Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are certain terms in the following theorem where I am finding difficulty to figure out. I need help.

Theorem. Let $\mathbb{C}_{r}^{m\times n}$ denote the set of all complex $m\times n$ matrices of rank $r$.

Let $A\in\mathbb{C}_{r}^{m\times n}$ , let $T$ be a subspace of $\mathbb{C^n}$ of dimension $s\leq r$, and let $S$ be a subspace of dimension $m-s$.

Then, $A$ has a $\{2\}$ - inverse X such that $R(X) = T$ and $N(X) = S$ iff $AT\oplus S$ = $C^{m}$, and $ P_{S}{^\perp} AT = S^{\perp}$(This too is not clear to me) where $P_{L.M}$ stands for the projection on to the space $L$ parallel to $M$.

in which case $X$ is unique. Where $\{2\}$ inverse of a matrix $A$ is the matrix $X$ satisfying the equation $XAX = A$ . $R(x)$, $N(X)$ denotes the range and null space of a matrix and $\oplus$ denotes the direct sum of subspaces.

I am confused with term $AT$.How to interpret $AT$ where $A$ is a $m\times n$ matrix while $T$ is a given subspace of $\mathbb{C^n}$. How can we multiply matrix with subspace? I try to construct such example but couldn't.I would be really great full fro any kind of help or support.Please accept my apology if my question is not suitable for this community. Here I am writing few lines of its proof.

Proof: Let the colums of $U\in \mathbb{C}_{s}^{n\times s}$ be a basis for $T$, and let the columns of $V^{*}\in\mathbb{C}_{s}^{m\times s}$ be the basis of $S^\perp$the. Then the columns of $AU$ span $AT$(How?). Since it follows from (direct sum $AT\oplus S$ = $C^{m}$) that $\dim AT = s$, $\mathrm{rank} AU = s$. A further consequence is that $AT\cap S = 0$. Moreover $s\times s$ matrix $VAU$ is nonsingular (i.e. of rank $s$).

Note that $AX$ is idempotent, $AT = R(AX)$, $S = N(X) = N(AX)$

I really need help here. A small hint will also work for me.

share|cite|improve this question
I'd imagine $AT$ is the set of vectors of the form $At$, where $t\in T$. – David Mitra May 14 '12 at 17:10
@DavidMitra Here i am using direct sum of $AT$ and $S$ that mean $AT$ must be subspace of $\mathbb C^{m}$. Is that possible in your consideration. Can i make such examples? – srijan May 14 '12 at 17:14
up vote 2 down vote accepted

Almost certainly, $AT$ is the set $\{ At\mid t\in T\}$. This will indeed be a subspace of $\Bbb C^m$. In particular, since $T$ is a subspace of $\Bbb C^n$, if $At_1$ and $At_2$ are in $AT$, then $At_1+At_2=A(t_1+t_2)$ is in $AT$. Also, for a scalar $\alpha$, we have $\alpha( At_1)=A(\alpha t_1)$ is in $AT$. (Or, just recall that matrix multiplication defines a linear mapping.)

This sort of notation is standard. If $A$ is a function, as we think of the matrix $A$ here, then $A(T)=AT$ is the image of $T$ under $A$.

share|cite|improve this answer
@ David Thanks you seems correct. I have edited my theorem. Can you take a look please? Your answer is helpful to me. – srijan May 14 '12 at 17:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.