Finding an $x \in \mathbb R$ so that $\|I - x A\|_2$ becomes minimal

Question

Let $A \in \mathbb R^{n \times n}$, and let $I$ denote the ($n \times n$-)identity matrix. Then I want to find a $x \in \mathbb{R}$, so that:

$$\|I - x A\|_2$$

becomes minimal (where $\|\cdot\|_2$ denotes the Euclidean matrix norm).

Let's assume $A ≠ 0$ to exclude the trivial case where any $x$ can be chosen. I think it's obvious that such a minimum exists, because $\|I - x A\|$ can be considered a real (continuous) function with respect to $x$ which is of course always positive, and $\lim_{\|x\| \to \infty} \|I - x A \|_2 = \infty$. So there must be (at least one) minimum, and it must somehow depend on $A$.

I'm not really sure on how to find it, though. I've tried to write out $\|I - x A\|_2 = \sqrt{\sum_{i=0}^n \sum_{j=0}^n |\delta_{ij}- x a_{ij}|^2}$ and thought about differentiating that thing, but from some of the squares we could pull out $x$ and from others (where $\delta_{ij} = 1$) we can't, so I'm afraid that $\frac d{dx}$ of that thing would quickly become very nasty, let alone setting it $= 0$ and solving it for $x$, so I frankly suspect there's an easier way. Could we somehow bring the eigenvalues, the spectral radius or something like that into the equation, or use one of the properties of the matrix norm?

Answer 1

Just minimise the square:

\begin{eqnarray} \|I-xA\|^2 &=& \sum_i \sum_j (\delta_{ij} -x \delta_{ij} A_{ij} + x^2 A_{ij}^2) \\ &=& n - x \sum_i A_{ii} + x^2 \sum_i \sum_j A_{ij}^2 \\ &=& n -x \operatorname{tr} A + x^2 \|A\|_F^2 \end{eqnarray}

Assuming that $A \neq 0$ we have (after differentiating with respect to $x$ and setting to zero): $x = { \operatorname{tr} A \over 2\|A\|_F^2 }$.

Answer 2

The norm can be express with the trace: $$ \forall B \in \mathcal{M}_n(\mathbb{R}) \quad ||B||_2^2 = \text{tr}(B^TB) $$ Using this: $$ ||I - xA||_2^2 = \text{tr}((I - xA)^T(I - xA)) $$ After expanding this you will find a 2nd degree polynomial in $x$. You know how to find the minimum of this.

Answer 3

Hint: By the Pytagoras Theorem for inner product spaces $$\|I- xA\|^2 \geq \| I -\proj_{A}(I) \|^2$$