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I was asked to find the inverse of matrix $A=\text{diag}(a_1,a_2,...,a_n)$

Let $B$ be the inverse of $A$, then we have to find $B$ such that $AB=I$ where $I$ is identity matrix of same order as of $A$.

I began by assuming that $B$ is diagonal matrix and by using the fact that the product of two diagonal matrices is itself a diagonal matrix with it's diagonal entries as product of corresponding diagonal entries of given matrices I found $B$ as

$$B=\text{diag}(a_1^{-1},a_2^{-1},...,a_n^{-1})$$

Here I assumed that $B$ will be a diagonal matrix. Can we prove that if product $AB=C$ is a diagonal matrix and $A$ is a diagonal matrix, then $B$ will necessary be a diagonal matrix?

If we pre multiply by inverse of $A$ , then we get $B=CA^{-1}$ and I can use the fact I mentioned above but for that I must also that $A^{-1}$ will be a diagonal matrix when $A$ is a diagonal matrix.

I searched the site and also found following answer If I have a diagonal matrix, is it necessarily the product of two other diagonal matrices?

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    $\begingroup$ You have found a matrix $B$ such that $AB=BA=I$. So you are done, $B$ is the inverse of $A$. (It does not matter how you found that matrix, by inspiration or a miracle or whatever) $\endgroup$
    – Martin R
    Oct 21 at 7:48
  • $\begingroup$ @MartinR That's true I don't doubt my proof but I wanted to tell source of my question and that's why wrote it $\endgroup$ Oct 21 at 7:50
  • $\begingroup$ Martin R wants to point out that the inverse is unique. Hence no matter which conditions you imposed on the matrix $B$, as soon as you found one which behaves like the inverse matix, it is the inverse. $\endgroup$
    – azimut
    Oct 21 at 7:55
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    $\begingroup$ @MartinR Ok so basically what I can do is that use the fact a matrix has unique inverse and show my verification which proofs that inverse of a diagonal matrix is itself a diagonal matrix $\endgroup$ Oct 21 at 7:56
  • $\begingroup$ @LalitTolani: Yes, exactly. Note that the counterexamples given in the answers below are all non-invertible matrices. $\endgroup$
    – Martin R
    Oct 21 at 7:58
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The answer to the question in the title is "no".

As a counter example, consider $0 \cdot A = 0$, where $0$ is the zero matrix (which is a diagonal matrix), and $A$ can be any (not necessarily diagonal) matrix.

However, if additionally the diagonal matrix in the product is invertible, the answer is "yes", see below.

Consider $A \cdot D_1 = D_2$ with diagonal matrices $D_1$ and $D_2$ such that $D_1$ is invertible. Then $A = D_2 \cdot D_1^{-1}$, and therefore $A$ is diagonal since the inverse of a diagonal matrix as well as the product of two diagonal matrices is again a diagonal matrix.

The other variant $D_1\cdot A = D_2$ is done analogously.

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  • $\begingroup$ Ok that's a trick counter example :-), what if I say that both are to be non-null matrices only $\endgroup$ Oct 21 at 7:48
  • $\begingroup$ @LalitTolani non-null doesn't help. It's not hard to come up with similar counter examples. What makes a difference is invertibility of the involved matrices, see my addition. $\endgroup$
    – azimut
    Oct 21 at 7:49
  • $\begingroup$ Why $D_2$ needs to be invertible? I mean if it is not invertible , then also A can be diagonal matrix, provide that it must be diagonal matrix $\endgroup$ Oct 21 at 8:04
  • $\begingroup$ @LalitTolani Since that is what you wrote :) "If product of two matrices is a diagonal matrix and one of them is diagonal matrix..." For completeness you should also consider the product the other way round, which is $D_1 \cdot A = D_2$ (which works in an analogous way). $\endgroup$
    – azimut
    Oct 21 at 8:17
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    $\begingroup$ @LalitTolani ah ok. You are right of course, that can be relaxed. I'll update my answer. $\endgroup$
    – azimut
    Oct 21 at 8:24
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Take $A=\left[\begin{array}{llll}1 & 0 \\ 0 & 0 \end{array}\right],$

$B=\left[\begin{array}{llll}1 & 0 \\ 1 & 0 \end{array}\right].$ Then $AB=A$.

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