Let $C$ be a fixed real $n\times n$ matrix, $X$ be an arbitrary real $n\times n$ matrix. Find the minimum value of:


When $n=1$ it's clear that the minimum value equal $|C|$, however it seems like with $n\ge2$ the minimum value is exactly zero, but I don't know how to prove it.

Also I have found out that this is relating to the minimum of $\det(I+Q^2)$ with $I$ is identity and $Q$ is arbitrary real matrix. If you can evaluate either:

  • $|\det(X+iC)|$

  • $det(I+Q^2)$

it would be appreciated!

  • $\begingroup$ Note that $\det(X+iC)\det(X-iC) = \det((X+iC)(X-iC)) = \det(X^2-C^2 + i[C,X])$, with $[C,X] = CX - XC \ne 0$ in general. So that second form is not equal to the original one. $\endgroup$ – mvw Jul 12 '14 at 12:45
  • $\begingroup$ It's just related, thks for pointing out anyway $\endgroup$ – Leaning Jul 12 '14 at 12:46

For any $C$, one can always find an $X$ such that $\det(X+iC)=0$ : if $C$ is not invertible, simply take $X=0$. If $C$ is invertible, take $X=CA$ where $A$ is any real matrix having a two-dimensional invariant subspace ${\sf span}(v_1,v_2)$ with $Av_1=v_2$, $Av_2=-v_1$ (so that $\pm i$ are eigenvalues of $A$). Then

$$ \det(X+iC)=\det(C(A+iI_n))=\det(C)\times 0=0. $$

Update I’ve just realized that my second case construction works in all cases, there’s no need to assume that $C$ is invertible.

  • $\begingroup$ @achillehui A few hours ago I proposed another, now deleted, simple and elegant (but completely wrong) answer :-) $\endgroup$ – Ewan Delanoy Jul 12 '14 at 12:31
  • $\begingroup$ @mvw corrected, thanks. $\endgroup$ – Ewan Delanoy Jul 12 '14 at 12:47
  • $\begingroup$ @EwanDelanoy The hardest bit for me to understand your argument is that it is always possible to provide a real matrix $A$ with eigenvalues $\pm i$. Your construction gives an $A$ with $A^2 v_1 = - v_1$ and $A^2 v_2 = - v_2$. So $A^2$ has the eigenvalue $-1$. I lack the argument how that implies $A$ has eigenvalue $i$. $\endgroup$ – mvw Jul 12 '14 at 13:54
  • $\begingroup$ @mvw Consider $Av$ where $v=v_1-iv_2$ $\endgroup$ – Ewan Delanoy Jul 12 '14 at 14:17
  • $\begingroup$ @EwanDelanoy Thanks. $\endgroup$ – mvw Jul 12 '14 at 14:45

If we assume true that the minimum is zero it means that we can minimize separately the real part and the imaginary part. Let n=2, knowing that $\det(X+iC)=\det(X)+i(\det(x_1,c_2)+\det(c_1,x_2))-\det(C)$ then it means that $\det(X)=\det(C)$ and $\det(x_1,c_2)=\det(x_2,c_1)$ where $x_i$ and $c_i$ represents the i-st columns of the matrices $X$ and $C$ respectively. If $\det(C)=0$ we would have that $X=C$ would be a minimizer with value $0$, so we can assume that $\det(C)\neq0$. We end the argument taking $X=(c_2,-c_1)$.

If $n>2$ we can note that taking $X=(c_2,-c_1,c_3,\dots,c_n)$ will give us: $$X+iC=(c_2+ic_1,-c_1+ic_2,\dots)=(c_2+ic_1,i(c_2+ic_1),\dots)$$ Which is clearly singular.

On a same note $\det(I+Q^2)=0$ if on some two dimensional space $Q$ is a rotation of a $\frac{\pi}{2}$ angle. Again if $n=2$ $$Q=\left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right)$$ then $Q^2=-I$. If $n>2$ then as written above it is needed that $\exists v_1, v_2\in\mathbb{R}^n$ s.t. $\|v_1\|=\|v_2\|$ and $Qv_1=v_2$ and $Qv_2=-v_1$.

  • $\begingroup$ $\det(X+iC)=\det(X)+i(\det(x_1,c_2)+\det(c_1,x_2))-\det(C)$ for $n=2$ $\endgroup$ – Lolman Jul 12 '14 at 10:25
  • $\begingroup$ I don't really understand your argument.. $\endgroup$ – Leaning Jul 12 '14 at 11:11
  • $\begingroup$ I think Ewan just explained it. Or are you talking about my comment about the determinant? $\endgroup$ – Lolman Jul 12 '14 at 12:29
  • $\begingroup$ Is it a problem of notation? $c_1$ is the first column of the $C$ matrix, the same goes for $x_1$. From the definition of the determinant follows the property that I wrote 2 hours ago. $\endgroup$ – Lolman Jul 12 '14 at 12:39
  • 1
    $\begingroup$ @Lolman I agree with your argument for $n=2$. But for $n > 2$ you would need to provide another expansion for $\det(X+iC)$. $\endgroup$ – mvw Jul 12 '14 at 13:23

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