# Number of multiplications and divisions for LU decomposition

#### mathmari

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Hey!!

Let $A$ a $n\times n$ matrix with known LU decomposition, let $u\in \mathbb{R}^n, v\in \mathbb{R}^{n+1}$.

Show that the number of multiplications and divisions that are needed to get a LU decomposition of the $(n+1)\times (n+1)$ matrix $$\begin{pmatrix}A & u \\ v^T\end{pmatrix}$$ is at most $O(n^2)$.

To get $U$, we have to eliminate $n$ terms below the main diagonal (which is the elements of $u^T$ except the last element of that row). Each elimination requires computing the row multiplier, which involves division by the pivotal element.

So we have $n$ divisions, or not?

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#### Klaas van Aarsen

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So we have $n$ divisions, or not?
Hey mathmari !!

Yep.
That leaves counting the number of multiplications that we need for each elimination, doesn't it?

#### mathmari

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That leaves counting the number of multiplications that we need for each elimination, doesn't it?
So for each element except the $(n+1)$th one of the last row we have to multiply the row multiplier with the respective row and subtract it from the last row, right?
Therefore we have to do $n$ multiplications, or not?

#### Klaas van Aarsen

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So for each element except the $(n+1)$th one of the last row we have to multiply the row multiplier with the respective row and subtract it from the last row, right?
Therefore we have to do $n$ multiplications, or not?
That sounds about right to find a single unknown entry, although a number of those multiplications are actually with zero.
Either way, for the worst case entry in row $n+1$ and column $n$ of the new $L$ matrix, we have indeed $n$ multiplications and $1$ division.

We still need to repeat for every entry in the extra row below the given L matrix, don't we?
And also for the extra column to the right of the given U matrix?

#### mathmari

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That sounds about right to find a single unknown entry, although a number of those multiplications are actually with zero.
Either way, for the worst case entry in row $n+1$ and column $n$ of the new $L$ matrix, we have indeed $n$ multiplications and $1$ division.

We still need to repeat for every entry in the extra row below the given L matrix, don't we?
And also for the extra column to the right of the given U matrix?
I got stuk right now. Could you explain that further to me? I haven't really understood that.

#### Klaas van Aarsen

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I got stuk right now. Could you explain that further to me? I haven't really understood that.
We want to find that the order of complexity is $O(n^2)$ don't we?
To find that, we need an algorithm to find the LU-decomposition of the matrix, don't we?

Let $A=L\cdot U$ be the given LU-decomposition of $A$.
Let $\tilde A=\begin{pmatrix}A & u \\ u^T\end{pmatrix}$ be the matrix to decompose.
And let $\tilde L,\,\tilde U$ be its LU-decomposition.

Can we write $\tilde L,\,\tilde U$ in terms of $L$, $U$, and $u$?

#### mathmari

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We want to find that the order of complexity is $O(n^2)$ don't we?
To find that, we need an algorithm to find the LU-decomposition of the matrix, don't we?

Let $A=L\cdot U$ be the given LU-decomposition of $A$.
Let $\tilde A=\begin{pmatrix}A & u \\ u^T\end{pmatrix}$ be the matrix to decompose.
And let $\tilde L,\,\tilde U$ be its LU-decomposition.

Can we write $\tilde L,\,\tilde U$ in terms of $L$, $U$, and $u$?
We have that $$\tilde{L}=\begin{pmatrix}L\\ 0 & 0 & \ldots & 0& \ell\end{pmatrix}$$ with $\ell>0$, or not?

#### Klaas van Aarsen

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We have that $$\tilde{L}=\begin{pmatrix}L\\ 0 & 0 & \ldots & 0& \ell\end{pmatrix}$$ with $\ell>0$, or not?
I don't think so.
If it would be, wouldn't we get that $u_1$ is always zero?

#### mathmari

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I don't think so.
If it would be, wouldn't we get that $u_1$ is always zero?
I got stuck right now. Why is $u_1$ always zero then?

#### Klaas van Aarsen

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I got stuck right now. Why is $u_1$ always zero then?
What do we get if we calculate:
$$\tilde{L} \cdot \tilde U =\begin{pmatrix} & \\ & \\ L\\ 0 & \ldots & 0& \ell \end{pmatrix}\cdot \begin{pmatrix} &&U& * \\ &&& \vdots \\ &&& * \\ 0 & \ldots & 0& * \end{pmatrix}$$
?
In particular for the entry in the bottom left corner?

Can it be equal to
$$\tilde A=\begin{pmatrix}A & u \\ u^T\end{pmatrix}$$
with $u_1\ne 0$?

#### mathmari

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What do we get if we calculate:
$$\tilde{L} \cdot \tilde U =\begin{pmatrix} & \\ & \\ L\\ 0 & \ldots & 0& \ell \end{pmatrix}\cdot \begin{pmatrix} &&U& * \\ &&& \vdots \\ &&& * \\ 0 & \ldots & 0& * \end{pmatrix}$$
?
In particular for the entry in the bottom left corner?

Can it be equal to
$$\tilde A=\begin{pmatrix}A & u \\ u^T\end{pmatrix}$$
with $u_1\ne 0$?
We have the following, or not?

$$\tilde{L} \cdot \tilde U =\begin{pmatrix} & & & 0\\ & & & 0\\ L & & & \vdots\\ \ell_1 & \ldots & \ell_n& \ell_{n+1} \end{pmatrix}\cdot \begin{pmatrix} &&U& u_1 \\ &&& \vdots \\ &&& u_n \\ 0 & \ldots & 0& u_{n+1} \end{pmatrix}=\begin{pmatrix} &&A& \ell_{11}u_1 \\ &&& \vdots \\ &&& \displaystyle{\sum_{i=1}^n\ell_{ni}u_i } \\ \ell_1u_{11} & \ldots & \displaystyle{\sum_{i=1}^n\ell_iu_{in}} & \displaystyle{\sum_{i=1}^{n+1}\ell_iu_i} \end{pmatrix}$$
where $L=(\ell_{ij})_{i,j\in \{1, 2, \ldots , n\}}$ and $U=(u_{ij})_{i,j\in \{1, 2, \ldots , n\}}$.

#### Klaas van Aarsen

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We have the following, or not?

$$\tilde{L} \cdot \tilde U =\begin{pmatrix} & & & 0\\ & & & 0\\ L & & & \vdots\\ \ell_1 & \ldots & \ell_n& \ell_{n+1} \end{pmatrix}\cdot \begin{pmatrix} &&U& u_1 \\ &&& \vdots \\ &&& u_n \\ 0 & \ldots & 0& u_{n+1} \end{pmatrix}=\begin{pmatrix} &&A& \ell_{11}u_1 \\ &&& \vdots \\ &&& \displaystyle{\sum_{i=1}^n\ell_{ni}u_i } \\ \ell_1u_{11} & \ldots & \displaystyle{\sum_{i=1}^n\ell_iu_{in}} & \displaystyle{\sum_{i=1}^{n+1}\ell_iu_i} \end{pmatrix}$$
where $L=(\ell_{ij})_{i,j\in \{1, 2, \ldots , n\}}$ and $U=(u_{ij})_{i,j\in \{1, 2, \ldots , n\}}$.
Yep.

It's a bit problematic to introduce $u_1,\ldots,u_{n+1}$ here though.
Doesn't that clash with the given vector $u$?

Anyway, we would get that for instance $\ell_1 u_{11}$ must be equal to the first entry of the vector $u$, mustn't it?
If that first entry of $u$ is non-zero, then $\ell_1$ cannot be zero can it?

#### mathmari

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It's a bit problematic to introduce $u_1,\ldots,u_{n+1}$ here though.
Doesn't that clash with the given vector $u$?
Ok, it is better to use other letters.

Anyway, we would get that for instance $\ell_1 u_{11}$ must be equal to the first entry of the vector $u$, mustn't it?
If that first entry of $u$ is non-zero, then $\ell_1$ cannot be zero can it?
So have:
$$\tilde{L} \cdot \tilde U =\begin{pmatrix} & & & 0\\ & & & 0\\ L & & & \vdots\\ \alpha_1 & \ldots & \alpha_n& \alpha_{n+1} \end{pmatrix}\cdot \begin{pmatrix} &&U& \beta_1 \\ &&& \vdots \\ &&& \beta_n \\ 0 & \ldots & 0& \beta_{n+1} \end{pmatrix}=\begin{pmatrix} &&A& \ell_{11}\beta_1 \\ &&& \vdots \\ &&& \displaystyle{\sum_{i=1}^n\ell_{ni}\beta_i } \\ \alpha_1u_{11} & \ldots & \displaystyle{\sum_{i=1}^n\alpha_iu_{in}} & \displaystyle{\sum_{i=1}^{n+1}\alpha_i\beta_i} \end{pmatrix}$$
where $L=(\ell_{ij})_{i,j\in \{1, 2, \ldots , n\}}$ and $U=(u_{ij})_{i,j\in \{1, 2, \ldots , n\}}$.

The result must be of the form $\tilde A=\begin{pmatrix}A & u \\ u^T\end{pmatrix}$, so it must hold \begin{align*}&\ell_{11}\beta_1=\alpha_1u_{11} \\ &\ldots \\ & \displaystyle{\sum_{i=1}^n\ell_{ni}\beta_i }=\displaystyle{\sum_{i=1}^n\alpha_iu_{in}} \end{align*}

What do we get from that?

#### Klaas van Aarsen

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The result must be of the form $\tilde A=\begin{pmatrix}A & u \\ u^T\end{pmatrix}$, so it must hold \begin{align*}&\ell_{11}\beta_1=\alpha_1u_{11} \\ &\ldots\end{align*}

What do we get from that?
From the first equation we get $$\ell_{11}\beta_1=\alpha_1u_{11} = u_1 \implies \begin{cases}\alpha_1 = \frac {u_1}{u_{11}} \\ \beta_1=\frac{u_1}{\ell_{11}}\end{cases}$$ don't we?

#### Klaas van Aarsen

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I have just noticed a vector $v$ in the problem statement:
Let $A$ a $n\times n$ matrix with known LU decomposition, let $u\in \mathbb{R}^n, v\in \mathbb{R}^{n+1}$.

Show that the number of multiplications and divisions that are needed to get a LU decomposition of the $(n+1)\times (n+1)$ matrix $$\begin{pmatrix}A & u \\ u^T\end{pmatrix}$$ is at most $O(n^2)$.
Is there a typo?
Shouldn't that $v$ be used somewhere?

#### mathmari

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I have just noticed a vector $v$ in the problem statement:

Is there a typo?
Shouldn't that $v$ be used somewhere?
Ahh yes, it is a typo. It should be:
 $$\begin{pmatrix}A & u \\ v^T\end{pmatrix}$$

So from $$\tilde{L} \cdot \tilde U =\begin{pmatrix} & & & 0\\ & & & 0\\ L & & & \vdots\\ \alpha_1 & \ldots & \alpha_n& \alpha_{n+1} \end{pmatrix}\cdot \begin{pmatrix} &&U& \beta_1 \\ &&& \vdots \\ &&& \beta_n \\ 0 & \ldots & 0& \beta_{n+1} \end{pmatrix}=\begin{pmatrix} &&A& \ell_{11}\beta_1 \\ &&& \vdots \\ &&& \displaystyle{\sum_{i=1}^n\ell_{ni}\beta_i } \\ \alpha_1u_{11} & \ldots & \displaystyle{\sum_{i=1}^n\alpha_iu_{in}} & \displaystyle{\sum_{i=1}^{n+1}\alpha_i\beta_i} \end{pmatrix}$$ what do we get?

#### Klaas van Aarsen

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We can deduce the values of the $\alpha_i$ and $\beta_i$ can't we?
Once we have those, we have the LU-decomposition of $\tilde A$.

#### mathmari

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We can deduce the values of the $\alpha_i$ and $\beta_i$ can't we?
Once we have those, we have the LU-decomposition of $\tilde A$.
Let $v=(v_1, \ldots , v_{n+1})$ and $u=(x_1, \ldots, x_{n+1})$.

We have that \begin{align*}&\alpha_1u_{11}=v_1 \Rightarrow \alpha_1=\frac{v_1}{u_{11}} \\ &\ldots \\ &\sum_{i=1}^n\alpha_iu_{in}=v_n \Rightarrow \alpha_n=\frac{v_n-\sum_{i=1}^{n-1}\alpha_iu_{in}}{u_{nn}}\\ &\sum_{i=1}^{n+1}\alpha_i\beta_i=v_{n+1}=x_{n+1} \Rightarrow a_{n+1}\beta_{n+1}=v_{n+1}-\sum_{i=1}^{n}\alpha_i\beta_i \ \text{ and } \ a_{n+1}\beta_{n+1}=x_{n+1}-\sum_{i=1}^{n}\alpha_i\beta_i\\ &\ell_{11}\beta_1=x_1 \Rightarrow \beta_i=\frac{x_1}{\ell_{11}} \\ &\ldots \\ &\sum_{i=1}^n\ell_{ni}\beta_i=x_n \Rightarrow \beta_n=\frac{x_n-\sum_{i=1}^{n-1}\beta_i\ell_{ni}}{\ell_{nn}}\end{align*}

Is everything correct?

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#### Klaas van Aarsen

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I see an $u_n$ that I think should be a $v_n$.
And I also see an $u_{n+1}$ that should be a $v_{n+1}$.

Other than that it looks good to me!

#### mathmari

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I see an $u_n$ that I think should be a $v_n$.
And I also see an $u_{n+1}$ that should be a $v_{n+1}$.

Other than that it looks good to me!
I changed it!

So in total we do $\displaystyle{2\cdot \sum_{i=1}^ni=2\cdot \frac{n(n+1)}{2}=n(n+1)}$ mutliplications and $\displaystyle{2n}$ divisions, or not?

#### Klaas van Aarsen

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I changed it!

So in total we do $\displaystyle{2\cdot \sum_{i=1}^ni=2\cdot \frac{n(n+1)}{2}=n(n+1)}$ mutliplications and $\displaystyle{2n}$ divisions, or not?
Yep.

#### mathmari

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Great!! Is this a formal proof to write the matrices and the formulas that we get? Or do we have to do also something else/more?

#### Klaas van Aarsen

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Great!! Is this a formal proof to write the matrices and the formulas that we get? Or do we have to do also something else/more?
We have identified an algorithm with order $O(n^2)$ that solves the problem.
So that qualifies as a proof that the order of the desired LU-decomposition is at most $O(n^2)$, doesn't it?