# Does the form $XAX^\top$ have a name?

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.

• 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.
– Byag
Commented Sep 24, 2019 at 0:43
• Commented Sep 24, 2019 at 5:38
• @TravisWillse That should be an answer. This is the key term
– a06e
Commented Feb 3 at 18:12

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 $$\forall v \in V, = $$ (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 $$\forall v \in V, < Tv, u> = $$) 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.