Homework Help: Find a 3x3 Matrix such that....

1. Jun 23, 2015

Jtechguy21

1. The problem statement, all variables and given/known data

Find a non zero matrix(3x3) that does not have in its range. Make sure your matrix does as it should.

3. The attempt at a solution

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 Im looking for.

2. Jun 23, 2015

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).

3. Jun 23, 2015

Ray Vickson

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 $\vec{x} \in R^3$.

4. Jun 23, 2015

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?

5. Jun 23, 2015

Ray Vickson

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.

Last edited: Jun 23, 2015
6. Jun 23, 2015

WWGD

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

7. Jun 23, 2015

Jtechguy21

Thanks for explaining that to me.

8. Jun 23, 2015

Jtechguy21

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.

9. Jun 23, 2015

Jtechguy21

thanks for your input. I had a feeling thats 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?

10. Jun 23, 2015

Ray Vickson

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?

11. Jun 23, 2015

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

Last edited: Jun 23, 2015