Matematické Fórum

stuart clark · 07. 03. 2016 11:57

If $A$ is a matrix of order $2\times 3$ and $B$ is a matrix of order $3\times 2$ and $\det(AB)=4\;,$ Then $\det(BA)=$

vanok · 07. 03. 2016 12:39

Hi ↑ stuart clark:,
$\det(AB)=4\neq 0\;,$ , give $rg (A)=rg(B)=2$ ,
so $rg(BA) \leq 2$ and $\det(BA)=0$

stuart clark · 07. 03. 2016 13:02

Thanks ↑ vanok:, would you like to explain me in detail

Pavel · 08. 03. 2016 00:25

↑ stuart clark:

Let us define matrices

$\mathbf A= \begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23} \end{pmatrix}, \quad \mathbf B= \begin{pmatrix} b_{11} & b_{12}\\ b_{21} & b_{22}\\ b_{31} & b_{32} \end{pmatrix}, \quad \mathbf C=\mathbf A\cdot\mathbf B= \begin{pmatrix} c_{11} & c_{12}\\ c_{21} & c_{22} \end{pmatrix}.$

Let us assume that $\text{rank}\,(\mathbf A)\neq 0\neq \text{rank}\,(\mathbf B)$ . Using vector notation we have

$(c_{11},c_{12})&=a_{11}(b_{11},b_{12})+a_{12}(b_{21},b_{22})+a_{13}(b_{31},b_{32})\\ (c_{21},c_{22})&=a_{21}(b_{11},b_{12})+a_{22}(b_{21},b_{22})+a_{23}(b_{31},b_{32}).$

If $\text{rank}\,(\mathbf B)=1$ then the vectors $(c_{11},c_{12})$ , $(c_{21},c_{22})$ would be linearly dependent and $\det(\mathbf A\cdot\mathbf B)=0$ which is not true.

Considering the fact that $\text{rank}\,(\mathbf B)<3$ we have $\boldsymbol{\text{rank}\,(\mathbf B)=2}$ .

On the other hand we have

$\begin{pmatrix} c_{11}\\ c_{21} \end{pmatrix} &=b_{11} \begin{pmatrix} a_{11}\\ a_{21} \end{pmatrix} + b_{21} \begin{pmatrix} a_{12}\\ a_{22} \end{pmatrix} + b_{31} \begin{pmatrix} a_{13}\\ a_{23} \end{pmatrix}\\ \begin{pmatrix} c_{12}\\ c_{22} \end{pmatrix} &=b_{12} \begin{pmatrix} a_{11}\\ a_{21} \end{pmatrix} + b_{22} \begin{pmatrix} a_{12}\\ a_{22} \end{pmatrix} + b_{32} \begin{pmatrix} a_{13}\\ a_{23} \end{pmatrix}$

If $\text{rank}\,(\mathbf A)=1$ then the vectors $\begin{pmatrix} c_{11}\\ c_{21} \end{pmatrix}$ , $\begin{pmatrix} c_{12}\\ c_{22} \end{pmatrix}$ would be linearly dependent and $\det(\mathbf A\cdot\mathbf B)=0$ .

Therefore $\boldsymbol{\text{rank}\,(\mathbf A)=2}$ .

Next let us define a matrix $\mathbf D$ such that

$\mathbf D=\mathbf B\cdot\mathbf A= \begin{pmatrix} d_{11} & d_{12} & d_{13}\\ d_{21} & d_{22} & d_{23}\\ d_{31} & d_{32} & d_{33} \end{pmatrix}$

Similarly as in the previous cases we have

$(d_{11},d_{12},d_{13})&=b_{11}(a_{11},a_{12},a_{13})+b_{12}(a_{21},a_{22},a_{23})\\ (d_{21},d_{22},d_{23})&=b_{21}(a_{11},a_{12},a_{13})+b_{22}(a_{21},a_{22},a_{23})\\ (d_{31},d_{32},d_{33})&=b_{31}(a_{11},a_{12},a_{13})+b_{32}(a_{21},a_{22},a_{23}).$

From the fact that $\text{rank}\,(\mathbf A)=2$ we deduce that the vectors $(d_{11},d_{12},d_{13})$ , $(d_{21},d_{22},d_{23})$ , $(d_{31},d_{32},d_{33})$ are linearly dependent which implies $\color{blue}\det(\mathbf B\cdot\mathbf A)=0$

vanok · 08. 03. 2016 17:04 — Editoval vanok (08. 03. 2016 17:06)

Hi ↑ stuart clark:,
As regards possible complements for the given solution, there is 2 ways.
The first one is to take advantage your first course of linear algebra (normally made in first year of university); and specially call back it two theorems
The determinant of a matrix M, of type (n, n) is no zero iff his rank $rg(M)$ is n.
And also
$rg (AB) \leq min (rg (A), rg (B))$ A,B matrices of type (m,n),(n,m) resp.
and simply to apply.
(It is my method).

Other method, (for the high of grammar schools) is the one used by ↑ Pavel: but which is applicable only to this particular situation.

Pavel · 08. 03. 2016 23:49

↑ vanok:

The method, I described former, can be used also for matrices of general type $m\times n$ , resp. $n\times m$ with $m\neq n$ not only in this particular case.

vanok · 09. 03. 2016 08:26

↑ Pavel:,
A lot of courage for your next exercise with n = 100 and m = 200.

Pavel · 09. 03. 2016 16:38

↑ vanok:

Let A be a 100x200 matrix, B be a 200x100 matrix such that det(AB)=4. Prove that det(BA)=0.

Solution:

Let $\mathbf A=\left(a_{i,j}\right)_{\substack{1\leq i\leq 100\\1\leq j\leq 200}}$ , $\mathbf B=\left(b_{i,j}\right)_{\substack{1\leq i\leq 200\\1\leq j\leq 100}}$ , $\mathbf C=\mathbf A\cdot\mathbf B=\left(c_{i,j}\right)_{\substack{1\leq i\leq 100\\1\leq j\leq 100}}$ and $\mathbf D=\mathbf B\cdot\mathbf A=\left(d_{i,j}\right)_{\substack{1\leq i\leq 200\\1\leq j\leq 200}}$ . Let us denote

$\vec a_{i*}&:=(a_{i,1},a_{i,2},\dots,a_{i,200})\\ \vec a_{*j}&:=(a_{1,j},a_{2,j},\dots,a_{100,j})\\ \vec b_{j*}&:=(b_{j,1},b_{j,2},\dots,b_{j,100})\\ \vec b_{*i}&:=(b_{1,i},b_{2,i},\dots,b_{200,i})\\ \vec c_{i*}&:=(c_{i,1},c_{i,2},\dots,c_{i,100})\\ \vec c_{*i}&:=(c_{1,i},c_{2,i},\dots,c_{100,i})\\ \vec d_{j*}&:=(d_{j,1},d_{j,2},\dots,d_{j,200})\\ \vec d_{*j}&:=(d_{1,j},d_{2,j},\dots,d_{200,j}),\qquad i=1,\dots,100,\ j=1,\dots,200.$

Then the following identity holds true for $i=1,\dots,100$ :

$\vec c_{i*}=\sum_{k=1}^{200}a_{i,k}\cdot \vec b_{k*}$

Among 200 100-dimensional vectors $\vec b_{j*}$ , $j=1,\dots,200$ there exist at most 100 vectors $\vec b_{{j_t}*}$ , $t=1,\dots,100$ that are linearly independent. If $\text{rank}\,(\mathbf B)\leq 99$ then all the vectors $\vec c_{1*},\vec c_{2*},\dots,\vec c_{100*}$ would be linearly dependent and $\det(\mathbf A\cdot\mathbf B)=0$ . Hence $\text{rank}\,(\mathbf B)=100$ .

There is a similar identity for the vectors $\vec c_{*i}$ with $i=1,\dots,100$ :

$\vec c_{*i}=\sum_{k=1}^{200}b_{k,i}\cdot \vec a_{*k}$

For the same reasons as in the previous case we can deduce that if $\text{rank}\,(\mathbf A)\leq 99$ then all the vectors $\vec c_{*1},\vec c_{*2},\dots,\vec c_{*100}$ would be linearly dependent and $\det(\mathbf A\cdot\mathbf B)=0$ . Hence $\text{rank}\,(\mathbf A)=100$ .

We also have for $j=1,\dots,200$

$\vec d_{j*}=\sum_{k=1}^{100}b_{j,k}\cdot \vec a_{k*}.$

Due to the fact that $\text{rank}\,(\mathbf A)=100$ we infer that the vectors $\vec d_{j*}$ , $j=1,\dots,200$ , are linearly dependent.

Thus

$\color{blue}\det(\mathbf B\cdot\mathbf A)=0$

If you wish to examine another case, e.g. $n=1000$ and $m=2000$ , I will be glad to show you.

vanok · 09. 03. 2016 16:54

And it not much costs you to generalize your results and to put you to the same level that I.
Then well continuation.
( I notice that in analysis you used by the very complicated theorems and that in algebra on the contrary you always want to take back, algebraic demonstrations for every result)

stuart clark

Thanks ↑ Pavel:↑ Vanokl:

Matematické Fórum

#1 07. 03. 2016 11:57

Given det(AB)=4, Then det(BA), where A and B are not square matrix

#2 07. 03. 2016 12:39

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#3 07. 03. 2016 13:02

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#4 08. 03. 2016 00:25

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#5 08. 03. 2016 17:04 — Editoval vanok (08. 03. 2016 17:06)

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#6 08. 03. 2016 23:49

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#7 09. 03. 2016 08:26

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#8 09. 03. 2016 16:38

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#9 09. 03. 2016 16:54

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

#10 13. 03. 2016 22:12 — Editoval stuart clark (13. 03. 2016 22:14)

Re: Given det(AB)=4, Then det(BA), where A and B are not square matrix

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