Find the parameter values that make ##L## surjective

In summary, Based on the given matrix representation of a linear mapping, we are asked to find the values of ##a,b,c## that make the linear map surjective. To do so, we can use the theorem that states a linear map is surjective if and only if it is injective, which is equivalent to being invertible with a non-zero determinant. Therefore, we need to find the values of ##a,b,c## that make the determinant of the matrix non-zero. By observing the symmetry of the matrix, we can see that if any two of ##a,b,c## are equal, the determinant will be zero. To simplify the problem, we can reorganize the determinant into a block form with a 3 x
  • #1
JD_PM
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TL;DR Summary
I am wondering if there is a theorem to check whether a linear map is surjective or not in function of the value of the parameters.
I am given the following matrix representation of a linear mapping

$$L_{\alpha}^{\beta}=
\begin{pmatrix}
1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 & 0 & 1 \\
0 & a & 0 & b & 0 & c \\
a & 0 & b & 0 & c & 0 \\
0 & a^3 & 0 & b^3 & 0 & c^3 \\
a^3 & 0 & b^3 & 0 & c^3 & 0 \\
\end{pmatrix}
$$

Where ##a,b,c \in \Re##

And I am asked to find the values that make L surjective.

Well, if the question was about what values of ##a,b,c## make ##L## injective I'd go for the rank-nullity theorem and see for what values I get that the nullity of ##L## is zero (as there's a theorem stating that if the nullity of ##L = 0 \rightarrow L## is injective).

My question is: is there an analogous theorem to check surjectivity?

I am aware of the basic definition of a surjective map: L is surjective if for every element ##y## in the codomain ##Y## there is at least one element ##x## in the domain ##X##.
 
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  • #2
A linear map on ##\mathbb{R}^n## is surjective iff it is injective!
 
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  • #3
Oh so if I find the values for which ##L## is injective then I am done.
 
  • #4
JD_PM said:
Oh so if I find the values for which ##L## is injective then I am done.
Yes, surjective iff injective iff invertible iff non-zero determinant.
 
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  • #5
Alright so it is simply about getting for what values of ##a,b, c## the determinant (which is a monster btw) is zero.

Thus we know that for those values the linear map is not injective and for the rest of values it is.

Thus, based on the theorem you stated, for those values the linear map is not surjective and for the rest of values it is (please correct me if I am wrong).

I found a proof of the theorem you stated (which relies on other theorems):

Let's assume that ##L## is injective. There is a theorem stating that if a linear map is injective then ##ker(L) =[0]##. This implies that the ##dim(kerL)=0##. Based on the dimension theorem (##dim(V)=dim(kerL)+ dim(ImL)##) we get that ##dim(V) = dim(ImL)##. Once here we know that there is a theorem that states ##dim(U)=dim(V) \iff U=V##. Thus we get ##Im(L) = V##, which is the definition of surjectivity.

QED.
 
  • #6
JD_PM said:
Alright so it is simply about getting for what values of ##a,b, c## the determinant (which is a monster btw) is zero.

The monster can be tamed easily enough! Focus on the symmetry.
 
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  • #7
Alright so after some thought I only came up with the brute method

FullSizeRender (51).jpg

FullSizeRender (52).jpg

FullSizeRender (53).jpg


Mmm yikes! Am I now supposed to check all parameter values for which this 20-terms equation becomes zero? (the only apparent case is when ##a=b=c=0##)

I guess there's a much neater way of solving this...
 
  • #8
First, you can see by symmetry of the original matrix that if any two of ##a,b,c## are equal then the determinant is zero.

I reorganised the determinant into a block form, with a 3 x 3 matrix in the top left and the same 3 x 3 matrix in the bottom right, and everything else zero. That reduces the problem to a 3 x 3 determinant.
 
  • #9
PeroK said:
First, you can see by symmetry of the original matrix that if any two of ##a,b,c## are equal then the determinant is zero.

Mmm why? I am afraid I do no see it... I guess that it has to do with a property of the determinant; I know that if any row/column of the determinant is zero then the determinant is zero.

PeroK said:
I reorganised the determinant into a block form, with a 3 x 3 matrix in the top left and the same 3 x 3 matrix in the bottom right, and everything else zero. That reduces the problem to a 3 x 3 determinant.

Could you please show/explain how to do so? :)
 
  • #10
JD_PM said:
Mmm why? I am afraid I do no see it... I guess that it has to do with a property of the determinant; I know that if any row/column of the determinant is zero then the determinant is zero.

If ##a = b##, then the second and fourth columns are equal. By a simple column operation of subtracting one from the other you get a zero column.
 
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  • #11
JD_PM said:
Could you please show/explain how to do so? :)

1) Swap the columns until the first row is ##1 \ 1 \ 1 \ 0 \ 0 \ 0##.

2) Swap the rows until the second row is ##a \ b \ c \ 0 \ 0 \ 0## and the third row is ##a^3 \ b^3 \ c^3 \ 0 \ 0 \ 0##.
 
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What does it mean for a function to be surjective?

A function is surjective if every element in the codomain has at least one preimage in the domain. In other words, every element in the output has at least one corresponding input.

How do you find the parameter values that make a function surjective?

To find the parameter values that make a function surjective, you need to set up an equation or system of equations and solve for the values that satisfy the condition of surjectivity. This may involve using algebraic manipulation, substitution, or graphing techniques.

Why is it important to find the parameter values that make a function surjective?

It is important to find the parameter values that make a function surjective because it ensures that every element in the codomain has at least one corresponding input. This helps to guarantee that the function is well-defined and has a meaningful output for every input.

Can a function be surjective for some parameter values but not others?

Yes, a function can be surjective for some parameter values but not others. This depends on the specific behavior of the function and how the parameter values affect its output. It is possible for a function to be surjective for some values and not for others, or for a function to be surjective for all values of the parameter.

Are there any techniques or strategies for finding the parameter values that make a function surjective?

Yes, there are various techniques and strategies for finding the parameter values that make a function surjective. These may include using algebraic methods such as solving equations or systems of equations, using graphing techniques to analyze the behavior of the function, or using specific properties of the function to determine the necessary parameter values.

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