Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ be a symmetric invertible matrix, $A^T=A$, $A^{-1}A = A A^{-1} = I$ Can it be shown that $A^{-1}$ is also symmetric?

I seem to remember a proof similar to this from my linear algebra class, but it has been a long time, and I can't find it in my text book.

share|improve this question

6 Answers 6

up vote 30 down vote accepted

In fact, $(A^T)^{-1}=(A^{-1})^T$. Indeed, $A^T(A^{-1})^T=(A^{-1}A)^T=I$.

share|improve this answer

Yes.

$$ AB=BA=I\quad\Rightarrow\quad B^TA^T=A^TB^T=I\quad\Rightarrow\quad B^TA=AB^T=I $$

share|improve this answer

Another way to see that is to recall the formula $$A^{-1} = \frac{1}{\det(A)} \mathrm{Adj}(A)^T$$ and to note that the adjoint matrix of a symmetric matrix is by construction symmetric.

share|improve this answer

Yes. The inverse $A^{-1}$ of invertible symmetric matrix is also symmetric:

\begin{align} A & = A^T \tag{Assumption: $A$ is symmetric}\\ \\ A^{-1} & = (A^T)^{-1} \quad\quad\quad\quad\tag{$A$ invertible $\implies A^T = A$ invertible}\\ \\ A^{-1} & = (A^{-1})^T \tag{identity: $(A^T)^{-1} = (A^{-1})^T$} \\ \\ & {\large \therefore}\quad \text{If $A$ is symmetric and invertible, then $A^{-1}$ is symmetric} \end{align}

share|improve this answer

You can't use the thing you want to prove in the proof itself, so the above answers are missing some steps. Here is a more complete proof. Given A is nonsingular and symmetric, show that $ A^{-1} = (A^{-1})^T $:

$$ I = I^T $$

since $ AA^{-1} = I $,

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

since $ (AB)^T = B^TA^T $,

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

since $ AA^{-1} = A^{-1}A = I $, we rearrange the left side

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

since $ A = A^T $, we substitute the right side

$$ A^{-1}A = (A^{-1})^TA $$ $$ A^{-1}A(A^{-1}) = (A^{-1})^TA(A^{-1})$$ $$ A^{-1}I = (A^{-1})^TI $$ $$ A^{-1} = (A^{-1})^T $$

and we are done.

share|improve this answer

We have two properties to use: $A^{T}=A$ and $A^{-1} $exist, here we go!

$$ A^{-1}A=I $$since $A^{-1}$exist $$ (A^{-1}A)^{T}=A^{T}(A^{-1})^{T}=A(A^{-1})^{T}=I^{T}=I $$since $A^{T}=A$ $$ A^{-1}A(A^{-1})^{T}=A^{-1}I $$left multiple by $A^{-1}$

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.