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?


2 Answers 2


If your matrix is real, $U$ and $V$ will be real.

From wikipedia:

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.

See Singular Value Decomposition (SVD) tutorial for explanations.


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