Ma...tri...ces.. (1 Viewer)

leehuan · Apr 13, 2016

I actually gave up on this question because I still couldn't see anything but looking at it again...

leehuan said:
${ \left[ A\left( BC \right) \right] }_{ ij }=\sum _{ k=1 }^{ n }{ { \left[ A \right] }_{ ik }{ \left[ BC \right] }_{ kj } } =\sum _{ k=1 }^{ n }{ { \left[ A \right] }_{ ik }\left( \sum _{ l=1 }^{ p }{ { \left[ B \right] }_{ kl }{ \left[C\right] }_{ lj } } \right) }$

${ \left[ \left( AB \right) C \right] }_{ ij }=\sum _{ l=1 }^{ p }{ { \left[ AB \right] }_{ il }{ \left[ C \right] }_{ lj } } =\sum _{ l=1 }^{ p }{ \left( \sum _{ k=1 }^{ n }{ { \left[ A \right] }_{ ik } } { \left[ B \right] }_{ kl } \right) { \left[ C \right] }_{ lj } }$

I'm not sure if this is allowed, hence why I'm asking... but would I obtain my required result by just changing the order of summation (with the commutative law of multiplication in R)...?

InteGrand · Apr 13, 2016

leehuan said:
I actually gave up on this question because I still couldn't see anything but looking at it again...

${ \left[ \left( AB \right) C \right] }_{ ij }=\sum _{ l=1 }^{ p }{ { \left[ AB \right] }_{ il }{ \left[ C \right] }_{ lj } } =\sum _{ l=1 }^{ p }{ \left( \sum _{ k=1 }^{ n }{ { \left[ A \right] }_{ ik } } { \left[ B \right] }_{ kl } \right) { \left[ C \right] }_{ lj } }$

I'm not sure if this is allowed, hence why I'm asking... but would I obtain my required result by just changing the order of summation (with the commutative law of multiplication in R)...?

$\noindent Yes the order of summation can essentially be changed due to the laws of arithmetic in $\mathbb{R}$ (or $\mathbb{C}$). Here's a general result: if $a_1,\ldots a_n$ are real numbers and $b_1,\ldots, b_m$ are real numbers (also works for complex numbers), then $\sum _{i=1} ^n \sum _{j=1} ^{m} a_i b_j = \sum _{j=1}^{m} \sum _{i=1} ^n a_i b_j$. In fact, both are just equal to $\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)$. To see this, note that$

$\begin{align*}\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right) &= \left(a_1 + a_2 + \cdots + a_n\right)\left( \sum _{j=1} ^{m}b_j\right) \\ &= a_1 \left( \sum _{j=1} ^{m}b_j\right) + a_2 \left( \sum _{j=1} ^{m}b_j\right) + \cdots + a_n \left( \sum _{j=1} ^{m}b_j\right) \quad (\text{distributive law}) \\ &= \sum _{i=1}^{n} a_i\left( \sum _{j=1} ^{m}b_j\right) \\ &= \sum _{i=1}^{n} \left( \sum _{j=1} ^{m} a_i b_j\right) \quad (\text{the }a_i \text{ can be moved inside, by distributivity}) \\ &\equiv \sum _{i=1}^{n} \sum _{j=1} ^{m} a_i b_j .\end{align*}$

$\noindent In a similar way, we can show that $\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)= \sum _{j=1} ^{m} \sum _{i=1} ^n a_i b_j$. To do this, expand the $b$ sum instead of the $a$ sum that we did above. Alternatively, note that $\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)=\left( \sum _{j=1} ^{m}b_j\right) \left(\sum _{i=1} ^n a_i \right) $ by commutativity, and then by the result proved above, $\left( \sum _{j=1} ^{m}b_j\right) \left(\sum _{i=1} ^n a_i \right)=\sum _{j=1} ^{m} \sum _{i=1} ^n a_i b_j$, so $\sum _{j=1} ^{m} \sum _{i=1} ^n a_i b_j=\left(\sum _{i=1} ^n a_i \right) \left( \sum _{j=1} ^{m}b_j\right)$ too.$

laters · Apr 19, 2016

If AB=BA (A, B square) does that necessitate that AB=BA=I? As in, if given AB=BA can I deduce that A and B are inverses?

InteGrand · Apr 19, 2016

laters said:
If AB=BA (A, B square) does that necessitate that AB=BA=I? As in, if given AB=BA can I deduce that A and B are inverses?

No. A counterexample is for instance if A is the zero matrix.

However, if we are given that A and B are square matrices with AB = I, then it turns out that yes, this implies that B and A are inverses and BA = I also.

leehuan · Apr 19, 2016

We request at least that B can't be the 0 matrix or the identity matrix itself

InteGrand · Apr 19, 2016

leehuan said:
We request at least that B can't be the 0 matrix or the identity matrix itself

Still no. A classic counterexample would be that if A and B are both diagonal matrices, then AB = BA, but it need not be the identity matrix.

If A and B are both diagonal matrices, then it is easy to verify that AB is the diagonal matrix whose diagonal elements are the corresponding diagonal elements of A multiplied by those of B. Since multiplication of scalars from the field is commutative, we can switch the order of these elements in AB and then see that this is also BA.

$\noindent E.g. Suppose $A = \begin{pmatrix}2 & 0 \\ 0 & -1\end{pmatrix}$ and $B = \begin{pmatrix}-7 & 0 \\ 0 & \pi \end{pmatrix}$. It is easily checked then that $AB = BA = \begin{pmatrix}-14 & 0 \\ 0 & -\pi\end{pmatrix} \neq I$.$

InteGrand · Apr 19, 2016

Of course another easy type of counterexample is if A = B. Then clearly AB = BA, but it's easy to see that this need not equal I.

leehuan · Apr 19, 2016

InteGrand said:
Still no. A classic counterexample would be that if A and B are both diagonal matrices, then AB = BA, but it need not be the identity matrix.

If A and B are both diagonal matrices, then it is easy to verify that AB is the diagonal matrix whose diagonal elements are the corresponding diagonal elements of A multiplied by those of B. Since multiplication of scalars from the field is commutative, we can switch the order of these elements in AB and then see that this is also BA.

$\noindent E.g. Suppose $A = \begin{pmatrix}2 & 0 \\ 0 & -1\end{pmatrix}$ and $B = \begin{pmatrix}-7 & 0 \\ 0 & \pi \end{pmatrix}$. It is easily checked then that $AB = BA = \begin{pmatrix}-14 & 0 \\ 0 & -\pi\end{pmatrix} \neq I$.$

Oh I wasn't intending to imply that they were the only solutions. I knew there were more counterexamples but couldn't figure out what at the time.

leehuan · May 17, 2016

In general, which pairs of matrices satisfy commutation again? I forgot

InteGrand · May 17, 2016

leehuan said:
In general, which pairs of matrices satisfy commutation again? I forgot

$\noindent Diagonal matrices are an important case where they always commute. I.e. if $D_1,D_2$ are same-sized (so that they are compatible) diagonal matrices, then $D_1 D_2 = D_2 D_1$.$

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