Which part of an input survives?

Source scope: §3.13, PDF pages 21, 22. Numerical examples, experiments, added derivations, and practice are teaching supplements unless identified as source results.

Split an input into visible and invisible parts

For $A=\begin{bmatrix}1&1\\0&0\end{bmatrix}$, the vectors $(1,1)$ and $(1,-1)$ are perpendicular. The first affects the output; the second vanishes. In fact $(3,1)=(2,2)+(1,-1)$ and $A(3,1)^T=(4,0)^T$. Losing information means losing the second component, not losing the whole input.

Theorem 4 (Fundamental Theorem of Linear Algebra). For $A\in\mathbb R^{m\times n}$:

1. $\operatorname{null}(A)=\operatorname{range}(A^T)^\perp$.
2. $\mathbb R^n=\operatorname{null}(A)\oplus\operatorname{range}(A^T)$, an orthogonal direct sum.
3. $\dim\operatorname{range}(A)+\dim\operatorname{null}(A)=n$.
4. For an SVD $A=U\Sigma V^T$ of rank $r$, these are orthonormal bases:

An empty list is a basis for the zero subspace. The source notes that the name “Fundamental Theorem” has no universally agreed scope. Here it means these four statements. Footnote 7: part 3 alone is often called the rank–nullity theorem.

Why these spaces fit together

The source proves parts 1–3. Regard each row of $A$ as a column vector $a_i\in\mathbb R^n$. Then

The row span is the range of $A^T$ because its columns are exactly these rows. This proves part 1. The finite-dimensional orthogonal-complement result gives $S\oplus S^\perp=\mathbb R^n$ with $S=\operatorname{range}(A^T)$, proving part 2. Dimensions add in a direct sum. Since row rank equals column rank, part 3 follows.

Teaching supplement for part 4. Express $x=Vc$. Orthogonality makes this a one-to-one change of coordinates. $Ax=U\Sigma c$ vanishes exactly when the first $r$ coefficients vanish; every remaining coordinate is free. Hence the last $n-r$ columns of $V$ span the nullspace. The nonzero images $Av_i=\sigma_i u_i$ span the range. Apply the same argument to $A^T=V\Sigma^TU^T$ for the other two spaces. The source gives the table without a separate proof of this part.

What is unique?

Part 2 says every input can be written

The vectors $A^Tv$ and $w$ are unique because they are orthogonal components. The coefficient vector $v$ need not be unique: adding anything in $\operatorname{null}(A^T)$ leaves $A^Tv$ unchanged. This clarifies the source’s phrase “written uniquely.”

Apply the theorem to $A^T$ to get $\mathbb R^m=\operatorname{range}(A)\oplus\operatorname{null}(A^T)$ and $r+\dim\operatorname{null}(A^T)=m$. Input and output decompositions live in different spaces.

Transfer

If $A$ is $3\times5$ with rank two, its four subspace dimensions, in the table’s order, are $2,2,1,3$. The two nullspaces have different dimensions because one lives in the output space and the other in the input space.

Explore the relationship