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

[JD_PM]

TL;DR

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 $X$.

 

[PeroK]

A linear map on $\mathbb{R}^n$ is surjective iff it is injective!

 

[JD_PM]

Oh so if I find the values for which $L$ is injective then I am done.

 

[PeroK]

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.

 

[JD_PM]

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.

 

[PeroK]

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.

 

[JD_PM]

Alright so after some thought I only came up with the brute method

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...

 

[PeroK]

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.

 

[JD_PM]

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.

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? :)

 

[PeroK]

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.

 

[PeroK]

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 $a^3 \ b^3 \ c^3 \ 0 \ 0 \ 0$.