Proving determinant properties (1 Viewer)

leehuan · Apr 19, 2016

Let U₁ and U₂ be two nxn row-echelon matrices. Prove that det(U₁)det(U₂)=det(U₁U₂)

So far...

$\\$Let the components of ${U}_{1}, U_{2}, U_{1}U_{2}$ respectively be ${u}_{ij},{v}_{ij},{a}_{ij}\\$ for some integers $i, j \in [1,n]$

$\\$Since ${U}_{1}$ and ${U}_{2}$ are upper triangular:$\\\left| { U }_{ 1 } \right| =\prod _{ j=1 }^{ n }{ { u }_{ jj } } \\ \left| { U }_{ 2 } \right| =\prod _{ j=1 }^{ n }{ { v }_{ jj } }$

$\\$Now:$, \\ { a }_{ jj }=\sum _{ k=1 }^{ n }{ { u }_{ jk }{ v }_{ kj } } =\sum _{ k=1 }^{ j }{ { u }_{ jk }{ v }_{ kj } } \\$ as if $j<k\le n$ the term vanishes$\\\therefore \left| { U }_{ 1 }{ U }_{ 2 } \right| =\prod _{ j=1 }^{ n }{ { a }_{ jj } } =\prod _{ j=1 }^{ n }{ \sum _{ k=1 }^{ j }{ { u }_{ jk }{ v }_{ kj } } }$

I could be wrong, but if I'm right what now?

InteGrand · Apr 19, 2016

leehuan said:
Let U₁ and U₂ be two nxn row-echelon matrices. Prove that det(U₁)det(U₂)=det(U₁U₂)

So far...

$\\$Let the components of ${U}_{1}, U_{2}, U_{1}U_{2}$ respectively be ${u}_{ij},{v}_{ij},{a}_{ij}\\$ for some integers $i, j \in [1,n]$

$\\$Since ${U}_{1}$ and ${U}_{2}$ are upper triangular:$\\\left| { U }_{ 1 } \right| =\prod _{ j=1 }^{ n }{ { u }_{ jj } } \\ \left| { U }_{ 2 } \right| =\prod _{ j=1 }^{ n }{ { v }_{ jj } }$

$\\$Now:$, \\ { a }_{ jj }=\sum _{ k=1 }^{ n }{ { u }_{ jk }{ v }_{ kj } } =\sum _{ k=1 }^{ j }{ { u }_{ jk }{ v }_{ kj } } \\$ as if $j<k\le n$ the term vanishes$\\\therefore \left| { U }_{ 1 }{ U }_{ 2 } \right| =\prod _{ j=1 }^{ n }{ { a }_{ jj } } =\prod _{ j=1 }^{ n }{ \sum _{ k=1 }^{ j }{ { u }_{ jk }{ v }_{ kj } } }$

I could be wrong, but if I'm right what now?

As you said, the matrices are upper triangular.

It's easy to show by induction and expanding down the first column that the determinant of an upper-triangular (in fact, any triangular) matrix is the product of the diagonal elements (realised now that you knew this as you wrote it in part of your steps).

It's also well-known and easy to show that the product of two upper triangular matrices is again an upper-triangular matrix, and the diagonal elements of this are the product of the corresponding diagonal elements of the original two matrices.

These two facts will imply the result.

InteGrand · Apr 19, 2016

After reading through your solution carefully (before I'd only glanced at it), I see you were almost there.

To finish it off, notice that u_jk = 0 whenever k < j, due to triangular matrix condition.

This means the only potentially non-zero term in the sum at the end will be when k = j, i.e. u_jjv_jj.

$\noindent So the determinant becomes $|U_1 U_2|=\prod _{j=1} ^{n} u_{jj}v_{jj}$.$

$\noindent Now, just like how sums can be split if the summand has an addition, a product can be split if the `productand' (lol) has a multiplication (this follows from associativity of multiplication, like how associativity of addition lets us split sums). So we get $|U_1 U_2| = \prod _{j=1} ^{n} u_{jj} \times \prod _{j=1} ^{n} v_{jj} = |U_1| |U_2 |$.$

leehuan · Apr 19, 2016

InteGrand said:
As you said, the matrices are upper triangular.

It's easy to show by induction and expanding down the first column that the determinant of an upper-triangular (in fact, any triangular) matrix is the product of the diagonal elements (realised now that you knew this as you wrote it in part of your steps).

It's also well-known and easy to show that the product of two upper triangular matrices is again an upper-triangular matrix, and the diagonal elements of this are the product of the corresponding diagonal elements of the original two matrices.

These two facts will imply the result.

Would the induction method have been easier/quicker to what I attempted?

InteGrand · Apr 19, 2016

leehuan said:
Would the induction method have been easier/quicker to what I attempted?

The induction method was just to prove the thing about determinant of triangular matrices being the product of diagonal elements. If you're allowed to assume that, then you don't need to prove it, so no induction needed. Then the only thing needed to be proved was that the diagonal elements of the U₁U₂ matrix are the product of the corresponding diagonal elements of U₁ and U₂, which is essentially what you did (although looking back at it now I think you assumed U₁U₂ would be triangular. It's easy to show this though; show that [U₁U₂]_ij = 0 whenever i > j, to show that it is upper triangular. This can be shown by considering a sum again and noticing how terms will vanish.).

leehuan · Apr 19, 2016

InteGrand said:
The induction method was just to prove the thing about determinant of triangular matrices being the product of diagonal elements. If you're allowed to assume that, then you don't need to prove it, so no induction needed. Then the only thing needed to be proved was that the diagonal elements of the U₁U₂ matrix are the product of the corresponding diagonal elements of U₁ and U₂, which is essentially what you did.

Oh I see. Yeah I made an assumption on the determinant of triangular matrices. If I had to prove it then I would've done the rigorous computation of a determinant but then said all the 0s in each row (or column) vanish away a ton of terms in the one go.

leehuan · Apr 21, 2016

This question was marked X so I reckon it's out of the syllabus, but it was still a part of the homework.

$$Prove that a square unitary matrix $Q$ has determinant$\\ \left| Q\right| = {e}^{i\theta}$

$\\$I got to $\\ \left| Q \right| \left| \bar { { Q }^{ T } } \right| =1\\ $which I'm not sure if implies $ \left| Q \right| \left| \bar { { Q } } \right| =1$

But I don't know how to proceed from there anyway.

InteGrand · Apr 21, 2016

leehuan said:
This question was marked X so I reckon it's out of the syllabus, but it was still a part of the homework.

$$Prove that a square unitary matrix $Q$ has determinant$\\ \left| Q\right| = {e}^{i\theta}$

$\\$I got to $\\ \left| Q \right| \left| \bar { { Q }^{ T } } \right| =1\\ $which I'm not sure if implies $ \left| Q \right| \left| \bar { { Q } } \right| =1$

But I don't know how to proceed from there anyway.

$\noindent Recall that the determinant of the conjugate of a matrix is the conjugate of the determinant of that matrix. (Also transpose and conjugate are interchangeable for a matrix). Also, determinant of a transpose is the determinant of the original matrix.$

$\noindent Hence $\det \left(\overline{Q} ^{T}\right) = \det \left(\overline{Q}\right) = \overline{\det \left(Q\right)}$.$

$\noindent So $\det \left(Q\right)\overline{\det \left(Q\right)} = 1$, since you showed that the two matrices' determinants have product 1.$

$\noindent So $\left | \det \left(Q\right)\right | = 1$ (using $z\overline{z} = |z|^2$ from HSC Complex Numbers). (Note $\det\left(Q\right)$ will in general be a complex number here, not necessarily real.)$

$\noindent This means that $\det \left(Q\right) = \mathrm{e}^{i \theta}$ for some real $\theta$.$

Proving determinant properties (1 Viewer)

leehuan

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