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If $M\in S_n(\mathbb C)$ is a symmetric matrix of rank two then it can be decomposed as the sum of two rank one matrices

$ M=uv^T+vu^T,\text{ with } u,v\in\mathbb C^n. $

For given $M$ how do I compute $u,v$ ?

Cellular Sheaf
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    I am not sure at all we can obtain such a decomposition. Don't you think it is $M=uu^T+vv^T$ ? – Jean Marie May 19 '17 at 13:59
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    https://math.stackexchange.com/a/425412/35416 uses such a decomposition in step 4 of the instructions, for the specific case of $n=3$. There the adjunct can be used as a hint for the antisymmetric matrix needed to compute what would be $2uv^T$ in your notation. – MvG Jun 26 '17 at 00:43
  • So instead of think of unlabeled points as in my case, you are talking about degenerate quadrics that are products of two hyperplanes (lines in your case)? Does that mean that any symmetric matrix of rank two represents a quadric which is a product of two hyperplanes? And basically I am searching for these? – Cellular Sheaf Jun 27 '17 at 15:48

2 Answers2

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Disclaimer: As @MvG has rightfully remarked under the form of a comment situated below, the property was asked in a first version for real valued entries. He has given a case where such a decomposition can exist for complex entries.

I think it is still of interest to see that the property is false for matrices with real entries. Here is a counterexample.

Let $M$ be the $n \times n$ matrix with entries $0$ except $M_{11}=M_{22}=1$.

It is a symmetrical rank 2 matrix.

Let us assume that $M$ has a decomposition as given in the question. Its upper left $2 \times 2$ block, call it $N$, will be:

$$N:=\begin{pmatrix}a\\b\end{pmatrix}\begin{pmatrix}c & d\end{pmatrix}+\begin{pmatrix}c\\d\end{pmatrix}\begin{pmatrix}a & b\end{pmatrix}=\begin{pmatrix}2ac & (ad+bc)\\ (ad+bc) & 2bd\end{pmatrix}$$

($a,b$ and $c,d$ are the first coeff. of $u$ and $v$ resp.). Thus

$$\det(N)=4abcd-(ad+bc)^2=-(ad-bc)^2 \leq 0$$

But the upper-left $2 \times 2$ block of $M$ is $I_2$ with $\det(I_2)=1$.

Contradiction.

Jean Marie
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  • Sorry I forgot to include that it is required of M to have rank two. Then the above condition is true and X and Y are the row and column span of M – Cellular Sheaf May 20 '17 at 15:30
  • I don't understand your remark: you had already said that $M$ has rank two. – Jean Marie May 20 '17 at 18:26
  • In your example $4abcd$ is not zero, but one. Thus the determinant is not zero. Or am I mistaken. – Cellular Sheaf May 21 '17 at 12:11
  • I don't unerstand why $4abcd-(ad+bc)^2=-(ad+bc)^2$. You example states that $N=I_2$ then $2ac$ and $2bd$ are equal to one, hence $4abcd=1$. – Cellular Sheaf May 21 '17 at 12:29
  • No, it is $4abcd-(ad+bc)^2=-(ad\color{red}{ - }bc)^2$ with a minus sign. – Jean Marie May 21 '17 at 12:44
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    Ah okay. my bad so it's only true in algebraic closed field. Which is fine with me. Any how to obtain the decomposition. – Cellular Sheaf May 21 '17 at 13:04
  • @Cellular, isn't this already a convincing proof that asking for a decomposition is unreasonable in general? – J. M. ain't a mathematician May 21 '17 at 13:08
  • There is actually a geometric construction behind that question. Consider two points $u,v$ in projective space $\mathbb P^{n-1}$. However if you forget their labeling, then this configuration of two unlabeled points is a point on the chow variety $Sym_2(\mathbb P^{n-1})$. For this simple case the chow variety defined by the 3-minors of $n \times n$ a generic symmetric matrix, these are all $n\times n$ symmetric matrices of rank two and smaller. It is obvious to go from two labeled points to their unlabeled point on the chow variety. Then $M=uv^T+vu^T$. However I am unsure how to go back – Cellular Sheaf May 21 '17 at 20:16
  • I have some faint notions about these kind of varieties, but not enough... We can agree on the fact that a matrix of the form $uv^T+vu^T$ is a rank 2 matrix, but that the converse is not true : a rank 2 matrix is not necessarily of this form. – Jean Marie May 21 '17 at 20:28
  • Yes, over the real this holds true – Cellular Sheaf May 22 '17 at 11:37
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    Your answer does not seem to apply to the question as it currently stands, i.e. formulated over complex numbers, since $u^T=\frac1{\sqrt2}(1,i)$ and $v^T=\frac1{\sqrt2}(1,-i)$ lead to $uv^T+vu^T=I_2$. It took me some reading through the comments and the history of the question to realize that it was originally formulated over the reals. While I'm no fan of questions rendering existing answers as invalid, I do think the question as it is now is more interesting. Would you mind adding a disclaimer to your answer to point out what you're actually answering? – MvG Jun 26 '17 at 00:54
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Going off the comment by @Cellular Sheaf referencing degenerate quadrics, I'm going to answer the question in the context of finding a pair of hyperplanes that represent solutions to the symmetric quadratic form $x^TMx=0$.

For the case of real symmetric $M$, there exists an eigendecomposition $M = Q^TDQ$ where is $Q$ is orthogonal consisting of the eigenvectors of $M$ and $D$ is the diagonal matrix of eigenvalues. For vectors $x$ and $y$ related by $y=Qx$, it can be seen that $x^TMx = y^TDy$. Since $M$ has rank 2, $D$ only has two nonzero elements which we can write as $\lambda_1$ and $\lambda_2$ leading to the following: $$y^T D y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}^T \begin{bmatrix} \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix} & \mathbf{0}_{2 \times (n-2)} \\ \mathbf{0}_{(n-2) \times 2} & \mathbf{0}_{(n-2) \times (n-2)} \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} = 0$$ $$\lambda_1y_1^2 + \lambda_2y_2^2=0$$ $$y_1 = \pm\sqrt{\frac{-\lambda_2}{\lambda_1}}y_2$$ $$y_1 \mp \sqrt{\frac{-\lambda_2}{\lambda_1}}y_2 = 0$$ Since $y_1$ and $y_2$ are just the linear products of $x$ and a row of $Q$, the last equation gives the two hyperplanes we were seeking. If $M$ is indefinite, $\lambda_1$ and $\lambda_2$ have opposite signs, so the square root will be real and we would have two real hyperplanes. Otherwise, we would have two complex hyperplanes.

For the case of symmetric complex matrices, $M$ is no longer necessarily diagonalizable, so the eigendecomposition approach from before is not guaranteed to lead to a linear constraint (likely will result in a quadric). If we assume the specific case that $M$ is diagonalizable such that $M=Z^TDZ$ for complex orthogonal $Z$ (i.e. $Z^TZ=I$) and diagonal matrix $D$, then we can once again recover the two complex hyperplanes using the same approach as before.

Crystal Math
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