# Is this vector in the image of the matrix?

• arhzz
In summary: We were given an exercise to solve, in summary, the problem involves finding the image of a given matrix and choosing a value for y so that the u vector is in the image. Since the vectors in the matrix are linearly dependent, we can choose either one. However, the vector (2,y) is not in the image. The question is whether we can take multiples of a vector to add it to the image, and if not, is it possible that there is no way to choose y for the vector to be in the image?
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!

Last edited:
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!

Delta2
Orodruin said:
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).

Delta2
arhzz said:
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.
arhzz said:
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##.
arhzz said:
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!

Delta2 and FactChecker
Mark44 said:
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 !

## What is the definition of "image of a matrix"?

The image of a matrix is the set of all possible outputs that can be obtained by multiplying the matrix by any valid input vector. It represents the range of the linear transformation defined by the matrix.

## How can I determine if a vector is in the image of a matrix?

To determine if a vector is in the image of a matrix, you can multiply the vector by the matrix and see if the resulting vector is a valid output. If it is, then the original vector is in the image of the matrix.

## What does it mean if a vector is not in the image of a matrix?

If a vector is not in the image of a matrix, it means that the vector cannot be obtained as an output of the linear transformation defined by the matrix. This could be due to various reasons such as the vector being outside the range of the transformation or the matrix not being able to map the vector to a valid output.

## Can a matrix have multiple images?

No, a matrix can only have one image. The image of a matrix is unique and represents the entire range of the linear transformation defined by the matrix. However, different matrices can have the same image if they define the same linear transformation.

## How can I use the image of a matrix in practical applications?

The image of a matrix can be used in various applications such as data compression, image processing, and machine learning. It helps in understanding the range of a linear transformation and can be used to solve systems of linear equations and find solutions to optimization problems.

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