# Linear Algebra - Find matrix A such that Ax is perpendicular to [1 2 3] for all x

1. Feb 12, 2012

### skyflashings

Find a nonzero 3x3 matrix A such that Ax is perpendicular to [1,2,3] for all x in R3

A vector perpendicular to [1, 2, 3] could be, say, [-2, 1, 0]. But what would be a general approach to finding a solution that would satisfy all x? Thanks!

2. Feb 12, 2012

### xaos

what definition of perpendicular are you using?

3. Feb 12, 2012

### SammyS

Staff Emeritus
[-2, 1, 0] is perpendicular to [1, 2, 3]. However there are many others which are perpendicular to [1, 2, 3] and are also linearly independent of [-2, 1, 0] .

I suppose you need to take the scalar product of Ax and [-2, 1, 0] and set it to zero.

4. Feb 13, 2012

### Ray Vickson

If $\vec{x} = (x_1, x_2, \ldots, x_n)^T$ and A has columns $\vec{v}_1, \ldots, \vec{v}_n,$ then
$$A \vec{x} = x_1 \vec{v}_1 + x_2 \vec{v}_2 + \cdots + x_n \vec{v}_n.$$ Since you want this to be perpendicular to [1,2,3] for all x, all the vectors $\vec{v}_i$ must be perpendicular to [1,2,3].

PS: Does anyone know how to get the "\boldmath" command to work in the version of tex used here?

RGV

5. Feb 13, 2012

### Deveno

i don't think such an A can be found whose determinant is non-zero.

therefore, one may choose some of the columns of A to be 0 columns, which makes the problem rather simple.

6. Feb 13, 2012

### lanedance

$$\textbf{A}$$
does \textbf{} help?

7. Feb 13, 2012

### lanedance

the set of vectors perpindicular to (1,2,3)^T will form a plane through the origin, in fact they form a vector subspace

If you could find an operator that maps any vector onto that plane, then you would have a suitable operator...

8. Feb 13, 2012

### Ray Vickson

Let's try it: $\textbf{A}.$ So, yes, it works. Thank you.

RGV