3

If $X$ is a singular matrix, then is $X'X$ singular? And why?

(' means transpose)

Thanks in advance!

Ian
  • 1,391
  • 1
    What are your thoughts? In this case it might help to explore equivalent conditions to singularity of a matrix. – Jonathan Y. Oct 15 '13 at 05:45

5 Answers5

5

$Xv=0$ implies $X^T X v = 0$.

$X^T X v = 0$ implies $v^T X^T X v = \|X v\|^2 = 0$, which in turn implies $X v=0$.

copper.hat
  • 172,524
4

Note that $rank(X'X)\leq\min\{rank(X'),rank(X)\}<n$, which implies $X'X$ is still singular.

Shuchang
  • 9,800
2

We know that a matrix is sigular is equivalent to the determinant being zero. Therefore from your statement we have $\det(X)=0$. We also know $$\det(X^{\prime})=\det(X),$$ and $$\det(AB)=\det(A)\det(B).$$ So we have $$\det(XX^{\prime})=\det(X)\det(X^{\prime})=0.$$ Which is equivalent to saying $XX^{\prime}$ is singular.

triomphe
  • 3,848
  • For a discussion on the $\det$ properties please refer http://en.wikipedia.org/wiki/Determinant. Thanks – triomphe Oct 15 '13 at 05:53
2

Since $X$ is singular, there exists a non-zero vector $v$ such that $Xv=0$.

Now, $$X^TXv=X^T(Xv)=X^T0=0,$$ which implies $v$ is also in the nullspace of $X^TX$. Thus, the matrix is singular.

Daryl
  • 5,598
  • Forgive the naivety, but isn't matrix multiplication noncommutative? And hence (') will not necessarily equal '(), right? I feel like I must be missing something, because I know your conclusion is correct. But I feel commutativity is necessary to say that your algebra implies " is also in the nullspace of '". – kdbanman Oct 23 '19 at 06:52
  • @kdbanman Yes. It’s non-commutative, but it is associative. So $AB\neq BA$, but $(AB)C=A(BC)$. – Daryl Oct 23 '19 at 21:17
  • 1
    I... am not clever. Thank you. The property I was looking for was indeed associativity, which matrix multiplication definitely has. – kdbanman Oct 24 '19 at 04:44
2

$X \, \text{is singular} \Leftrightarrow \det(X) = 0, \tag{1}$

$\det (X) = 0 \Rightarrow \det (X^TX) = \det (X^T) \det (X) = 0, \tag{2}$

$\det (X^TX) = 0 \Rightarrow X^TX \text{is singular}; \tag{3}$

OR

one might observe that $X$ singular implies the existence of a vector $v \ne 0$ such that $Xv = 0$, whence $X^TXv = 0$, whence $X^TX$ is singular! QED.

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
  • 71,180