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Some Background: There are many simple results in matrix theory which mimic and/or stem from analogous facts which hold for real or complex numbers. For example, complex numbers of the form $1 + ri$, $r \in \Bbb R$, are always invertible in $\Bbb C$: $(1 + ri)^{-1} = (1 - ri)/(1 + r^2)$. This problem shows how the invertibility of $1 + ri$ generalizes to allow $r$ to be replaced by a symmetric real matrix $A$, thus verifying a useful property which carries over to the matrix case. Note Added by Robert Lewis, 30 August 2018, 9:58 AM PST.

I have prove it in $2 \times 2$ case. but to generalize it $n \times n$ case.

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  • $\begingroup$ Please include your proof for the $2\times 2$ case, so we can see where it might generalize. $\endgroup$ Aug 30 '18 at 15:45
  • $\begingroup$ Is $i=\sqrt {-1}$ here? What if we take $A=i\times I$? $\endgroup$
    – lulu
    Aug 30 '18 at 15:45
  • $\begingroup$ I edited your post a little to make the $\LaTeX$ work. Remember to put "\$" around your math! Thus "\$\theta\$" yields $\theta$! Cheers! $\endgroup$ Aug 30 '18 at 15:49
  • $\begingroup$ Perhaps by saying $A$ is "symmetric" you mean that it is real symmetric. Otherwise I do not think you would have been able to prove it even in the $2\times 2$ case. If you did prove that special case, it would improve your Question to include at least a sketch of what you did. $\endgroup$
    – hardmath
    Aug 30 '18 at 15:51
  • $\begingroup$ Also, in re. $\LaTeX$: "\$\times\$ is $\times$! so "\$ 2 \times 2 \$" gives $2 \times 2$! $\endgroup$ Aug 30 '18 at 15:54
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I guess that $A$ is a real symmetric, so it can be written as $P^TDP$ for an orthogonal matrix $P$ and a diagonal $D$. So $B=I+iA=PP^T+iPDP^T=P(I+iD)P^T$. $I+iD$ is obviously a regular diagonal matrix so it has a inverse $D'$. Now $PD'P^T$ is the inverse of $B$.

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$I+iA = i (A -iI)$ is invertible if and only if $i$ is not an eigenvalue of $A$. But the eigenvalues of real symmetric matrices are real.

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  • $\begingroup$ how you can say the if and only if condition please say Robert Israel $\endgroup$
    – Priya Dey
    Aug 30 '18 at 16:27
  • $\begingroup$ Definition of eigenvalue. $\endgroup$ Aug 30 '18 at 17:17
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This is arguably the closest you can get to the proof that $1+i r$ is invertible with $r$ with $r$ a real number:

The inverse of $B = I+A$ is given by $$B^{-1} = (I+A^2)^{-1} (I-i A)$$ provided that $(I+A^2)$ is invertible.

The question thus reduces to the showing that $(I+A^2)$ is invertible. But this is the case as $$I +A^2 = I + A^T A = I+ M.$$ Since $M$ is positive semidefinite, $I+M$ is positive definite; thus invertible.

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  • $\begingroup$ Why is this "the closest you can get etc."? Inquiring minds want to know . . . $\endgroup$ Aug 30 '18 at 17:40
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    $\begingroup$ @RobertLewis: I understand that there is a certain level of arbitrariness in the statement "closest" that is why I have edited the answer. However, in the argument with number we need to show that $1+r^2$ is invertible whereas here we need to show that $I+A^2$ is invertible. In the former case, we argue that $1+r^2$ is positive, here we argue that $I+A^2$ is positive definite (I cannot see how one can get closer to the argument with numbers). $\endgroup$
    – Fabian
    Aug 30 '18 at 17:47
  • $\begingroup$ Fair enough. True enough. A judgment call, for sure. Cheers! $\endgroup$ Aug 30 '18 at 18:03
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One doesn't really need the machinery of eigenvalues, eigenvectors, and diagonalization to get at this, viz:

I assume $A$ is a real symmetric matrix, that is $A \in M_n(\Bbb R)$ and

$A^T = A; \tag 1$

that is the meaning of "symmetric" in the usual parlance, as I understand it.

So if

$B = I + iA, \tag 2$

then we have the hermitian adjoint of $B$:

$B^\dagger = (B^T)^\ast = I - iA^T, \tag 3$

where ${}^\ast$ denotes the entry-wise complex conjugate of a matrix, thus:

$B^\ast = [b_{ij}^\ast]; \tag 4$

then

$B^\dagger B = (I - iA^T)(I + iA) = (I - iA) (I + iA)$ $= I^2 + iA - iA -(iA)^2 = I + A^2 = I + A^TA; \tag 5$

then for any $n$-vector $x$,

$\langle x, B^TBx \rangle = \langle x, (I + A^TA)x \rangle = \langle x, x \rangle + \langle x, A^TA x \rangle; \tag 6$

now

$\langle x, A^TA x \rangle = \langle Ax, Ax \rangle = \Vert Ax \Vert ^2 \ge 0, \tag 7$

so

$\langle x, B^\dagger B x \rangle = \langle x, x \rangle + \langle x, A^TAx \rangle \ge \langle x, x \rangle; \tag 8$

it follows that $B$ cannot be singular; if it were, then for some $y \ne 0$ we would have

$By = 0; \tag 9$

thus, via (8), a contradiction arises:

$\langle y, B^\dagger By \rangle = 0 \Longrightarrow \Longleftarrow \langle y, y \rangle > 0; \tag{10}$

it follows that $B$ is non-singular, $\ker B \ne 0$, and hence $B$ is invertible.

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