# Let $A_{n\times n}$ be a real matrix. Is it true that $I+A^TA$ is invertible?

Let $$A_{n\times n}$$ be a real matrix. Is it true that $$I+A^TA$$ is always invertible?

Let $$M = I+A^tA$$. Take $$v$$ in the kernel of $$M$$. Now,

$$0 = [I+A^tA]v,$$

and so we get $$A^tAv = -v$$. Right multiplying by $$v^t$$ once again, we have $$v^tA^tAv = -v^tv$$ and so $$0 \leq \|Av\|^2 = (Av)^t(Av) = v^tA^tAv = -v^tv = -\|v\|^2 \leq 0$$

which proves that $$v = 0$$. Thus, $$M$$ is injective and since it is square, it is invertible.

• No need to consider eigenvalues and positive semidefinite matrices. +1 Commented Oct 26, 2018 at 8:56

If $$I+A^TA$$ is not invertible, then there exists a non-null vector $$v$$ such that:

$$(I+A^TA)v = 0.$$

This means that:

$$(I+A^TA)v = 0 \Rightarrow v + A^TAv = 0 \Rightarrow A^TAv = -v.$$

In other words, $$v$$ is an eigenvector of $$A^TA$$ with eigenvalue $$\lambda = -1$$.

Since $$A^TA$$ is by definition semidefinite positive, then there is no negative eigenvalue. That is, it is impossible that $$I+A^TA$$ is not invertible.

If $$I + A^TA$$ only has positive eigenvalues, then $$I + A^TA$$ is invertible.

We just need to show that $$A^TA$$ only has non-negative eigenvalues, which is the same as saying that $$A^TA$$ is positive semi-definite. $$A^TA$$ is of such type if

$$x^TA^TAx \geq 0\ \forall x$$

notice that $$Ax$$ is a vector and $$x^TA^TAx = (Ax)^TAx = (Ax, Ax) = ||Ax||^2\geq 0$$

hence $$A^TA$$ is positive semi-definite. Now assume $$\lambda$$ is an eigenvalue for $$A^TA$$ and $$A^TAv = \lambda v$$ for some $$v$$, then

$$(I + A^TA)v = Iv + A^TAv = v + \lambda v = (1 + \lambda)v$$

hence the eigenvalues of $$I + A^TA$$ are the same as the ones for $$A^TA$$ but shifted one unit. Hence, only if $$A^TA$$ had $$-1$$ as eigenvalue the matrix $$I +A^TA$$ wouldn't be invertible. But $$-1$$ cannot be an eigenvalue of $$A^TA$$, hence we have proved that $$I + A^TA$$ is invertible.

Yes. Indeed, $$A^TA$$ is a symmetric matrix, and is positive, so $$I + A^T A$$ is strictly positive, and in particular is invertible.

Yes, because $$A^TA$$ is symmetric positive semi-definite and therefore $$I+A^TA$$ can't have $$0$$ as an eigenvalue.

• Probably a typo, but I think you mean eigenvalue, not eigenvector. Commented Oct 26, 2018 at 9:00
• @F.Carette Indeed.
– user562983
Commented Oct 26, 2018 at 9:13