Finding a Nonzero 3x3 Matrix A Such That Ax is Perpendicular to [1,2,3]

[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!

 

[xaos]

what definition of perpendicular are you using?

 

[SammyS]

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, 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.

 

[Ray Vickson]

[-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.

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

 

[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.

 

[lanedance]

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

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

 

[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...

 

[Ray Vickson]

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

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

RGV