Monday, 12 September 2016

If $A$ is a symmetric and regular matrix, then $A^{-1}$ is symmetric.

Theorem: Suppose $A$ is a symmetric and regular matrix, then $A^{-1}$ is symmetric too.

Proof: $A^{-1}$ exists as A is a regular matrix.

So, we have: $AA^{-1} = I_n$ and we have to show that ${(A^{-1})}^{T} = A^{-1}$

Now, transposing everything, we obtain, because ${I_n}^{T} = I_n$ and $(AB)^{T} = B^TA^T$:

$(A^{-1})^TA^T = I_n$, but we also have:  $A^{-1}A = I_n$

So: $(A^{-1})^TA^T = A^{-1}A$

But $A$ is symmetric, thus $A^T = A$

So: $(A^{-1})^TA = A^{-1}A$

And by the theorem: $AB = CB$ and $B$ regular $\Rightarrow A = C$, we get:

$(A^{-1})^T = A^{-1}$

And this is exactly the definition of a symmetric matrix.

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