# Does a Singular Value Decomposition of a real matrix ensure that the unitary matrices U and V are real?

I am answering this simple question using SVD (yes I am aware this is not the optimal way to show this but my question is about this particular line of reasoning). I am including the question here just for context - not for an answer.

Q: Let $$A$$ be an invertible $$n \times n$$ real matrix and let $$\lambda$$ be a real eigenvalue of $$A$$. Show that $$\frac{1}{\lambda}$$ is an eigenvalue of $$A^{-1}$$.

If $$A$$ is an invertible square matrix then it has a singular value decomposition of the form \begin{align*} A = UDV^T \quad \text{(SVD)} \end{align*} where $$U, V^T$$ are unitary matrices such that the columns of $$U$$ are the singular vectors of $$A$$ and the columns of $$V^T$$ are the singular vectors of $$A^T$$. Additionally, $$D \in R^{n\times n}$$ where $$D_{i,i} = \sigma_i$$ and $$\sigma_i$$ is the $$i$$th largest singular value of $$A$$. Computing the inverse of this decomposition yields \begin{align} (UDV^T)^{-1} &{}= (V^T)^{-1}D^{-1}U^{-1}\\ &{}= (V^T)^T D^{-1} U^{T}\\ &{}= VD^{-1}U^{T}\\ &{}= A^{-1} \end{align} which gives a new singular value decomposition where the columns of $$V$$ are the singular vectors of $$A$$ and the columns of $$U^T$$ are the singular vectors of $$A^T$$. Additionally, now $$D^{-1}$$ contains the singular values of $$A^{-1}$$ in ascending order on its diagonal.

Note that (2) is derived from the definition of unitary matrices. Given unitary matrix $$M$$, it turns out that $$M^{-1} = M^T$$ if $$M$$ is real.

We know this is a valid singular value decomposition of $$A^{-1}$$ for two reasons. First, the singular vectors of $$A$$ must be equal to the singular vectors of $$A^{-1}$$ because the set of eigenvectors of $$A$$ and $$A^{-1}$$ are the same. (This is because by definition the eigenvectors must define the same span). Second, the singular values of $$D^{-1}$$ while inverted in ordering still correspond to the correct singular vector of $$A^{-1}$$.

Finally, once we accept that $$VD^{-1}U^{T}$$ is a valid singular value decomposition of $$A$$ then we can be certain that any eigenvalue $$\lambda$$ of $$A$$ must exist as a singular value $$\lambda^2$$ on the diagonal of $$D$$, and since the inverse of a diagonal matrix $$D$$ simply inverts each value on the diagonal, we know with certainty that the singular value $$\lambda^{-2}$$ exists on $$D^{-1}$$ and therefore $$\lambda^{-1}$$ is an eigenvalue of $$A^{-1}$$.

My question: I use the fact that a real unitary matrix $$M$$ has the property that $$M^{-1} = M^T$$, however, I am not sure I can make this claim. Is it sufficient that the problem statement says $$A$$ is an invertible $$n \times n$$ real matrix to assume that $$V$$ and $$U$$ in its singular value decomposition are real? Or is it possible that (say, if the span of $$A$$'s eigenvectors does not cover the entire codomain of $$A$$), that $$U$$ and $$V$$ could be complex?

If your matrix is real, $$U$$ and $$V$$ will be real.
the singular value decomposition of an $$m × n$$ real or complex matrix $$M$$ is a factorization of the form $$U Σ V^∗$$, where $$U$$ is an $$m × m$$ real or complex unitary matrix, $$Σ$$ is an $$m × n$$ rectangular diagonal matrix with non-negative real numbers on the diagonal, and $$V$$ is an $$n × n$$ real or complex unitary matrix. If $$M$$ is real, $$U$$ and $$V^T = V^∗$$ are real orthogonal matrices.
According to this, if your matrix is real, so are $$U$$ and $$V$$.
$$U, V, D$$ matrices for a real matrix $$A$$ are real matrices.