# Matrix version of Pythagoras theorem

Can I find a solution for $C_{n\times n}$, explicitly, for the given $A_{n\times n}$ and $B_{n\times n}$ such that $AA^{T} + BB^{T} = CC^{T}$? Here $A^{T}$ denotes the transpose of $A$ and all the matrices ($A$, $B$, and $C$) are real.

I think there is a solution for $C$ as there are $n^2$ unknowns (the $n^2$ components $C$) and $n^2$ equations (comparing the components of LHS and RHS).

Note: If it can not be solved (explicitly), then one can assume mild conditions (like positive definiteness, symmetry) on $A$ and $B$.

• What is $A'$ ? The transpose of $A$ ? – Peter May 20 '18 at 18:18

This can always be done.

The following is, by and large, written under the assumption that the matrices are taken over the complex field, assuming that

$A' = A^\dagger, \tag 1$

that is, $A'$ denotes the adjoint matrix of $A$, which of course is the transpose of the complex conjugate:

$A^\dagger = (\bar A)^T; \tag 2$

all the key steps of the argument, however, are valid when restricted solely to the reals, provided we limit the unitary matrix $U$ of (10), which diagonalizes $AA^\dagger + BB^\dagger$, to be in fact orthogonal; that such an orthogonal $U$ exists in the real case is a well-known result; then we of course also have $\bar A = A$, so (2) becomes $A^\dagger = A^T$.

By using the notation (1) we may write the given equation as

$AA^\dagger + BB^\dagger = CC^\dagger; \tag 3$

we are given $A$ and $B$ and we seek $C$ satisfying (3).

Now for any complex matrix such as $A$ we observe that $AA^\dagger$ is self-adjoint:

$(AA^\dagger)^\dagger = (A^\dagger)^\dagger A^\dagger = AA^\dagger, \tag 4$

since

$(A^\dagger)^\dagger = A; \tag 5$

furthermore, $AA^\dagger$ is positive semidefinite, viz,

$x^\dagger (AA^\dagger x) = (x^\dagger A)(A^\dagger x) = (A^\dagger x)^\dagger (A^\dagger x) \ge 0. \tag 6$

Now the matrix $AA^\dagger + BB^\dagger$ also enjoys these two properties, self-adjointness and positive semi-definiteness, as may easily be see in a fashion similar to (4)-(6):

$(AA^\dagger + BB^\dagger)^\dagger = (AA^\dagger)^\dagger + (BB^\dagger)^\dagger = AA^\dagger + BB^\dagger, \tag 7$

where we have deployed (4), and also

$x^\dagger ((AA^\dagger + BB^\dagger)x) = x^\dagger(AA^\dagger x + BB^\dagger x) = x^\dagger(AA^\dagger x) + x^\dagger (BB^\dagger x) \ge 0, \tag 8$

since each term on the right of this equation is non-negative by way of (6).

At this point we pause to note once again that if $A$ and $B$ are real matrices, everything we have done so far still binds; in the real case, we simply have $\bar A = A$ and hence

$A^\dagger = (\bar A)^T = A^T \tag 9$

etc.

The matrix $AA^\dagger + BB^\dagger$, being self-adjoint, may diagonalized by a unitary matrix $U$, as is well-known; that is, there is a complex matrix $U$ such that

$U^\dagger U = UU^\dagger = I, \tag{10}$

and

$U(AA^\dagger + BB^\dagger)U^\dagger = D, \tag{11}$

where $D$ is a diagonal matrix; furthermore, $D$ is itself is self-adjoint:

$D^\dagger = (U^\dagger)^\dagger(AA^\dagger + BB^\dagger)^\dagger U^\dagger = U(AA^\dagger + BB^\dagger)U^\dagger = D \tag{12}$

(where we have made use of (7)), which implies that all the entries of $D$ are real; obviously, the off-diagonal elements, being $0$, are real; furthermore, the diagonal elements of $D$ satisfy

$D_{ii} = (D^\dagger)_{ii} = ((\bar D)^T)_{ii} = \bar D_{ii}, \tag{13}$

hence they are also all real, and since it follows from (8) and (11) that

$x^\dagger D x = x^\dagger U^\dagger (AA^\dagger + BB^\dagger)Ux = (Ux)^\dagger (AA^\dagger + BB^\dagger)Ux \ge 0; \tag{14}$

we thus find that $D$ is positive semi-definite, and from here it is but an easy step to see that the diagonal entries of $D$ obey

$D_{ii} \ge 0. \tag{15}$

By virtue of (15) we may define the diagonal matrix $\sqrt D$ such that

$(\sqrt D)_{ij} = [\delta_{ij} (\sqrt D)_{ii}]; \tag{16}$

it is clear that

$(\sqrt D)^\dagger = \sqrt D, \tag{17}$

and furthermore that

$(\sqrt D)^2 = D; \tag{18}$

thus, from (11),

$(U^\dagger \sqrt D U)(U^\dagger \sqrt D U) = U^\dagger \sqrt D UU^\dagger \sqrt D U = U\dagger \sqrt D I \sqrt D U = U^\dagger (\sqrt D)^2 U = U^\dagger D U$ $= U^\dagger (U(AA^\dagger + BB^\dagger)U^\dagger)U = (U^\dagger U)(AA^\dagger + BB^\dagger)(U^\dagger U)$ $= I(AA^\dagger + BB^\dagger)I = AA^\dagger + BB^\dagger; \tag{19}$

if we now set

$C = U^\dagger \sqrt D U, \tag{20}$

we see that $C$ is self-adjoint:

$C^\dagger = (U^\dagger \sqrt D U)^\dagger = U^\dagger (\sqrt D)^\dagger (U^\dagger)^\dagger = U^\dagger \sqrt D U = C, \tag{21}$

and (19) yields

$C^\dagger C = C^2 = ( U^\dagger \sqrt D U)^2 = AA^\dagger + BB^\dagger, \tag{22}$

as desired. We note that the procedure followed above provides a de facto recipe for explicitly computing $C$.

I think you are saying: if we know A and B can we find a C such that the equation holds.

A Hermitian Positive Definite matrix M has a Cholesky decomposition $M=CC^{\ast}$. So, if $AA^T+BB^T$ is positive definite symmetric (implying real), then a C exists that satisfies the equation.

A sufficient condition is that A and B are both positive definite.

• I think $AA^{T}$ and $BB^{T}$ are positive semi-definite matrices by construction. And the sum of two positive semi-definite matrices is also positive semi-definite. Hence I can find a $C$ using the Cholesky's decomposition of $(AA^{T} + BB^{T})$. Is the sufficient condition of positive definiteness of $A$ and $B$ necessary? – Shanks May 20 '18 at 21:05
• @Shanks I don't know if its enough to be positive semidefinite to guarantee a Cholesky Decomposition. That is, I don't know if a positive semi-definite matrix always has a Cholesky decomposition. – NicNic8 May 20 '18 at 22:16
• I think this will be a useful link: eprints.ma.man.ac.uk/1193/1/covered/MIMS_ep2008_56.pdf – Shanks May 20 '18 at 22:47

We assume that the matrices are real.

About a (false) remark by the OP; in fact, we consider a system of only $\approx n^2/2$ equations in $n^2$ unknowns, because $S=AA^T+BB^T$ is symmetric. Thus, there are an infinity of solutions (except when $A=B=0$).

In general, $\ker(A^T)\cap \ker(B^T)=\{0\}$ and $S$ is symmetric $>0$; note that there are two particular solutions that are interesting:

The symmetric $>0$ matrix $C=S^{1/2}$ and the triangular matrix $C$ obtained by the standard Choleski method.

Exceptionally, $S$ is not invertible; in this case, we can again calculate the above two matrices $C$, the second one, using the extended Choleski method. About that, the OP gave a good reference (dated 1990) by Higham. Higham is a very good mathematician who also writes very well; he has written a book on this subject that I recommend everyone read:

Handbook of writing for the mathematical sciences, Siam.

Our friend @Robert Lewis also writes perfectly; I do not know if he read this book...

Remark. Higham, as many other researchers, uses the expression "positive semidefinite". In my opinion, it's a fault of English; we should say "non-negative semidefinite"...