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

In summary, the conversation discusses finding a nonzero 3x3 matrix A that satisfies the condition Ax is perpendicular to [1,2,3] for all x in R3. It is suggested to take the scalar product of Ax and [-2, 1, 0] and set it to zero, and to find an operator that maps any vector onto the plane formed by vectors perpendicular to [1,2,3].
  • #1
skyflashings
15
0
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!
 
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  • #2
what definition of perpendicular are you using?
 
  • #3
skyflashings said:
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.
 
  • #4
SammyS said:
[-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 [itex] \vec{x} = (x_1, x_2, \ldots, x_n)^T [/itex] and A has columns [itex] \vec{v}_1, \ldots, \vec{v}_n, [/itex] then
[tex] A \vec{x} = x_1 \vec{v}_1 + x_2 \vec{v}_2 + \cdots + x_n \vec{v}_n. [/tex] Since you want this to be perpendicular to [1,2,3] for all x, all the vectors [itex]\vec{v}_i[/itex] 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
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
Ray Vickson said:
PS: Does anyone know how to get the "\boldmath" command to work in the version of tex used here?

[tex] \textbf{A}[/tex]
does \textbf{} help?
 
  • #7
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
lanedance said:
[tex] \textbf{A}[/tex]
does \textbf{} help?

Let's try it: [itex] \textbf{A}.[/itex] So, yes, it works. Thank you.

RGV
 

1. What is a nonzero 3x3 matrix?

A nonzero 3x3 matrix is a matrix with 3 rows and 3 columns, where at least one of the elements is not equal to zero. It is typically denoted as A or [A].

2. How do you determine if a matrix is perpendicular to a vector?

To determine if a matrix A is perpendicular to a vector [x,y,z], we can use the dot product. If the dot product of A and [x,y,z] is equal to 0, then A is perpendicular to [x,y,z].

3. What does it mean for a matrix to be perpendicular to a vector?

When a matrix A is perpendicular to a vector [x,y,z], it means that the columns of A are orthogonal to the vector [x,y,z]. This means that the dot product of each column of A with [x,y,z] is equal to 0.

4. Can a matrix be perpendicular to more than one vector?

Yes, a matrix can be perpendicular to more than one vector. This means that the columns of the matrix are orthogonal to each of the vectors. However, there may not be a unique matrix that is perpendicular to multiple vectors.

5. How can I find a nonzero 3x3 matrix A such that Ax is perpendicular to [1,2,3]?

To find a nonzero 3x3 matrix A such that Ax is perpendicular to [1,2,3], we can use the dot product condition mentioned in question 2. We can set up a system of equations and solve for the elements of A. Alternatively, we can use matrix operations to manipulate a matrix that is already known to be perpendicular to [1,2,3] and create a new nonzero matrix A that also satisfies the condition.

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