# I got some problems with ker and dim in matrices

• transgalactic
Thank you. In summary, solving a matrix without any parameters involves making equations out of each line and solving for the values of variables that will satisfy all the equations. In the case of parameters, we look for values that will make the rows all zeros. The solution to the matrix represents the vectors of the kernel and the values of variables that will satisfy the equations represent the image of the matrix. The maximal values of the dimensions of the kernel and image cannot be determined without solving the equations.
transgalactic
http://img505.imageshack.us/my.php?image=52555095wf0.jpg

ker:
this is how we solve a question without any parameter in the matrix:

we take each line make and equation out of it and put =0 in the end of each line
than we find the solution to this matrix
(we find the value for X value for Y value for Z of course there might be more than one vector)
and the solution represents the vectors of the ker

i asked previosly what should i do in case of parameters
the answer that i got was
"find the values of parameters for which the rows will be all zeros"

why arent we solving it?

regarding the Im:

we put =b1 =b2 =b3 in the end of each line
and solve
i don't know how

i got here a simple parameter matrix can u please write the full solution
of maximal value of dim(ker A) and the maximal value of dim(Im A)

i simplified the matrix from the original form to the simplest form

i heard many version many explanations to this sort of problem
can you write an answer to this ?

Last edited:

Hello ker,

Thank you for your forum post. I would like to provide some insights and clarification on how to solve a matrix without any parameters.

Firstly, to solve a matrix without any parameters, we follow the steps you mentioned in your post. We take each line and make an equation out of it, and then add =0 at the end of each line. This is because we are trying to find the values of variables (in this case, X, Y, and Z) that will satisfy all the equations, hence making the matrix equal to zero.

Next, we solve the equations to find the values of X, Y, and Z. This will give us the solution to the matrix, which represents the vectors of the kernel (ker). As you mentioned, there may be more than one vector that satisfies the equations, so we can have multiple solutions for the kernel.

Now, let's discuss what to do in case of parameters in the matrix. In this scenario, we are dealing with a system of equations with variables and parameters. To find the kernel, we need to find the values of parameters for which the rows will be all zeros. This means that we are looking for the values of parameters that will make the equations redundant, resulting in a matrix with only zero values. This will give us the solution to the kernel.

Moving on to the Im (image) of the matrix, we use a similar approach as finding the kernel. We put =b1 =b2 =b3 at the end of each line, as we are trying to find the values of variables that will satisfy the equations and make the matrix equal to the given values of b1, b2, and b3. Again, we solve the equations to find the values of variables, which will give us the solution to the image of the matrix.

Finally, you asked for the maximal value of dim(ker A) and dim(Im A). This refers to the dimension of the kernel and the image of the matrix, respectively. The maximal value of the dimension of the kernel is the number of free variables in the solution to the kernel, and the maximal value of the dimension of the image is the number of pivot columns in the matrix. These values will depend on the specific matrix and cannot be determined without solving the equations.

I hope this helps clarify the process of solving a matrix without parameters. If you have any further questions or need clarification, please let me know.