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Consider unit vectors $u_i\in \mathbb{R}^n,\ m>n$ s.t. $A=[u_1\cdots u_m]$ has rank $n$.

If $c_i>0$ and $c=(c_1,\cdots, c_m),\ C={\rm diag}\ (c_1,\cdots, c_m)$ s.t. $$Ac=0,$$ then $$ ACA^T =\sum_{i=1}^m c_i u_iu_i^T = aI_n $$ where $T$ is a transpose, $a>0$ is some constant, and $I_n$ is identity on $\mathbb{R}^n$. How can we prove this ?

Thank you in advance.

[Add] Equivalent condition on $u_i$ is that origin is in the interior of convex hull of $u_i$.

(1) If $e_i$ is orthonormal set, then we have $\{e_i\}\cup \{ -e_i \}$ That is $$ \sum_{i=1}^n \frac{1}{2} \{ e_i\otimes e_i + (-e_i)\otimes (-e_i) \} = I_n $$

(2) If unit vectors $u_i\in \mathbb{R}^2$ forms vertices of equilateral triangle, then $\sum_{i=1}^3 \frac{2}{3}u_i\otimes u_i $ is $I_3$.

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  • $\begingroup$ Are some of the $u_i$ also orthogonal to each other? That is, does it hold that $u_j^T u_j = 0$ for $j=1,\dots,n$ or some other indexing set of cardinality $n$ ? $\endgroup$ – PseudoRandom Jun 30 '17 at 11:29
  • $\begingroup$ Yes. I add some more explanation. $\endgroup$ – HK Lee Jun 30 '17 at 11:48
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    $\begingroup$ Take $n=2$, $m=3$, $A=\begin{bmatrix}1&0&-1/\sqrt{2}\\0&1&-1/\sqrt{2}\end{bmatrix}$, $c=[1/2,1/2,1/\sqrt{2}]^T$. I have unit columns in $A$, $Ac=0$, $c>0$ entrywise, but $ACA^T=\frac{1}{2}I_2+\frac{1}{2\sqrt{2}}[1,1][1,1]^T\neq a I_2$. Did I miss anything? $\endgroup$ – Algebraic Pavel Jun 30 '17 at 12:18
  • $\begingroup$ No. you are right. Thank you very much. $\endgroup$ – HK Lee Jun 30 '17 at 12:26
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You can already find in the comments a great counter-example, but I wanted to add a consideration regarding the special case: $m=n+1$.

Let $u_i \in \mathbb{R}^{n}$, with $i=1,\dots,n$, be orthonormal vectors. Then, matrix $A$ can be written as $$ A = \begin{bmatrix}U & v \end{bmatrix} $$ where $U$ is an orthogonal matrix, $U^TU = U U^T = I_n$, and $v$ is a unit vector. Consider now a single element of $w = Ac$, say $$ w_j = \sum_{i=1}^{n+1} a_{ji} c_i = \sum_{i=1}^{n} a_{ji}c_i + v_j c_{n+1} = 0 $$ Now, $\sum_{i=1}^{n} a_{ji}c_i$ cannot be zero, since $c_i \neq 0$ and the submatrix $U$ of $A$ has linearly independent rows (and columns). It also follows that $v_j \neq 0$. So, we can write $$ v_j = - \frac{\sum_{i=1}^{n} a_{ji}c_i}{c_{n+1}} $$ If $U$ and $C$ are given, $v$ must be picked that way to satisfy $Ac=\mathbf{0}$. Back to the original expression, we have that $$ ACA^T = [U \ v] \begin{bmatrix} D & 0_{(n \times 1)} \\ 0_{(1 \times n)} & c_{n+1} \end{bmatrix} \begin{bmatrix} U^T \\ v^T\end{bmatrix} = UDU^T + c_{n+1} vv^T $$ which is in general not equal to $a I_{n+1}$.

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