3
$\begingroup$

I'm reviewing linear algebra for my exams this year, and I just encountered this problem.

For an arbitrary matrix, $\boldsymbol{A} \in \mathcal{R}^{m \times n}$, prove there must be a unique matrix $\boldsymbol{P} \in \mathcal{R}^{n \times m}$ matching the following 4 equations. $$ APA = A \\ PAP = P \\ (AP)^T = AP \\ (PA)^T = PA $$

Things will be easy if $A$ is invertible, but when $A$ is not invertible I have no ideas how to do it.

$\endgroup$
3
  • $\begingroup$ Is there anything like, $m<n$? $\endgroup$ Mar 25, 2019 at 7:38
  • 1
    $\begingroup$ @SujitBhattacharyya no. for arbitrary m, n and A. $\endgroup$
    – Shuumatsu
    Mar 25, 2019 at 8:56
  • 1
    $\begingroup$ this is the Moore-Penrose inverse $\endgroup$
    – daw
    Mar 27, 2019 at 20:44

2 Answers 2

2
$\begingroup$

There is no point in doing this only for linear maps between spaces of the form $\Bbb R^k$, so I will generalise to assuming $V,W$ are Euclidean vector spaces, and $A:V\to W$ and $P:W\to V$ are $\Bbb R$ linear maps, and instead of transpose I will take the adjoint map operation defined in this setting (and which like transpose reverses the direction of the map).

The first equation $APA=A$ implies both $PAPA=PA$ and $APAP=AP$, so $PA$ is a projector (an endomorphism equal to its own square) on $V$ and $AP$ is a projector on$~W$. (So far the second equation $PAP=P$ could have been used instead just as well.)

It is well known (and elementary) that projectors are diagonalisable with eigenvalues in $\{0,1\}$, so they are entirely determined by their eigenspaces for $\lambda=0$ and for $\lambda=1$, which is will designate by their kernel respectively image (because they are). Clearly $\def\im{\operatorname{im}}\im(PA)\subseteq\im(P)$, but since $PA$ acts as the identity on $\im(P)$ by the second equation, one also has the opposite inclusion, and $\im(PA)=\im(P)$. Similarly $\ker(PA)\supseteq\ker(A)$ is obvious, while if $v\in\ker(PA)$ then $A(v)=APA(v)=0$ by the first equation, which shows $\ker(PA)\subseteq\ker(A)$ and we can conclude $\ker(PA)=\ker(A)$. On the other side ($W$ instead of $V$) the situation is symmetric, and we conclude $\im(AP)=\im(A)$ and $\ker(AP)=\ker(P)$. So without using the last two equations, we have that $V=\ker(A)\oplus\im(P)$ and $W=\ker(P)\oplus\im(A)$, and by restriction $A$ and $P$ define linear maps between $\im(P)$ and $\im(A)$, which the equations say are inverses of each other; also both $A$ and $P$ vanish on the complementary direct summand (which is in both cases in fact their kernel).

Finally the last two equations say that $PA$ and $AP$ are symmetric endomorphisms, so their eigenspaces are mutually orthogonal: the mentioned direct sums are orthogonal direct sums.

Now if only $A$ is given, then we know the subspaces $\ker(A)\subseteq V$ and $\im(A)\subseteq W$. Our argument shows that any solution $P$ to this problem must have $\im(P)=\ker(A)^\perp$ and $\ker(P)=\im(A)^\perp$. Since $\ker(A)^\perp$ is a complementary subspace to $\ker(A)$, the restriction $f$ of $A$ to $\ker(A)^\perp$ is injective, with $\im(f)=\im(A)$, and the restriction of $P$ to $\im(A)$ can be taken to be $f^{-1}:\im(A)\to\ker(A)^\perp$. Since $P$ must vanish on $\ker(P)$ this completely determines $P$, and the map so obtained satisfies the requirements.

$\endgroup$
1
$\begingroup$

The matrix $A$ induces a decomposition $\ker A\oplus(\ker A)^\perp$ in the domain and $\mathrm{im}A\oplus(\mathrm{im}A)^\perp$ in the codomain. The core of the matrix is the square matrix $U:(\ker A)^\perp\to\mathrm{im}A, x\mapsto Ax$.

Since $AP$ is to act as an 'left-identity' on $A$ we have to define $Px=U^{-1}x$ for $x\in \mathrm{im}A$. Similarly $PA$ acts as right-identity for $A$, so $Px=0$ for $x\in(\mathrm{im}A)^\perp$.

Thus $P=\begin{pmatrix}U^{-1}&0\\0&0\end{pmatrix}$ with respect to bases for the above spaces. Then $AP$ and $PA$ are idempotent symmetric matrices. The definitions for $Px$ are forced, so this makes $P$ unique.

Edit: If you prefer using algebra only: Every matrix can be written as $Q^*UR$ where $Q,R$ are projection matrices of the type $[I,O]$. Note that $QQ^*=I$, $RR^*=I$. The equation $APA=A$ implies $Q^*URPQ^*UR=Q^*UR$; so multiplying by $Q$ and $R^*$ gives $RPQ^*=U^{-1}$.

Edit: Example to illustrate the above. Take $$A=\begin{pmatrix}-1&-1&2&3\\0&0&1&1\\2&2&1&-1\end{pmatrix}.$$ Its nullspace or kernel consists of the plane spanned by the vectors $u_3=(1,-1,0,0)$, $u_4(0,1,-1,1)$. Take the perpendicular space $(\ker A)^\perp$ spanned by, say, $u_1=(1,1,1,0)$, $u_2=(0,0,1,1)$. The image space $\mathrm{im}A$ is the plane spanned by $v_1=Au_3=(0,1,5)$ and $v_2=Au_4=(5,2,0)$. Its perpendicular space is spanned by $v_3=(2,-5,1)$ (using cross product). So the matrix of $A$ using the basis $u_i$ for the domain and $v_j$ for the codomain looks like $$\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&0\end{pmatrix}$$ So the required matrix $P=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\\0&0&0\end{pmatrix}$ with respect to the same bases in reverse. Hence $P=\frac{1}{150}\begin{pmatrix}-2 & 5 & 29 \\ -2 & 5 & 29 \\ 24 & 15 & 27 \\ 26 & 10 & -2 \\\end{pmatrix}$ with respect to the standard bases. You can check that this matrix has the desired properties.

The proof is based on the rank-nullity formula, in essence. The dimension of $(\ker A)^\perp=\dim V_1-\mathrm{nullity}(A)=\mathrm{rank}(A)=\dim(\mathrm{im}A)$. Hence every matrix $A:V_1\to V_2$ can be decomposed into three parts, where the first part $R:V_1\to(\ker A)^\perp$ is a projection, the second $U:(\ker A)^\perp\to\mathrm{im}(A)$ is invertible (I called this the core but it is not a standard name), and the third $Q^*:\mathrm{im}(A)\to V_2$ is the embedding of the image subspace into the codomain.

$\endgroup$
2
  • $\begingroup$ sry, I didn't understand your solution. Is it beyond the confines of the linear algebra that one acquires at the undergraduate level? At least my textbook doesn't cover some of the terms you mentioned(or it may be using different terms). I searched the core and domain of matrix on google, but there are few results useful. And what's this 𝑄∗𝑈𝑅 decomposition method called, I'd like to check it in case I missed something? $\endgroup$
    – Shuumatsu
    Mar 26, 2019 at 12:39
  • $\begingroup$ I edited the solution with an example. A matrix can be thought of as a function from a vector space to another, called the domain and codomain spaces. Everything in the solution should be covered at undergrad level, in particular the rank-nullity formula, and how to convert a matrix with respect to different bases. $\endgroup$ Mar 27, 2019 at 17:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .