What is the form of that matrix?

[mathmari]

Hey!

Calculate the Cholesky decomposition of the matrix, the only non-vanishing elements are the diagonals $1,2,3, \lambda$ and all under and upper secondary diagonal elements are $1$.

For which $\lambda$ is the matrix singular?

Could you please explain the form of the Matrix?

Does the matrix on the diagonal have $1,2,3, \lambda$ ? So do we have a $4 \times 4$ matrix? Or are these the only non zero diagonal elements? :unsure:

 

[I like Serena]

Hey mathmari!

I find it confusing as well.
My best guess is that they mean 'diagonal elements' when they write 'diagonals'.
In that case the matrix would be:
\begin{bmatrix}1&1\\ 1&2&1\\&1&3&1\\&&1&\lambda\end{bmatrix}
At least we can find an answer if that is the case.

 

[mathmari]

I find it confusing as well.
My best guess is that they mean 'diagonal elements' when they write 'diagonals'.
In that case the matrix would be:
\begin{bmatrix}1&1\\ 1&2&1\\&1&3&1\\&&1&\lambda\end{bmatrix}
At least we can find an answer if that is the case.

For that matrix we have:
\begin{align*}&\ell_{11}=\sqrt{a_{11}}=\sqrt{1}=1 \\ & \ell_{21}=\frac{1}{\ell_{11}}\left (a_{21}-\sum_{k=1}^{0}\ell_{2k}\ell_{jk}\right )=\frac{1}{1}\left (1-0\right )=1 \\ &\ell_{22}=\sqrt{a_{22}-\sum_{k=1}^{1}\ell_{2k}^2}=\sqrt{2-\ell_{21}^2}=\sqrt{2-1^2}=\sqrt{2-1}=\sqrt{1}=1 \\ & \ell_{31}=\frac{1}{\ell_{11}}\left (a_{31}-\sum_{k=1}^{0}\ell_{3k}\ell_{1k}\right )=\frac{1}{1}\left (0-0\right )=0 \\ & \ell_{32}=\frac{1}{\ell_{22}}\left (a_{32}-\sum_{k=1}^{1}\ell_{3k}\ell_{2k}\right )=\frac{1}{1}\left (1-\ell_{31}\ell_{21}\right )=\frac{1}{1}\left (1-0\cdot 1\right )=1 \\ & \ell_{33}=\sqrt{a_{33}-\sum_{k=1}^{2}\ell_{3k}^2}=\sqrt{3-\left (\ell_{31}^2+\ell_{32}^2\right )}=\sqrt{3-\left (0^2+1^2\right )}=\sqrt{3-1}=\sqrt{2} \\ & \ell_{41}=\frac{1}{\ell_{11}}\left (a_{41}-\sum_{k=1}^{0}\ell_{4k}\ell_{1k}\right )=\frac{1}{1}\left (0-0\right )=0 \\ & \ell_{42}=\frac{1}{\ell_{22}}\left (a_{42}-\sum_{k=1}^{1}\ell_{4k}\ell_{2k}\right )=\frac{1}{1}\left (0-\ell_{41}\ell_{21}\right )=\frac{1}{1}\left (0-0\cdot 1\right )=0 \\ & \ell_{43}=\frac{1}{\ell_{33}}\left (a_{43}-\sum_{k=1}^{2}\ell_{4k}\ell_{3k}\right )=\frac{1}{\sqrt{2}}\left (1-\left (\ell_{41}\ell_{31}+\ell_{42}\ell_{32}\right )\right )=\frac{1}{\sqrt{2}}\left (1-\left (0\cdot 0+0\cdot 1\right )\right ) =\frac{1}{\sqrt{2}} \\ & \ell_{44}=\sqrt{a_{44}-\sum_{k=1}^{3}\ell_{4k}^2}=\sqrt{\lambda-\left (\ell_{41}^2+\ell_{42}^2+\ell_{43}^2\right )}=\sqrt{\lambda-\left (0+0+\frac{1}{2}\right )}=\sqrt{\lambda-\frac{1}{2}}\end{align*}

So we get the matrix
\begin{equation*}\tilde{L}=\begin{pmatrix}1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & \sqrt{2} & 0 \\ 0 & 0 & \frac{1}{\sqrt{2}} & \sqrt{\lambda-\frac{1}{2}}\end{pmatrix}\end{equation*}

Is everything correct? :unsure:

 

[I like Serena]

It looks correct to me and I verified that $\tilde L\tilde L^*$ is indeed the original matrix. (Nod)

 

[mathmari]

It looks correct to me and I verified that $\tilde L\tilde L^*$ is indeed the original matrix. (Nod)

Great!

As for the second part of the question. Can we check the condition that the matrix is singular from the matrix $\tilde{L}$ ? Or do we check that directly at the matrix $A$ ? :unsure:

 

[I like Serena]

A matrix is singular if its determinant is zero.
And also if it's the product of matrices where one of those matrices is singular.
Can we find the determinant of $\tilde L$?

 

[mathmari]

A matrix is singular if its determinant is zero.
And also if it's the product of matrices where one of those matrices is singular.
Can we find the determinant of $\tilde L$?

$\tilde{L}$ is singular when $\lambda=\frac{1}{2}$. So for this valus $A$ is also singular, right? :unsure:

 

[I like Serena]

Yep. (Nod)

And if $\lambda\ne \frac 12$ then $\tilde L$ is not singular, so $\tilde L \tilde L^*=A$ is not singular either.

 

[mathmari]

Yep. (Nod)

And if $\lambda\ne \frac 12$ then $\tilde L$ is not singular, so $\tilde L \tilde L^*=A$ is not singular either.

Great! Thanks a lot! (Happy)