6
$\begingroup$

The pattern $XAX^\top$ where $A$ and $X$ are matrices, arises in various areas and applications. I am aware that if $A$ has special properties (definiteness, ...), then $XAX^\top$ becomes more interesting.

Does this pattern have a name? Is it studied formally in any way?

I have noticed that some matrix functions $f$ such as the Matrix Exponential can benefit from $f(XAX^\top)=Xf(A)X^\top$. When and why does this occur?

I have also noticed that when $X$ has special properties such as orthogonality, other interesting features arise.

I am asking for more experienced people to guide me towards a more formal direction of finding the answers to my questions. Thank you.

$\endgroup$
3
  • 1
    $\begingroup$ Nice question. I've also noticed it, is a really common form. Just want to say, when $x$ is a vector, it can describe also what is called a "quadratic form". And also I've seen this product in several demonstrations of propositions and so on. $\endgroup$
    – Byag
    Sep 24, 2019 at 0:43
  • 5
    $\begingroup$ Relevant: en.m.wikipedia.org/wiki/Matrix_congruence $\endgroup$ Sep 24, 2019 at 5:38
  • $\begingroup$ @TravisWillse That should be an answer. This is the key term $\endgroup$
    – a06e
    Feb 3 at 18:12

1 Answer 1

4
$\begingroup$

Let $T$ be some linear operator on a finite-dim Real inner-product space $V$, with $dim V = n$. Let the matrix $A$ you mentioned be with respect to some orthonormal basis of $V$, $u_1,\dots,u_n$. Let $B$ be another matrix realization of $T$ but with respect to another orthonormal basis of $V$, $v_1,\dots v_n$. Define a linear operator, $S$, by $Sv_i = u_i, i=1,\dots n$. Since $S$ maps an orthonormal basis to another orthonormal basis, it can be shown, that $S$ is an isometry/orthogonal operator. This means that $SS^T = I$. Define $X$ to be the matrix realization of $S$ with respect to $u_1,\dots u_n$, then $B = XAX^T$, $X$ acts as a "change-of-basis" matrix for $A$. Here we can see that we get the property that you mentioned above. This is for functions that are powers of matrices, or in general consider the Taylor expansion of a function. $\forall m \in \mathbb{Z^+}, f(C) = C^m \implies f(B) = f(XAX^T) = (XAX^T)^m = XA^mX^T$. You can try this with the Taylor expansion for the matrix-exponential. Now $A$ may not have any special properties which allows us to compute $A^m$ more easily than $B^m$ in which case this is somewhat pointless for computing $f$ more easily. But this form is called the "change-of-basis" from $u_1, \dots, u_n$ to $v_1, \dots, v_n$ for $A$.

But due to the Spectral Theorem, if $T$ is a self-adjoint operator, meaning $\forall v \in V, <Tv,u> = <v, Tu>$ (a positive-(semi)definite operator is also self-adjoint), there exists an orthonormal basis of $V$ consisting of eigenvectors of $T$, call it $w_1,\dots, w_n$. With respect to this basis, a matrix realization of $T$, $A$ is diagonal. Let $B$ be another matrix realization of $T$ with respect to some other orthonormal basis, call it $z_1,\dots z_n$. Then the operator,$S$,now defined by $Sw_i = u_i, i=1,\dots, n$ is again, an isometry. A matrix for $S$ with respect to $u_1, \dots, u_n$ is orthogonal. So we get the same form $B = XAX^T$. However, this time $A$ is a diagonal matrix, which makes powers of it easy compute. Thus the function $f$ defined above, now becomes easier to compute with $A$: $f(B) = f(XA^mX^T) = XA^mX^T$, where $A^m$ is taking the power of a diagonal matrix. $XAX^T$ is then called the eigen decomposition or spectral decomposition of $B$.

Note: the reason I specified real inner product space, is this spectral decomposition exists for normal operators on complex spaces, and there does exists a decomposition (given $B$ is with respect to an orthonormal basis), $A$ with respect to an orthonormal basis of $V$ eigenvectors of $B$. $X$ defined as mentioned earlier. $B = XAX^{-1}$, where $X$ is an isometry/unitary. However, $X^{-1}$ does not necessarily equal $X^T$. $X^{-1}$ is in fact the adjoint (denoted $T^*$, and defined by the relation $\forall v \in V, < Tv, u> = <v, T^*u>$) of $X$, or conjugate-transpose (only since $X$ is with respect to an orthonormal basis of $V$).

Thus this form, or in the more general case $XAX^{-1}$, $A$ being diagonal, $X^{-1}$ mapping a basis of $V$ to a basis of $V$ consisting of eigenvectors of $A$, makes computing functions on self-adjoint in the real case and normal operators in the complex case easy.

$\endgroup$

You must log in to answer this question.

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