MHB Is $s$ the unique vector that spans the solution space $L(A,0)$?

[mathmari]

Hey!

For a field $K$ and $1<n\in \mathbb{N}$ let $A\in K^{(n-1)\times n}$ aa matrix with rank $n-1$. For a row vector $z\in K^{1\times n}$ let $\left (\frac{A}{z}\right )\in K^{n\times n}$ be the matrix that we get if we add as the $n$-th row of the matrix $A$ the vector $z$.

To show that there is a column vector $s=(s_1, \ldots , s_n)^T$ such that for each row vector $z=(z_1, \ldots , z_n)$ it holds $$\det \left [\left (\frac{A}{z}\right )\right ]=\sum_{i=1}^nz_is_i=z\cdot s$$ we consider the laplace formula for the clculation of the determinant. We expand for the last row. For $i\in \{1, \ldots , n\}$ let $A_i$ be the submatrix of$A$ if we remove the $i$-th columnn and $s_i:=(-1)^{i+n}\det A_i$.

Then we get $$\det \left [\left (\frac{A}{z}\right )\right ]=\sum_{i=1}^nz_is_i$$

Is this correct? (Wondering) We wcould also for an other row or column, right?I want to show that the vector $s$ spans the solution space $L(A,0)$ of the linear system of equations $A\cdot x=0$ as a $K$-vector space.

How could we do that? Could you give me a hint? (Wondering)

 

[I like Serena]

To show that there is a column vector $s=(s_1, \ldots , s_n)^T$ such that for each row vector $z=(z_1, \ldots , z_n)$ it holds $$\det \left [\left (\frac{A}{z}\right )\right ]=\sum_{i=1}^nz_is_i=z\cdot s$$ we consider the laplace formula for the clculation of the determinant. We expand for the last row. For $i\in \{1, \ldots , n\}$ let $A_i$ be the submatrix of$A$ if we remove the $i$-th columnn and $s_i:=(-1)^{i+n}\det A_i$.

Then we get $$\det \left [\left (\frac{A}{z}\right )\right ]=\sum_{i=1}^nz_is_i$$

Is this correct?

Hey mathmari!

Yep.

We wcould also for an other row or column, right?

Didn't we already do it for all rows and columns? (Thinking)

I want to show that the vector $s$ spans the solution space $L(A,0)$ of the linear system of equations $A\cdot x=0$ as a $K$-vector space.

How could we do that? Could you give me a hint?

What is the rank of the solution space $L(A,0)$?
Is $\mathbf s$ unique and non-zero?
What is $A\cdot\mathbf s$? (Wondering)

 

[mathmari]

Didn't we already do it for all rows and columns? (Thinking)

What do you mean? (Wondering)

What is the rank of the solution space $L(A,0)$?
Is $\mathbf s$ unique and non-zero?
What is $A\cdot\mathbf s$? (Wondering)

The solution space is equal the kernel of$A$, right?

$s$ is uniquely defined.

We have that $\text{Rank}(A)=n-1$ so from the formula of dimensions we get that $\dim (\ker(A))=1$. That means that the basis of the solution space contains only one element.

If $z$ is an arbitrary row of $A$, $z=a_i, \forall i$, then $a_i\cdot s=0$. That means that $A\cdot s=0$, so $s\in \ker (A)$, and so $s$ is contained in the soution space.

From that we get the desired result, right? (Wondering)

 

[I like Serena]

What do you mean?

Didn't we find each of the $s_i$ by working through each of the columns?
And by using all rows to find the sub determinant $\det(A_i)$? (Wondering)

The solution space is equal the kernel of$A$, right?

$s$ is uniquely defined.

We have that $\text{Rank}(A)=n-1$ so from the formula of dimensions we get that $\dim (\ker(A))=1$. That means that the basis of the solution space contains only one element.

If $z$ is an arbitrary row of $A$, $z=a_i, \forall i$, then $a_i\cdot s=0$. That means that $A\cdot s=0$, so $s\in \ker (A)$, and so $s$ is contained in the soution space.

From that we get the desired result, right?

Suppose $\mathbf s$ is the zero vector. Then all these statements are true, but we still don't get the desired result do we? (Wondering)

How did you find that $\mathbf s$ is unique?
You did find that there is at least one $\mathbf s$, but there could still be more, couldn't there? (Wondering)