# Why the transpose of a singular matrix is singular? [closed]

I have to prove this lemma without using the concept of rank neither the concept of determinant:

$A$ is a singular matrix iff $A^T$ is singular

Unfortunately i've only found proofs that contains rank and determinant. Can you help me ?

• What's your definition of singular matrix? Jul 24, 2018 at 16:05
• To me, singular matrix = Columns are linear dipendent. I want to prove that that if columns are linear dipendent also rows are. Jul 24, 2018 at 16:10
• $\det A=\det A^T$. Jul 24, 2018 at 17:13
• @Sorfosh Singular matrices are not invertible. Jul 24, 2018 at 19:56
• For a proof using the definition you mentioned above, have a look here. Jul 25, 2018 at 2:36

Assume for contradiction that $A^T$ was invertible, then there would be a matrix $B$ with $BA^T=I$. But that means $I=I^T=(BA^T)^T=AB^T$, so $B^T$ would be an inverse for $A$, which is impossible.
Formally, a singular matrix $A$ is one for which there does not exist another matrix $B$ with $AB=BA=I$.
The statement here can be proven through the contrapositive: if $A$ is not singular, there exists some $B$ with $AB=I$. Transposing this gives $B^TA^T=I$, so $A^T$ is not singular. Thus if $A^T$ is singular, $A$ is singular. Replacing $A$ with $A^T$ in the last sentence gives the other direction, so the original statement is established.
If $A$ were singular, then the kernel (null space) of $A$ has nonzero vectors in it. That is, $Ax = 0$ admits nontrivial (nonzero) solutions. Now take the transpose on each side. Then if $x$ is nonzero, so is $x^T$ (clearly), and so $x^T A^T = 0^T$ also admits nonzero solutions, so $A^T$ is singular.
• How do you know that $x^TA^T=0$, $x\ne0$ implies that there exists $y\ne0$ with $A^Ty=0$? Jul 24, 2018 at 16:46
• @DavidC.Ullrich Sean Roberson did not directly claim that. If $x^TA^T=0^T$ has a nontrivial solution, then clearly its rows are not linearly independent, so it is singular. Jul 24, 2018 at 21:48