Proving that determinants aren't linear transformations?

[Eclair_de_XII]

Homework Statement

"Determine whether the function $T:M_{2×2}(ℝ)→ℝ$ defined by $T(A)=det(A)$ is a linear transformation.

Homework Equations

$det(A)=\sum_{i=1}^n a_{ij}C_{ij}$

The Attempt at a Solution

I'm assuming that it isn't a linear transformation because $det(A+B)≠det(A)+det(B)$. So I have here:

$det(A)=\sum_{i=1}^n a_{ij}C_{ij}$
$det(B)=\sum_{i=1}^n b_{ij}C_{ij}$

$det(A)+det(B)=\sum_{i=1}^n (a_{ij}C_{ij}+b_{ij}C_{ij})$

Meanwhile, I have...

$det(A+B)=\sum_{i=1}^n (a_ij+bij)(C_{ij}+D_{ij})$

I'm fairly certain that the equation directly above is incorrect; perhaps that of the one above that one, too.

 

[fresh_42]

What do you get for simple examples like $A=\lambda I$ and $B=-A$ before you try to prove the general case?

 

[member 587159]

Homework Statement

"Determine whether the function $T:M_{2×2}(ℝ)→ℝ$ defined by $T(A)=det(A)$ is a linear transformation.

Homework Equations

$det(A)=\sum_{i=1}^n a_{ij}C_{ij}$

The Attempt at a Solution

I'm assuming that it isn't a linear transformation because $det(A+B)≠det(A)+det(B)$. So I have here:

$det(A)=\sum_{i=1}^n a_{ij}C_{ij}$
$det(B)=\sum_{i=1}^n b_{ij}C_{ij}$

$det(A)+det(B)=\sum_{i=1}^n (a_{ij}C_{ij}+b_{ij}C_{ij})$

Meanwhile, I have...

$det(A+B)=\sum_{i=1}^n (a_ij+bij)(C_{ij}+D_{ij})$

I'm fairly certain that the equation directly above is incorrect; perhaps that of the one above that one, too.

You are correct. It is not a linear transformation because $det(A + B) \neq det(A) + det(B)$ (a counterexample is easy to find, so if you do this this is sufficient to prove the statement, there is really no need to use these definitions). Moreover, $det(\lambda A) = \lambda^n det(A)$ if $A$ is a $n \times n$ matrix, which means $det(\lambda A) \neq \lambda det(A)$ in general.

 

[Eclair_de_XII]

(a counterexample is easy to find, so if you do this this is sufficient to prove the statement, there is really no need to use these definitions)

I see. In any case, I'm just going to do the problem like this and hope that my teacher accepts it:

"Let $A=[I_n]$ and $B$ be an $n×n$ matrix such that $ent_{ij}(B)=β≠0$ iff $i=j=1$ and $ent_{ij}(B)=0$ otherwise. Then $det(A)=1$ and $det(B)=0$. However, $det(A+B)=1+β≠1=det(A)+det(B)$. Therefore, the determinant is not a linear transformation from $M_{n×n}(ℝ)→ℝ$."

 

[member 587159]

I see. In any case, I'm just going to do the problem like this and hope that my teacher accepts it:

"Let $A=[I_n]$ and $B$ be an $n×n$ matrix such that $ent_{ij}(B)=β≠0$ iff $i=j=1$ and $ent_{ij}(B)=0$ otherwise. Then $det(A)=1$ and $det(B)=0$. However, $det(A+B)=1+β≠1=det(A)+det(B)$. Therefore, the determinant is not a linear transformation from $M_{n×n}(ℝ)→ℝ$."

Counterexamples are more elegant imo, and you are sure there is no flaw in the reasoning, but your approach is fine.

 

[Eclair_de_XII]

Thanks...