Suppose A is an m x n matrix, such that ker(A) = 0. prove that $A^TA$ is invertible

Question

Ok I did most of the work but i need someone to check my work.

1) Since we know that $Ker(A) = 0$, then this implies that $Ker(A^TA) = 0$

2) if the kernel of $A^TA$ is $0$, then $dim(ker(A^TA)) = 0$

3) Now, if we use the rank nullity theorem,

$Rank(A^TA) + dim(ker(A^TA)) = n$

$n + 0 = n$

4) We can clearly see that $Rank(A^TA)$ is equal to n

5) According to the invertible matrix theorem, one of its conditions clearly states that if a matrix has full rank, then it is invertible.

Thus we conclude that $A^TA$ is invertible

Is what i just did enough to prove that $A^TA$ is invertible? did i make a mistake in my logic somewhere?

Answer 1

Another way to see it:

$A^TA$ is invertible if and only if

$\ker A^TA = \{0\}. \tag 1$

If

$x \in \ker A^TA, \tag 2$

then

$A^TAx = 0; \tag 3$

thus

$x^TA^TAx = 0, \tag 4$

or

$(Ax)^T(Ax) = 0, \tag 5$

which implies

$Ax = 0, \tag 6$

or

$x \in \ker A = \{0\}; \tag 7$

thus

$x = 0, \tag 8$

so (1) binds.

Answer 2

That looks fine to me. The other minor piece is that $A^{T}A$ is square. So you have a square matrix with full rank, so it's invertible.

A worse way to prove this is using the singular value decomposition of $A = U\Sigma V^{T}$ where $U$ and $V$ are orthogonal and $\Sigma$ is diagonal with all positive entries. Thus $A^{T}A = V \Sigma^2 V^{T}$ which is invertible since all the eigenvalues $\Sigma_{ii}^2$ are nonzero.