Find a 3x3 Matrix such that....

[Jtechguy21]

Homework Statement

Find a non zero matrix(3x3) that does not have

in its range. Make sure your matrix does as it should.

The Attempt at a Solution

[/B]
I know a range is a set of output vectors, Can anyone help me clarify the question?
I'm just not sure specifically what its asking of me, in regards to the matrix I am looking for.

 

[HallsofIvy]

You are asked to find a 3 by 3 matrix, A, such that Ax is NOT equal to \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} for any x. This does not have a single correct aswer. The trivial solution, the zero matrix, would be such a matrix because the zero matrix times any x is the 0 vector- but that is specifically excluded- you must give a "non-zero" matrix. There are still an infinite number of such matrices. What about a matrix that makes the first component of any vector 0? Or the second or third component? (But not all three).

 

[Ray Vickson]

Homework Statement

Find a non zero matrix(3x3) that does not have

in its range. Make sure your matrix does as it should.

The Attempt at a Solution

[/B]
I know a range is a set of output vectors, Can anyone help me clarify the question?
I'm just not sure specifically what its asking of me, in regards to the matrix I am looking for.

You are being asked to find a $3 \times 3$ matrix $A$ for which the vector $\vec{r} = \langle 1,2,3 \rangle^T$ (transpose) is an impossible output; that is, you want$A \vec{x} \neq \vec{r}$ for ALL input vectors $\vec{x} \in R^3$.

 

[Jtechguy21]

We haven't learned about tranpose yet in class(but i just briefly looked it up right now)

What is the purpose of taking the transpose of the the column vector 1 2 3 in this case?

So I find this A= 3x3 matrix that when I multiply it by any column vector(x,y,z) the ouput does not equal r<1,2,3>T is this correct?

 

[Ray Vickson]

We haven't learned about tranpose yet in class(but i just briefly looked it up right now)

What is the purpose of taking the transpose of the the column vector 1 2 3 in this case?

So I find this A= 3x3 matrix that when I multiply it by any column vector(x,y,z) the ouput does not equal r<1,2,3>T is this correct?

Is that not what I just said?

Anyway, writing a transpose simplifies typing. Instead of taking 3 lines to write the column vector like this:
\left[ \begin{array}{c}1\\2\\3 \end{array} \right]
we can write it in one line, like this: $(1,2,3)^T$ or $\langle 1,2,3 \rangle^T$, as the transpose of a row-vector.

 

[WWGD]

You can use rank-nullity: just use a matrix that "kills" the vector $(1,2,3)^T$, say.

 

[Jtechguy21]

Is that not what I just said?

Anyway, writing a transpose simplifies typing. Instead of taking 3 lines to write the column vector like this:
\left[ \begin{array}{c}1\\2\\3 \end{array} \right]
we can write it in one line, like this: $(1,2,3)^T$ or $\langle 1,2,3 \rangle^T$, as the transpose of a row-vector.

Thanks for explaining that to me.

 

[Jtechguy21]

You can use rank-nullity: just use a matrix that "kills" the vector $(1,2,3)^T$, say.

Sounds interesting. Unfortunately we have not covered this yet(2nd week in) but I am watching a youtube videos to see how this "Killing" things works.

 

[Jtechguy21]

You are asked to find a 3 by 3 matrix, A, such that Ax is NOT equal to \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} for any x. This does not have a single correct aswer. The trivial solution, the zero matrix, would be such a matrix because the zero matrix times any x is the 0 vector- but that is specifically excluded- you must give a "non-zero" matrix. There are still an infinite number of such matrices. What about a matrix that makes the first component of any vector 0? Or the second or third component? (But not all three).

thanks for your input. I had a feeling that's exactly what it meant, but sometimes i doubt my self with these kind of things. and when I was trying to understand the problem I figured there was more than one answer(you confirmed it)

When you refer to x, does that mean the column vector with x,y,z (any x ,y,z, values) which when multiplied by Matrix A does not equal specified range?

 

[Ray Vickson]

Sounds interesting. Unfortunately we have not covered this yet(2nd week in) but I am watching a youtube videos to see how this "Killing" things works.

I don't know if you have had the following material yet, but in 2d or 3d it is "obvious" from diagrams: given a fixed vector $\vec{u}$, any vector $\vec{v}$ can be decomposed into a component $\vec{v}_{||}$ parallel to $\vec{u}$ and a component $\vec{v}_{\perp}$ perpendicular to $\vec{u}$. (Think of decomposing a force in Physics into components parallel and perpendicular to some given direction.)

What would happen if, for every vector $\vec{v}$ you took the perpendicular component $\vec{v}_{\perp}$? Is the operation $\vec{v} \rightarrow \vec{v}_{\perp}$ a linear operation?

 

[WWGD]

My idea is this: you can use a matrix whose kernel is either 1- , 2- , or 3- dimensional ( if the kernel is 0-dimensional, the matrix is invertible, so $Ax=b$ has a solution for any b, including, in particular, $(1,2,3)^T$. If the kernel is 3-dimensional, you get the $0$-matrix ). So include , in each case, $(1,2,3)^T$ as a basis vector in the kernel of a matrix. You can also use the fundamental theorem of linear algebra:

https://en.wikipedia.org/wiki/Fundamental_theorem_of_linear_algebra