Questions regarding orthogonal subspaces of an $n \times m$ matrix

Question

For an $m \times n$ matrix $A$ of rank $r$,

Is it true that while the dimension of the null space and row space of $A$ add up to $n$, the union of the two vector spaces may not cover entire $R^n$? I think it is as geometrically in $R^3$, we could have one space be a plane while the other be a line that is orthogonal to that plane, but the union of the two would not cover the entirety of $R^3$.
If I take a basis of row space of $A$ ($\{v_1, v_2, .. v_r\}$) and a basis for the null space of $A$ ($\{n_1, ... n_{n-r}\}$), does vector space spanned by the union of the two basis equal $R^n$? I can see that it probably does geometrically, but is there an algebraic way to show this?

Check out $A = \begin{pmatrix}0&1\0&0\end{pmatrix}$. And you mean the sum, not the union... — amsmath, Aug 02 '19 at 06:15
The title is also misleading. There is no orthogonality involved here. — amsmath, Aug 02 '19 at 06:20
Do you mean $n \times m $ or $m \times n $? There is a contradiction there. — Javi, Aug 02 '19 at 07:37
The union of two vector spaces is almost never itself a vector space. — amd, Aug 03 '19 at 01:14

score 1 · Accepted Answer · answered Aug 02 '19 at 08:36

1

For any matrix $A_{m\times n}$, with rank $r \leq \min(m, n)$, one can show that $R(A^T) = N(A)^{\perp}$. You can see the proof here. It means row space of $A$ is orthogonal to it's null space, and $dim(R(A^T)+N(A)) = n$. i.e. the basis of row space of A, and null space of A, can span the entire $\mathbb R^n$. Any vector in $\mathbb R^n$ can be written as a linear combination of vectors in the basis of row space and null space of A. As for your first question, the union of two subspaces need not necessarily be a subspace. This is expanied in this thread here.

answered Aug 02 '19 at 08:36

Vamshi Kumar Kurva

507

Can I also approach it this way: the $n-r$ vectors in the basis for the null space of A is orthogonal to the $r$ vectors in the basis for the row space of A (assume both are subspaces of $R^n$ and $A$ has rank $r$). Hence every vector in the union of the two basis are linearly independent, and we have $n$ vectors in this union, so this set of vectors must span $R^n$? – Yandle Aug 02 '19 at 18:02
Yes. Union of set of basis vectors of row space and null space can span entire $\mathbb R^n$. This is not the same as saying union of row space and null space is $\mathbb R^n$. – Vamshi Kumar Kurva Aug 02 '19 at 18:57

Questions regarding orthogonal subspaces of an $n \times m$ matrix

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