Express the given matrix in the form ##L_1DU_1##

[chwala]

Homework Statement

This is my own question ...refreshing on matrices after a long time... i do not think there is more insight on this...i checked my working by multiplying the three matrices and confirmed that product is correct.

If there is anything new then i would appreciate to hear from you. Cheers!

$\begin{bmatrix} 4& 3 \\ 6 & -2 & \\ \end{bmatrix}$

Relevant Equations

understanding of matrix decomposition.

$$\begin{bmatrix} 4& 3 \\ 6 & -2 & \\ \end{bmatrix}= \begin{bmatrix} 1& 0 \\ a& 1& \\ \end{bmatrix}⋅ \begin{bmatrix} b& c \\ 0& d& \\ \end{bmatrix}$$ $b=4, ab=6,⇒b=1.5, d=-6.5, c=3$ $$\begin{bmatrix} 4& 3 \\ 6 & -2 & \\ \end{bmatrix} = \begin{bmatrix} 1& 0\\ \dfrac{3}{2} & 1 & \\ \end{bmatrix}⋅ \begin{bmatrix} 4& 3\\ 0 & \dfrac{-13}{2} & \\ \end{bmatrix} $$$$\begin{bmatrix} 4& 3 \\ 0 & \dfrac{-13}{2}& \\ \end{bmatrix} = \begin{bmatrix} e& 0\\ 0& f & \\ \end{bmatrix}⋅ \begin{bmatrix} 1& g\\ 0 & 1 & \\ \end{bmatrix} $$ It follows that, $e=4, f=\dfrac{-13}{2}$ therefore, $$\begin{bmatrix} 4& 3 \\ 6 & -2 & \\ \end{bmatrix} = \begin{bmatrix} 1& 0\\ 1.5 & 1 & \\ \end{bmatrix}⋅ \begin{bmatrix} 4& 0\\ 0 & -6.5 & \\ \end{bmatrix}⋅ \begin{bmatrix} 1& 0.75\\ 0 & 1 & \\ \end{bmatrix} $$

 

[fresh_42]

Not new, and probably not much to say about the matrix case. One can consider this decomposition in terms of linear transformations with certain properties of a vector space, and how this corresponds to a decomposition of that vector space. Of course, the two-dimensional case is not very enlightening.

However, there is a deeply dug relation between this decomposition and quantum physics.

\begin{matrix}
L D U &=&\text{ { lower unipotent } }&\cdot&\text{ { diagonal } }&\cdot&\text{ { upper unipotent } }&\in &\operatorname{GL(V)}\\
L +D+ U &=&\text{ { lower nilpotent } }&+&\text{ { diagonal } }&+&\text{ { upper nilpotent } }&\in &\operatorname{\mathfrak{gl}(V)}\\
X&=&\underbrace{\displaystyle{\sum_{\alpha \in \Delta^\mathbf{-}}y_\alpha E_\alpha }}_{\in \mathfrak{N^\mathbf{-}}}&+&\underbrace{\displaystyle{\sum_{\alpha \in \Delta}h_\alpha H_\alpha }}_{\in \mathfrak{h}}&+&\underbrace{\displaystyle{\sum_{\alpha \in \Delta^\mathbf{+}}x_\alpha E_\alpha }}_{\in \mathfrak{N^\mathbf{+}}}&\in&\mathfrak{gl(g)}\\
&&\text{ negative weights }&&\text{ semisimple }&&\text{ positive weights }&&\\
\text{operator}&:&\text{ lowering ladder }&&\text{ stationary }&&\text{ raising ladder }&&
\end{matrix}
Yes, there is a long way to go from your example above to quantum mechanics, and I left out several hundred pages of theory or roughly four textbooks, but in the end, it comes down to that list.

 

[Steve4Physics]

If there is anything new then i would appreciate to hear from you. Cheers!

Not sure if this answers what you are asking but there is an equivalent (but IMO quicker) procedure.

This starts by reducing the original matrix to row echelon form. You can easily find examples/explanations, e.g.

 

[chwala]

Not sure if this answers what you are asking but there is an equivalent (but IMO quicker) procedure.

This starts by reducing the original matrix to row echelon form. You can easily find examples/explanations, e.g.

I am following your video example; but there is a problem...from

$$\begin{bmatrix}
4& 3 \\
6 & -2 & \\
\end{bmatrix}$$

I used $-3R_1+2R_2$ to get;

$$\begin{bmatrix}
4& 3 \\
0 & -13 & \\
\end{bmatrix}$$

The next step here as suggested is to factor out the diagonals ...to get matrix $D$. When you pull out (factor out as suggested) you get;

$$\begin{bmatrix}
4& 0 \\
0 & -13 & \\
\end{bmatrix}$$

instead of

$$\begin{bmatrix}
4& 0 \\
0 & -6.5 & \\
\end{bmatrix}$$

I can tell that we have to divide row 2 by 2...but the problem is how would you tell that by using the indicated approach? I went ahead with the other steps i.e in getting matrix $U$ this was ok. The steps i followed were

$$\begin{bmatrix}
4& 3 \\
0 & -13 & \\
\end{bmatrix}$$

$R_1\div4$

$R_2\div(-13)$

to realize;

$$\begin{bmatrix}
1& 0.75 \\
0 & 1 & \\
\end{bmatrix}$$

I also found matrix $L$ i.e by using the identity matrix as suggested; using the reverse operation $3R_1+2R_2$ and $R_2\div 2$ we shall have;

$$\begin{bmatrix}
1& 0 \\
0 & 1 & \\
\end{bmatrix}=

\begin{bmatrix}
1& 0 \\
3 & 2 & \\
\end{bmatrix}=
\begin{bmatrix}
1& 0 \\
1.5 & 1 & \\
\end{bmatrix}
$$

The only hitch on this approach is on how to get the correct matrix for $D$. You may counter check this...thanks.

 

[Steve4Physics]

$$\begin{bmatrix}
4& 3 \\
6 & -2 & \\
\end{bmatrix}$$

I used $-3R_1+2R_2$ ...

Edit. Aplogies. You hadn't made any mistakes so my reply (below) is wrong. I've struck it through.

I think you have performed the wrong row operation and also got the arithmetic wrong!$$\begin{bmatrix}
4& 3 \\
6 & -2 & \\
\end{bmatrix}$$Replace row-2 by $R_2 +(-1.5R_1)$$$\begin{bmatrix}
4& 3 \\
{6+(-6)} & {-2+(-4.5)}& \\
\end{bmatrix}$$giving$$\begin{bmatrix}
4& 3 \\
0&-6.5& \\
\end{bmatrix}$$which has the required diagonal. values.

 

[chwala]

I think you have performed the wrong row operation and also got the arithmetic wrong!$$\begin{bmatrix}
4& 3 \\
6 & -2 & \\
\end{bmatrix}$$Replace row-2 by $R_2 +(-1.5R_1)$$$\begin{bmatrix}
4& 3 \\
{6+(-6)} & {-2+(-4.5)}& \\
\end{bmatrix}$$giving$$\begin{bmatrix}
4& 3 \\
0&-6.5& \\
\end{bmatrix}$$which has the required diagonal. values.

Ok thanks...I will look at it again.

 

[chwala]

I think you have performed the wrong row operation and also got the arithmetic wrong!$$\begin{bmatrix}
4& 3 \\
6 & -2 & \\
\end{bmatrix}$$Replace row-2 by $R_2 +(-1.5R_1)$$$\begin{bmatrix}
4& 3 \\
{6+(-6)} & {-2+(-4.5)}& \\
\end{bmatrix}$$giving$$\begin{bmatrix}
4& 3 \\
0&-6.5& \\
\end{bmatrix}$$which has the correct required diagonal values.

The row operation is correct...the only problem is that it's twice the one you've posted...what maybe I would want to know is why is it wrong?...my understanding of echelon matrices is that we can reduce rows and columns in whichever way as long as we are consistent in the arithmetic...

Not unless the approach requires $R_2$ not to have a coefficient...then that would be clear.

 

[Steve4Physics]

The row operation is correct...

Yes. My apologies. Your row reduction was correct. My bad. I’ve updated Post #5 to reflect this.

When I get chance I’ll go through your working and get back to you.

 

[Steve4Physics]

Not unless the approach requires $R_2$ not to have a coefficient...then that would be clear.

Yes - that seems correct. Though we should really say that the coefficient of $R_2$ must be 1.

This is more formally stated (though in the context of LU decomposition) here: https://en.wikipedia.org/wiki/LU_decomposition#Using_Gaussian_elimination

If that rule is not followed, I believe a modified procedure would be needed which would result in a valid alternative LDU decomposition - but such that L would not have 1s in its diagonal.

However, I must admit no claim to real expertise - just a dabbler. So I’ll leave it to the experts to provide any further thoughts on this.

Also, note that the video referenced in Post #3 was picked more or less at random. I’ve watched it through now and it’s not that good – the second example (factorisation of a 3x3 matrix) is rushed and has errors.

 

[Steve4Physics]

Yes - that seems correct. Though we should really say that the coefficient of $R_2$ must be 1.

This is more formally stated (though in the context of LU decomposition) here: https://en.wikipedia.org/wiki/LU_decomposition#Using_Gaussian_elimination

If that rule is not followed, I believe a modified procedure would be needed which would result in a valid alternative LDU decomposition - but such that L would not have 1s in its diagonal.

This may tidy up a loose end...

As suspected, using $-3R_1+2R_2$ leads to a different LDU factorisation:$$LDU = \begin{bmatrix}
1& 0 \\
1.5 & 0.5 & \\
\end{bmatrix}
\begin{bmatrix}
4& 0 \\
0 & -13 & \\
\end{bmatrix}
\begin{bmatrix}
1& 0.75 \\
0 & 1 & \\
\end{bmatrix}$$which gives the correct original matrix.

If you want to try this for yourself, note that $-3R_1+2R_2$ is equivalent to applying the matrix$$E = \begin{bmatrix}
1& 0 \\
-3 & 2 & \\
\end{bmatrix}$$If the original matrix is $A$ then $EA = U$ and it is not hard to show that $L = E^{-1}$.

(If this is unfamiliar, read about elimination matrices.)