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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.
Thanks

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    $\begingroup$ I'd imagine $AT$ is the set of vectors of the form $At$, where $t\in T$. $\endgroup$ May 14 '12 at 17:10
  • $\begingroup$ @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? $\endgroup$
    – srijan
    May 14 '12 at 17:14
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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 $A$.

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  • $\begingroup$ @ David Thanks you seems correct. I have edited my theorem. Can you take a look please? Your answer is helpful to me. $\endgroup$
    – srijan
    May 14 '12 at 17:55

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