# Assured non-zero orthogonal projection of a vector

This question is related to the Symmetrical and skew-symmetrical part of rotation matrix

Let's suppose that we have orthogonal projection matrix $P$ onto subspace generated by a unit vector $v$.

Questions:

• What is the simplest method to construct such a vector taking entries of $P$ (without deconstructing $P$ to $vv^T$) where we would be sure that projection of this vector onto $v$ gives a non-zero vector?

(I suspect that the vector made from diagonal entries of $P$ has such property but I can't prove this ... other simple methods should for selected entries of $P$ take into account only basic arithmetic operations like summing or multiplying these entries)

• Is it also a method of finding such vectors for matrices $P$ of projections onto higher dimension subspaces (for example onto plane) ?

Edit 1
Now I know that my assumption about diagonal entries was false: ${\begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} \\ -\dfrac{1}{2} & \dfrac{1}{2} \end{bmatrix}}$

Probably there is no such a train of linear operations which can lead for randomly chosen (and from the point of view of formula unknown) that guarantees obtaining a non-zero vector. Algorithm somewhere must have inserted "if" statement. The last hope is in the absolute value operator (it is non linear) but I have not find a formula. It remains only to take any non-zero (but for this we need "if" decision) column of $P=vv^T$ (case 1 dim. subspace) and normalize it to the unit vector.

Edit 2 ( 21 hours after)

Even for a case of dimension 2 is hard to find a solution and to prove that solution is impossible in the general case. Let's say that we want to construct vector from entries of projection matrix $P$

which for $P$ generated by a vector $v=[x \ y]^T$ are $x^2, y^2, xy$

i.e.

$vv^T= \begin{bmatrix} x^2 & xy \\ xy & y^2 \end{bmatrix}$

what means that the wanted vector has a form in the most general case

$b=[Q_1(x^2, y^2, xy),Q_2(x^2, y^2, xy)]^T$

where $Q_1,Q_2$ are some polynomials.
In this case we must have assured $xQ_1+yQ_2\neq{0}$ because $v$ and $b$ should not be orthogonal.

For a limited range of possible $x,y$ values it is possible to construct such a vector. Let's assume that $x,y$ are non-negative. In this case we can construct simple vector

$b=\begin{bmatrix} x^2+1 \\ y^2+1 \end{bmatrix}$

which gives $v^Tb = x^3+x + y^3+y > 0$ for any non-negative $x,y$ so the vectors $v$ and $b$ can't be orthogonal and it is assured non-zero projection of vector $b$ onto $v$. Such procedure can be expressed by a single formula

$b=(i^TPi +1)i+(j^TPj +1)j$

however its usage is not universal, it is dependent on the assumed restricted values of $x,y$, (in this case it can be easily extended for any dimension).

(I have described above procedure not to give a solution, because for restricted values of $x,y$ to non-negative numbers even the vector $b=[1,1]^T$ would be enough but to give a hint how the wanted general solution could be presented - this was, I think, unclear here for some users, maybe now it is more explicit what I expect from the solution.)

• I am not sure if this is the simplest method, but how about solving $(P - I)x = 0$ where $I$ denotes the identity matrix? The set of solutions is the space of eigenvectors to the eigenvalue $1$. Since $P$ is a projection matrix, this is its image. – Matthias Klupsch Jul 11 '16 at 8:02
• @Matthias Klupsch Yes, it is not so simple like taking some entries of $P$ and compose with them the wanted vector. Is it my assumption about taking diagonal entries of $P$ true or false ? It has an advantage that it eliminates the case of P with zeros in some columns.. I suppose (it's rather obvious) for projection matrix it is impossible to have all zeros on diagonal.. – Widawensen Jul 11 '16 at 8:11
• No, taking diagonal entries will not work in general. An example where this does not work is $P = \left(\begin{smallmatrix} 2 & -1 \\ 2 & -1 \end{smallmatrix}\right)$. However, you are right that nonzero projection matrices have to have a nonzero diagonal entry since their trace is the dimension of their image. – Matthias Klupsch Jul 11 '16 at 8:55
• @MatthiasKlupsch Your $P$ matrix is not a projection, I suppose, because it isn't of the form $vv^T$. .. ? – Widawensen Jul 11 '16 at 9:02
• What is a projection for you? For me it's a matrix $P$ with $P^2 = I$. – Matthias Klupsch Jul 11 '16 at 9:13

For one dimensional projection the solution is easy, in fact if your matrix $P$ is zero then your vector is $v=0$, now suppose that you have a non zero entries in $P$ say $a_{i,j}$ where $(P=(a_{i,j})_{i,j\leq n})$ so if you put $$v=\frac{1}{\sqrt{\sum_{k=1}^n a_{k,i}^2}}(a_{1,i}, a_{2,i}, \dots , a_{j,i}, \dots a_{n,i})$$ you can see that $v$ is a unit vector, and that $v$ answer to your question, in fact we have a 1 dimensional projection so the dimension of $Ran(P)$ is equal to one, that mean that : $$Ran(P)=\mathbb{R}v$$ where $v$ is any nonzero vector in the range of $P$, but we know that any column in a matrix is the image of the canonical basis.
• Yes, it is true, however it is not the exact answer for the question because you have assumed that we know which column is non-zero, it means we are able to use logic of "if-then type. I constrained solution only to the train of typical linear operations like summation, multiplication, scalar product, transpose etc.. by your formula starting from $P$ matrix we, without checking what are non-zero columns, could generate also zero vector ( or rather not generate because it be would division by 0) – Widawensen Jul 11 '16 at 17:24
• @Widawensen since $\operatorname{tr}P=\|v\|^2=1$, you only need to scan the main diagonal to find a non-zero column, but that doesn’t seem to be the sort of thing you’re looking for. – amd Jul 11 '16 at 19:11
• Of course you are right Hamza, maybe it is a little non-standard question because I wanted to impose some restrictions on the method and not to use "If-then" decision when constructing a wanted vector.....example.. Let[s suppose that we would want to calculate sum of diagonal entries of P, we can use always formula $sum=i^TPi+j^TPj+k^TPk.$ - it should be something like this formula. – Widawensen Jul 11 '16 at 19:58