true, how can I prove it?

Thanks :)

Answer 1

The answer is yes. This statement often goes under the name "row rank equals column rank". Knowing that, it is easy to search the internet for proofs, e.g.

http://homepage.mac.com/ehgoins/ma265/lecture_28.pdf

Also any reputable linear algebra text should prove this: it is indeed a rather important result.

Finally, since you said that you had only a substitute lecturer, I won't castigate him, but this would be a distressing lacuna of knowledge for someone who is a regular linear algebra lecturer.

Answer 2

Since you talked about reduced row-echelon form, I assume you know what elementary row and column operations are. The basic fact concerning these operations is the following:

Elementary (row or column) operations change neither the row rank nor the column rank of a matrix.

Now, given a nonzero matrix $A$, try the following:

1. Bring $A$ to its reduced row-echelon form $R$ using elementary row operations.
2. Bring $R$ to its reduced column-echelon form $B$ using elementary column operations.

Then $B$ is of the form $$ \begin{pmatrix} 1&&&0&\ldots&0\\ &\ddots&&\vdots&&\vdots\\ &&1&0&\ldots&0\\ 0&\ldots&0&0&\ldots&0\\ \vdots&&\vdots&\vdots&&\vdots\\ 0&\ldots&0&0&\ldots&0\\ \end{pmatrix}. $$ Now it is obvious that the row rank of $B$ is equal to the column rank of $B$ (which is equal to the number of ones in the above "reduced row-and-column-echelon form"). Hence the row rank of $A$ is equal to the column rank of $A$, i.e. the row rank of $A$ is equal to the row rank of $A^T$.

Answer 3

There are several simple proofs of this result. Unfortunately, most textbooks use a rather complicated approach using row reduced echelon forms. Please see some elegant proofs in the Wikipedia page (contributed by myself):

http://en.wikipedia.org/wiki/Rank_%28linear_algebra%29

or the page on rank factorization:

http://en.wikipedia.org/wiki/Rank_factorization

Another of my favorites is the following:

Define $\operatorname{rank}(A)$ to mean the column rank of A: $\operatorname{col rank}(A) = \dim \{Ax: x \in \mathbb{R}^n\}$. Let $A^{t}$ denote the transpose of A. First show that $A^{t}Ax = 0$ if and only if $Ax = 0$. This is standard linear algebra: one direction is trivial, the other follows from:

$$A^{t}Ax=0 \implies x^{t}A^{t}Ax=0 \implies (Ax)^{t}(Ax) = 0 \implies Ax = 0$$

Therefore, the columns of $A^{t}A$ satisfy the same linear relationships as the columns of $A$. It doesn't matter that they have different number of rows. They have the same number of columns and they have the same column rank. (This also follows from the rank+nullity theorem, if you have proved that independently (i.e. without assuming row rank = column rank)

Therefore, $\operatorname{col rank}(A) = \operatorname{col rank}(A^{t}A) \leq \operatorname{col rank}(A^{t})$. (This last inequality follows because each column of $A^{t}A$ is a linear combination of the columns of $A^{t}$. So, $\operatorname{col sp}(A^{t}A)$ is a subset of $\operatorname{col sp}(A^{t})$.) Now simply apply the argument to $A^{t}$ to get the reverse inequality, proving $\operatorname{col rank}(A) = \operatorname{col rank}(A^{t})$. Since $\operatorname{col rank}(A^{t})$ is the row rank of A, we are done.

Answer 4

Yes, it is a fact. This is true over any commutative field. See for instance the first chapter of Emil Artin, Galois Theory for a very elementary argument.

If you need to phrase that argument in more conceptual terms, consider the matrices as linear transformations. If A is the matrix, then let $A^t$ be the transpose, and then $A^tA$ and $A$ have the same domain, and use the fact that they have the same null space, and use the dimension theorem rank + nullity = dimension of the space.

Your argument is true for real matrices only. For complex matrices they may not have the same null space.

Answer 5

(1) If $f:V\to W$ is a linear map and $f^*:W^*\to V^*$ is its transpose, then we have a canonical isomorphism

$$\text{Im}(f^*)=\text{Im}(f)^*.$$

This can bee seen as follows:

(2) If $$ V\ \overset{p}{\twoheadrightarrow}A\ \overset{i}{\rightarrowtail}\ W\quad\text{and}\quad V\ \overset{q}{\twoheadrightarrow}B\ \overset{j}{\rightarrowtail}\ W $$ are two diagrams of linear maps such that

(a) $i$ and $j$ are injective, $p$ and $q$ are surjective,

(b) $i\circ p=j\circ q$,

then there is a unique linear map $\varphi:A\to B$ such that $\varphi\circ p=q$ and $j\circ\varphi=i$. Moreover $\varphi$ is bijective. The proof is easy.

To prove that (2) implies (1), note that the three diagrams
$$ V\ \overset{p}{\twoheadrightarrow}\ \text{Im}(f)\ \overset{i}{\rightarrowtail}\ W, $$ $$ W^*\ \overset{i^*}{\twoheadrightarrow}\ \text{Im}(f)^*\ \overset{p^*}{\rightarrowtail}\ V^*, $$ $$ W^*\ \overset{q}{\twoheadrightarrow}\ \text{Im}(f^*)\ \overset{j}{\rightarrowtail}\ V^*, $$ where $p,i,q,j$ are the obvious maps, satisfy (a). As $p^*\circ i^*=f^*=j\circ q$, we see that (2) implies (1).

Assume the rank of $f:V\to W$ is infinite. The Erdős-Kaplansky Theorem implies then

$$\text{rank}(f^*)=|K|^{\text{rank}(f)},$$

where $K$ is the ground field and $|X|$ is the cardinality of $X$ for any set $X$.

More precisely, the Erdős-Kaplansky Theorem says $$ \dim(V^*)=|K|^{\dim(V)} $$ whenever $V$ is infinite dimensional, or, equivalently $$ \dim(K^S)=|K^S|, $$ where $S$ is an infinite set and $K^S$ is the set of families $(a_s)_{s\in S}$ with $a_s$ in $K$. In words:

The dimension of an infinite dimensional dual vector space is equal to its cardinality.

For a proof of the Erdős-Kaplansky Theorem, please see this answer.