Is this vector in the image of the matrix?

[arhzz]

Homework Statement

Find the vectors in the image of the matrix

Relevant Equations

-

Hello!

I have this system here $$ \left[ \begin{matrix} -2 & 4 & \\\ 1 & -2 & {} \end{matrix} \right]x +\begin{pmatrix} 2 \\\ y \end{pmatrix}u $$ Now although the problem is for my control theory class,the background is completely math(as is 90% of control theory)

Basically what I need to do is to find the image of this matrix,and than chose y so that the u vector is in the image of the matrix.

So since the vectors in the matrix are linearly dependent we can choose either. so $$ im(A) = \begin{pmatrix} -2 \\\ 1 \end{pmatrix} $$

Now I need to chose y. Now 1 would be the obvious choice but the actual problem is the -2. Since the 2 in the vector is already given to me I cannot change it. And here is my question; let's say I choose y to be -1 that would give me the vector (2 -1),that is not in the image but if I multiply that vector by -1 I get the image vector.So my question is am I allowed to do that? Can i take multiples of a vector for them to be still in my image? If not is it possible that there is no way to choose y for the vector to be in the image of the matrix?

Thanks!

 

[Orodruin]

The image is an entire subspace (in essence, the linear hull of the columns), not a single vector. If it were a single vector you would not be allowed to pick one of two arbitrarily!

 

[arhzz]

The image is an entire subspace (in essence, the linear hull of the columns), not a single vector. If it were a single vector you would not be allowed to pick one of two arbitrarily!

Your definately right, I am not really good at the exact (mathematical) background of these (Engineering student).

 

[Mark44]

I have this system here $$ \left[ \begin{matrix} -2 & 4 & \\\ 1 & -2 & {} \end{matrix} \right]x +\begin{pmatrix} 2 \\\ y \end{pmatrix}u $$

I'm confused by your question. A "system" is typically a system of equations, and such a system can usually be written as a single matrix equation. Should the '+' in your expression above have been '='? That would have been an easy typo to make as both these characters are on the same keyboard key.

Now although the problem is for my control theory class,the background is completely math(as is 90% of control theory)

Basically what I need to do is to find the image of this matrix,and than chose y so that the u vector is in the image of the matrix.

So since the vectors in the matrix are linearly dependent we can choose either. so $$ im(A) = \begin{pmatrix} -2 \\\ 1 \end{pmatrix} $$

Is A the 2x2 matrix you showed? If so, the fact that the columns of the matrix are linearly dependent means that the nullspace does not consist solely of the zero vector. (The nullspace, or kernel, is the subspace of vectors $\vec x$ in $\mathbb R^2$ such that $A\vec x = \vec 0$.

Now I need to chose y. Now 1 would be the obvious choice but the actual problem is the -2. Since the 2 in the vector is already given to me I cannot change it. And here is my question; let's say I choose y to be -1 that would give me the vector (2 -1),that is not in the image but if I multiply that vector by -1 I get the image vector.So my question is am I allowed to do that? Can i take multiples of a vector for them to be still in my image? If not is it possible that there is no way to choose y for the vector to be in the image of the matrix?

Thanks!

 

[arhzz]

I'm confused by your question. A "system" is typically a system of equations, and such a system can usually be written as a single matrix equation. Should the '+' in your expression above have been '='? That would have been an easy typo to make as both these characters are on the same keyboard key.

Is A the 2x2 matrix you showed? If so, the fact that the columns of the matrix are linearly dependent means that the nullspace does not consist solely of the zero vector. (The nullspace, or kernel, is the subspace of vectors $\vec x$ in $\mathbb R^2$ such that ##A\vec x = \vec

By system is meant that the controller is described by the matrix and the vector, it is not the general sense that it is a system of linear equations, but how we can model the controller so that we can "use it". And you are right about the null space. Since the exam is coming up shortly we had a tutor show us how this is done. I did not actually write the entire problem down correctly, and hence missed a step completely which made the problem of the - sign infront of the 2. I am currently on my phone when I get home I will post the solution and where the error has come up. Thanks for the help !