MHB Show determinant operator Det not linear

[ognik]

Probably trivial, but for matrices with different ranks, Det is not closed for addition?

I think it is closed under multiplication?

So really I must show Det not closed under addition for square matrices of the same order...

$ D(A_n) = \sum_{j=1}^{n} a_{1j}C_{ij} $ and $ D(B_n) = \sum_{j=1}^{n} b_{1j}E_{ij} $, so $ D(A_n) + D(B_n) = \sum_{j=i}^{n} \left( a_{1j}C_{ij} + b_{1j}E_{ij} \right) $

Now $ D(A+B) =\sum_{j=1}^{n} \left( a_{1j} + b_{1j} \right) F_{1j} = $ (I'm not sure this bit is right either way?)

So $ D(A_n) + D(B_n) = D(A+B) $ only when A+B. Is that close?

 

[Sudharaka]

Probably trivial, but for matrices with different ranks, Det is not closed for addition?

I think it is closed under multiplication?

So really I must show Det not closed under addition for square matrices of the same order...

$ D(A_n) = \sum_{j=1}^{n} a_{1j}C_{ij} $ and $ D(B_n) = \sum_{j=1}^{n} b_{1j}E_{ij} $, so $ D(A_n) + D(B_n) = \sum_{j=i}^{n} \left( a_{1j}C_{ij} + b_{1j}E_{ij} \right) $

Now $ D(A+B) =\sum_{j=1}^{n} \left( a_{1j} + b_{1j} \right) F_{1j} = $ (I'm not sure this bit is right either way?)

So $ D(A_n) + D(B_n) = D(A+B) $ only when A+B. Is that close?

Hi ognik, :)

If the determinant operator is linear it should satisfy,

\[D(A+B)=D(A)+D(B)\]

for all matrices $A$ and $B$. So basically to show that the determinant operator is not linear you have to show a counterexample. Consider the matrices,

\[A=\begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}\mbox{ and }B=\begin{pmatrix}1 & 1 \\1 & 1\end{pmatrix}\]

and calculate $D(A+B)$ and $D(A)+D(B)$.

 

[ognik]

Thanks, that - and no doubt others - do the job.

But I was trying to use index notation and a generalised counter example - to practice index notation, would you (or anyone) mind critiquing my attempt?

 

[Sudharaka]

Thanks, that - and no doubt others - do the job.

But I was trying to use index notation and a generalised counter example - to practice index notation, would you (or anyone) mind critiquing my attempt?

What do you mean by a generalized counter example? It would still have to be a counter example isn't? And that means whatever approach you use you'll have to substitute specific values for $A$ and $B$.

 

[ognik]

Well, I've tried to use the cofactor definition and tried to show that the only way they are equal is if the cofactors $C_{ij} = F_{ij} = E_{ij} $ and that can't be because we started with different C and E so I guess its more a (forgotten the word) than a counter example ...and I called it generalised because it's not using specific numbers, so my terminology may be wrong?

 

[Sudharaka]

Well, I've tried to use the cofactor definition and tried to show that the only way they are equal is if the cofactors $C_{ij} = F_{ij} = E_{ij} $ and that can't be because we started with different C and E so I guess its more a (forgotten the word) than a counter example ...and I called it generalised because it's not using specific numbers, so my terminology may be wrong?

But $A$ and $B$ doesn't necessarily have to be different. For example a possible counter example is,

$A=B=\begin{pmatrix}1&0\\0&1\end{pmatrix}$

So even if you did prove that somehow you'll have to account for the case where $A$ and $B$ are equal.

 

[ognik]

Must.learn.to.use.counter.examples...:-)