Given a $4 \times 4$ symmetric matrix $A$, are there any matrices $B,C$ that: $BAC = I_{4}$ ?

I've thought of $B$ being a orthogonal matrix $P$ ($B=P$) and $ C = P^{T}$ so we get $PAP^{T} = \begin{bmatrix}\lambda_{1}&0&0&0\\0&\lambda_{2}&0&0\\0&0&\lambda_{3}&0\\0&0&0&\lambda_{4}\end{bmatrix} $
Where $\lambda_{i}$ with $i \in {1,2,3,4}$ are the eigenvalues of matrix $A$.

And then demanded that $\{\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}\} = \{1,1,1,1\}$ But I am not sure that this is correct.

Thank you for your help!

  • 1
    $\begingroup$ This is true if and only if $A$ is invertible, in which case you can take $B = I$ and $C = A^{-1}$. $\endgroup$ – Qiaochu Yuan Jun 11 '12 at 0:48
  • $\begingroup$ A is invertible ($det(A) != 0$) but is there something I can study about this? Or it's just a matter of picking the right matrices? $\endgroup$ – Chris Jun 11 '12 at 0:51
  • $\begingroup$ For any invertible $B$ you can find a $C$ such that the equality holds: $C=(BA)^{-1}$. $\endgroup$ – Generic Human Jun 11 '12 at 1:01
  • $\begingroup$ Maybe an interesting variant is to see if you can get $A=C$ and/or symmetric as well. $\endgroup$ – GEdgar Jun 11 '12 at 1:41
  • $\begingroup$ If $A$ is symmetric, then you can do this with $B = C^T$ if and only if all eigenvalues of $A$ are positive. See Sylvester's Law of Inertia. $\endgroup$ – Robert Israel Jun 11 '12 at 2:13

Expanding Qiaochu Yuan's comment, suppose that A is invertible. Then we can choose $B = I, C = A^{-1}$ so that $BAC = IAA^{-1} = I$.

On the other hand, if $C$ is not invertible, then its null space is non-trivial (this follows because its a square matrix), so there exists $x \neq 0$ such that $Cx = 0$. Hence for any matrices $A,B$ $$BACx = BA0 = 0$$

If also $BAC = I$, then there would exist $x \neq 0$ such that $Ix = x = 0$ which is a contradiction. Therefore, if the identity is to hold, then $C$ must be surjective.

Now if $A$ is not invertible then again its null space is non-trivial so there exists $x \neq 0$ such that $Ax = 0$ and since $C$ is surjective there exists $y \neq 0$ such that $Cy = x$. Hence

$$BACy = BAx = B0 = 0$$

which shows that we cannot have $BAC = I$.

Alternatively, if you have seen determinants, then $$\det(B)\det(A)\det(C) = \det(BAC) = \det(I) = 1$$ implies $A,B$ and $C$ cannot have determinant equal to $0$ and hence cannot be singular.


Multiplying a matrix left and right by freely chosen matrices can change it quite radically. Assuming you are considering matrices with entries in a field (e.g., real or complex matrices), the most useful fact to know here is that if one requires those multiplying matrices ($B,C$ in your question) to be invertible (and in particular square) then the only invariants are the size of the matrix, obviously, and its rank, that is the dimension of the space spanned by its columns. In other words for a matrix $A$ to be transformable into a matrix $A'$ by such operation is it necessary and sufficient that $A$ and $A'$ be of the same size and that they have the same rank. For a square matrix, the rank is equal to the size of the matrix if and only if the matrix is invertible (non-zero determinant). Therefore one can immediately answer your question with the additional requirement $B,C$ invertible as follows: since $I_4$ has rank $4$ it is necessary and sufficient that $A$ also be of rank $4$, that is invertible.

If you allow $B,C$ to be non-invertible, this can only lead to decreasing the rank of the product, which is not helpful if you are trying to obtain $I_4$ which has maximal rank, in other words if $A$ is not invertible then it will still be impossible to obtain $BAC=I_4$. So the answer to your question remains the same as with $B,C$ invertible. Note that $A$ being symmetric has no effect whatsoever on the answer.


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