# A mistake in the equation $H'H = \Sigma ^{-1}$

I can not find the mistake in this sequence of operations.

If $\Sigma$ is a positive definite matrix, we can write $\Sigma = C \Lambda C '$, where $C$ is orthonormal $(CC' = C'C = I)$ and $\Lambda$ is diagonal with all diagonal elements positive. Considering $\Lambda^{*}$ the diagonal matrix whose diagonal elements are the reciprocal square roots of the corresponding diagonal elements of $\Lambda$. Let $H = C \Lambda^{*} C'$. I would like to prove that $$H'H = \Sigma ^{-1}$$

First, $H' = H$ and $\Lambda = \hbox{diag}\{ d_1, ... , d_n\}$ and $\Lambda^{*} = \hbox{diag}\{ \sqrt{d_1}, ... , \sqrt{d_n}\}$. With this in hand, we have

$$H'H = H H = C \Lambda^{*} C' C \Lambda^{*} C' = C \Lambda^{*} I \Lambda^{*} C' = C \Lambda^{*}\Lambda^{*} C' = C \Lambda C' = \Sigma$$

Some ideia where is the error?

• I think you mean $\Lambda^{\star}=\text{diag}(1/\sqrt{d_1},...,1/\sqrt{d_n})$ – Lindon Sep 13 '17 at 3:55
• I am not native speaker english... so, what is the meaning of " reciprocal square root"?. For example, if $a> 0$, the reciprocal square root of a is $1/(a^{1/2})$? – orrillo Sep 13 '17 at 3:59
• @orrillo The reciprocal of a number $x$ means $\frac1x$. So, the reciprocal square root of a number $x$ means the reciprocal of the square root of $x$, i.e. $\frac1{\sqrt{x}}$. – user1551 Sep 13 '17 at 8:10

We have that: $\Sigma=C\Lambda C'$ so $$\Sigma^{-1}=(C\Lambda C')^{-1}=(C')^{-1}\Lambda^{-1}C^{-1}=C\Lambda^{-1}C'$$
If $\Lambda=diag\{d_1,...,d_n\}$ then $\Lambda^{-1}=diag\{\dfrac{1}{d_1},...,\dfrac{1}{d_n}\}$.
"The reciprocal square roots" of $a>0$ means that: $\dfrac{1}{\sqrt{a}}$
So $\Lambda^*=diag\{\dfrac{1}{\sqrt{d_1}},...,\dfrac{1}{\sqrt{d_n}}\}$. With this fact, we have:
$$HH'=HH=C\Lambda^* C'C\Lambda^* C'=C\Lambda^*\Lambda^* C'=C\Lambda^{-1} C'=\Sigma^{-1}$$