Representation-independent valuation of hypothetical outputs

A linear map can send different input vectors to the same output. If a proposed value is meant to describe the output, changing its input representation should not change that value. This section turns that requirement into a precise null-space test.

All vectors and matrices here are real, and input vectors may range over the whole input space. The scenario tables are hypothetical linear models. The argument concerns algebraic consistency; it does not assert empirical prices, positivity or a complete financial-market model.

1. Distinguish an input from its output

Let $A$ be an $m\times n$ matrix. An input $x\in\mathbb R^n$ produces $Ax\in\mathbb R^m$. The attainable output space is

$$\operatorname{im}A=\{Ax:x\in\mathbb R^n\},$$

and the null space is $N(A)=\{z\in\mathbb R^n:Az=0\}$. A null vector is an input change that leaves every output coordinate unchanged.

Suppose $x_0$ produces an attainable output $b$, so $Ax_0=b$. Then

$$Ax=b \quad\Longleftrightarrow\quad A(x-x_0)=0 \quad\Longleftrightarrow\quad x=x_0+z\text{ for some }z\in N(A).$$

This describes every representation of the same output. It also separates existence from uniqueness: an output outside $\operatorname{im}A$ has no representation, while a nonzero null vector makes every attainable output have multiple representations.

2. Work out a two-scenario table

Consider

$$A=\begin{pmatrix}1&0&1\\0&1&1\end{pmatrix}.$$

For $x=(x_1,x_2,x_3)^T$, multiplication gives

$$Ax=(x_1+x_3,x_2+x_3)^T.$$

Every $b=(b_1,b_2)^T$ is attainable: choose $x=(b_1,b_2,0)^T$. To find the entire representation set, put $x_3=t$. The equations force $x_1=b_1-t$ and $x_2=b_2-t$, so

$$x=(b_1-t,b_2-t,t)^T, \qquad t\in\mathbb R.$$

Setting $b=0$ gives $N(A)=\{t(-1,-1,1)^T:t\in\mathbb R\}$. In particular, output $(4,7)^T$ has all the representations $(4-t,7-t,t)^T$. These are equal-output inputs, although their three coordinates differ.

3. Prove the consistency criterion

Let $c\in\mathbb R^n$ be fixed, and assign input $x$ the linear value $c^Tx$. We call this rule representation-independent when

$$Ax=Ay\quad\Longrightarrow\quad c^Tx=c^Ty \qquad\text{for every }x,y\in\mathbb R^n.$$

The rule is representation-independent if and only if

$$c^Tz=0\qquad\text{for every }z\in N(A).$$

Necessity. If $z\in N(A)$, then $Az=A0=0$. Representation independence makes $c^Tz=c^T0=0$.

Sufficiency. If $Ax=Ay$, then $A(x-y)=0$. Thus $x-y\in N(A)$, and the stated condition gives $c^Tx-c^Ty=c^T(x-y)=0$.

The word “every” matters: testing one convenient pair of inputs is not enough. If $z_1,\ldots,z_k$ form a basis of $N(A)$, it does suffice to test those basis vectors, since linearity then gives zero on every $z=\sum_j a_jz_j$.

When the criterion holds, we may define a value directly on attainable outputs by

$$\nu(b)=c^Tx\quad\text{for any }x\text{ satisfying }Ax=b.$$

Existence of a representative follows from $b\in\operatorname{im}A$; independence of the choice follows from the criterion. Moreover $\nu$ is linear: if $Ax=b$ and $Ay=d$, then $A(\alpha x+\beta y)=\alpha b+\beta d$, giving $\nu(\alpha b+\beta d)=\alpha\nu(b)+\beta\nu(d)$.

4. Compare a consistent and an inconsistent rule

For the two-scenario table, a basis of the null space consists of $z=(-1,-1,1)^T$. The column $c=(2,3,5)^T$ passes the test because

$$c^Tz=-2-3+5=0.$$

Indeed, on an arbitrary representation of output $b$,

$$c^Tx=2(b_1-t)+3(b_2-t)+5t=2b_1+3b_2.$$

The parameter cancels. Thus $\nu(b)=2b_1+3b_2$, and $\nu((4,7)^T)=29$.

By contrast, $c'=(2,3,6)^T$ gives $(c')^Tz=1$. For the same output $(4,7)^T$, its value is $29+t$. For example, inputs $(4,7,0)^T$ and $(3,6,1)^T$ produce the same output but have values 29 and 30. This explicit pair witnesses the failure.

The criterion does not require nonnegative input weights or positive entries of $c$. Additional restrictions would define a different modeling question. Nor does representation independence mean that the input itself is unique: a well-defined output value can coexist with infinitely many input representations.

Exercises

P09. For the two-scenario matrix above, determine all columns $c=(c_1,c_2,c_3)^T$ that give representation-independent values. Express the induced value as a function of $b=(b_1,b_2)^T$.

P09a. Take $c=(4,-1,3)^T$ and output $b=(1,-2)^T$. Find all input representations, verify consistency and calculate the output value. Then show exactly what changes for $c'=(4,-1,4)^T$.

P09b. Now let $B=\begin{pmatrix}1&1\\2&2\end{pmatrix}$. Describe its attainable outputs and null space. Determine when the value rule $a x_1+d x_2$ depends only on $Bx$, and express the resulting value on every attainable output.

Full solutions

P09. Testing $(-1,-1,1)^T$ gives $-c_1-c_2+c_3=0$. Thus exactly the columns with $c_3=c_1+c_2$ work. For $Ax=b$,

$$c^Tx=c_1(x_1+x_3)+c_2(x_2+x_3)=c_1b_1+c_2b_2.$$

P09a. All representations are $(1-t,-2-t,t)^T$. Since $-4-(-1)+3=0$, the first rule is consistent. Direct substitution gives $4(1-t)-(-2-t)+3t=6$ for every $t$. With $c'$, the value is $6+t$, so it depends on the representation. The negative second entry of $c$ did not prevent the first rule from being algebraically well-defined.

P09b. Writing $s=x_1+x_2$, we have $Bx=(s,2s)^T$. These and only these outputs are attainable, and $N(B)=\{t(1,-1)^T:t\in\mathbb R\}$. The null-space test is $a-d=0$. If $a=d$, the input value is $a(x_1+x_2)=as$, so the induced value of $(s,2s)^T$ is $as$. If $a\ne d$, the inputs $(1,0)^T$ and $(0,1)^T$ have the same output $(1,2)^T$ but values $a$ and $d$, proving inconsistency.