MHB According to Buckingham Theorem the rank of A should be 2

[evinda]

Hello! (Wave)

A physical system is described by a law of the form $f(E,P,A)=0$, where $E,P,A$ represent, respectively, energy, pressure and surface area. Find an equivalent physical law that relates suitable dimensionless quantities.

That' what I have tried so far:

1st step:


Choice of quantities

Mass: $M$

Time: $T$

Length: $L$

So:

$$[E]=M L^2 T^{-2}$$
$$[P]=ML^{-1}T^{-2}$$
$$[A]=L^2$$2nd step:

Construction of dimonsionless quantitiesThe matrix of dimensions:$A=\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}$I tried to find the rank, determining the smallest $n$ for which $A^n=I$.$\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}=\begin{bmatrix}
-1 & -1 & 0\\
2 & 2 & 0 \\
8 & 2 &4
\end{bmatrix}$$\begin{bmatrix}
-1 & -1 & 0\\
2 & 2 & 0 \\
8 & 2 &4
\end{bmatrix}\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
2 & -1 &2
\end{bmatrix}=\begin{bmatrix}
1 & 1 & 0\\
-2 & -2 & 0 \\
12 & 0 &8
\end{bmatrix}$But I saw the solution and there should be only one dimensionless quantity, so according to Buckingham Theorem the rank of $A$ should be $2$.

Where is my mistake? (Thinking)

 

[I like Serena]

Hey! (Smile)

I tried to find the rank, determining the smallest $n$ for which $A^n=I$.

Isn't that the order of $A$ instead of the rank of $A$? (Wondering)
The rank of $A$ is indeed $2$.

 

[evinda]

Isn't that the order of $A$ instead of the rank of $A$? (Wondering)

What is the difference between order and rank? (Thinking)

The rank of $A$ is indeed $2$.

Because of the fact that there are two linearly dependent rows? (Thinking)

 

[I like Serena]

What is the difference between order and rank? (Thinking)

In algebra the order of an element $a$ is the lowest power $n$ such that $a^n=id$.

The rank of a matrix is the dimension of its range. (Nerd)

Because of the fact that there are two linearly dependent rows? (Thinking)

Yes. (Nod)

More specifically, there are 2 linearly independent column vectors.
The range of the matrix is the span of those 2 vectors, meaning that range has dimension 2. (Emo)

 

[evinda]

In algebra the order of an element $a$ is the lowest power $n$ such that $a^n=id$.

The rank of a matrix is the dimension of its range. (Nerd)

A ok... (Nod)

Yes. (Nod)

More specifically, there are 2 linearly independent column vectors.
The range of the matrix is the span of those 2 vectors, meaning that range has dimension 2. (Emo)

So could we say that the following two vectors
$\begin{pmatrix}
1\\
-2\\
-1
\end{pmatrix} , \begin{pmatrix}
0\\
0\\
2
\end{pmatrix}$ span the vector space of the matrix? (Thinking)

 

[I like Serena]

So could we say that the following two vectors
$\begin{pmatrix}
1\\
-2\\
-1
\end{pmatrix} , \begin{pmatrix}
0\\
0\\
2
\end{pmatrix}$ span the vector space of the matrix? (Thinking)

Yes. (Nod)