In a matrix $A$ $(m\times n)$ ,if $\Bbb{rank}A^T= dim(ker A)$, how to show that A must have an even number of columns?

Question

Question says itself. I was preparing for my Maths 2 exams tomorrow where this question came up. I looked up at lecture notes and all but it doesn't have anything related to this.

Answer 1

Hint:

• rank($A$)+nullity(A) = number of columns

Answer 2

By the rank-nullity theorem applied to the matrix $A$:

$\textrm{dim(ker A) + rank(A)} = n$ $\; \; \; \; $(i)

By the rank-nullity theorem applied to the matrix $A^t$:

$\textrm{dim(ker $A^t$) + rank($A^t$)} = m$ $\; \; \; \; $ (ii)

By the assumption stated in the problem, equation (ii) can be written as

$\textrm{dim(ker $A^t$)} + \textrm{dim(ker A)} = m$ $\; \; \; \; $ (iii)

Subtracting equation (iii) from equation (i) yields that

$\textrm{rank(A) - dim(ker $A^t$) } = n - m$ $\; \; \; \; $ (iv)

But by equation (ii), $\textrm{m - rank($A^t$) = dim(ker $A^t$)} $. Hence, equation (iv) becomes

$\textrm{rank(A) + rank($A^t$)} = n$ $\; \; \; \; $ (v)

But it is known that $\textrm{rank(A) = rank($A^t$)}$ for any matrix $A$, so we have that

$\textrm{rank(A)} = \frac{n}{2} $ $\; \; \; \; $ (vi)

Hence, $n$ must be even, for $\textrm{rank(A)} \in \mathbb{Z_{\geq 0}}$.