91Ó°ÊÓ

Let $\mathrm{A}={\mathrm{P}}^{\mathrm{T}}\mathrm{DP}$and ${\mathrm{Q}}^{\mathrm{T}}\mathrm{EQ}$be orthogonal diagonalizations of A and B.

Assume that the diagonal entries of D and E are in no increasing order. That is, ${\mathrm{d}}_1â‰\yen {\mathrm{d}}_2â‰\yen …â‰\yen {\mathrm{d}}_{\mathrm{n}}$and ${\mathrm{e}}_1â‰\yen {\mathrm{e}}_2â‰\yen …â‰\yen {\mathrm{e}}_{\mathrm{n}}$. We have ${\mathrm{A}}^3={\mathrm{P}}^{\mathrm{T}}{\mathrm{D}}^3\mathrm{P}$and ${\mathrm{B}}^3={\mathrm{Q}}^{\mathrm{T}}{\mathrm{E}}^3\mathrm{Q}$.

Now, ${\mathrm{A}}^3={\mathrm{B}}^3$implies that ${\mathrm{P}}^{\mathrm{T}}\mathrm{DP}={\mathrm{Q}}^{\mathrm{T}}{\mathrm{E}}^3\mathrm{Q}$. Rewrite this as ${\mathrm{D}}^3\left({\mathrm{QP}}^{\mathrm{T}}\right){\mathrm{E}}^3\left({\mathrm{QP}}^{\mathrm{T}}\right)$.

Let $\mathrm{R}={\mathrm{QP}}^{\mathrm{T}}$. Then, R is orthogonal and role="math" localid="1659678611118" ${\mathrm{D}}^3={\mathrm{R}}^{\mathrm{T}}{\mathrm{E}}^3\mathrm{R}$
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Since ${\mathrm{D}}^3$is diagonal, this means ${\mathrm{D}}^3$is diagonalization of ${\mathrm{E}}^3$. But since ${\mathrm{E}}^3$itself is diagonal and since the entries of ${\mathrm{D}}^3$and ${\mathrm{E}}^3$are in the non-increasing order, we get ${\mathrm{D}}^3={\mathrm{E}}^3$.

This implies ${\mathrm{d}}_{\mathrm{i}}^3={\mathrm{e}}_{\mathrm{i}}^3$for $1≤\mathrm{i}≤\mathrm{n}$which means that role="math" localid="1659678803573" ${\mathrm{d}}_{\mathrm{i}}={\mathrm{e}}_{\mathrm{i}}$for $1≤\mathrm{i}≤\mathrm{n}$.

This implies for which means that for .

Hence, $\mathrm{D}=\mathrm{E}$. Thus, the above equation becomes

${\mathrm{D}}^3={\mathrm{R}}^{\mathrm{T}}{\mathrm{D}}^3\mathrm{R}$

Let $\mathrm{R}=\left({\mathrm{r}}_{\mathrm{ij}}\right)$be the matrix mentioned above. Let X be a diagonal matrix and ${\mathrm{x}}_{\mathrm{i}}$be its the ${\mathrm{i}}^{\mathrm{th}}$diagonal entry. If the relation $\mathrm{X}={\mathrm{R}}^{\mathrm{T}}\mathrm{XR}$holds, then we can infer a bit more about R.

$$\begin{gathered} \mathrm{X}={\mathrm{R}}^{\mathrm{T}}\mathrm{XR}⇔\mathrm{RX}=\mathrm{XR} \\ ⇔{\mathrm{x}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{ij}}={\mathrm{x}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{ji}}\left(\mathrm{after}\mathrm{expanding}\mathrm{the}\mathrm{matrices}\right) \\ ⇔{\mathrm{r}}_{\mathrm{ij}}={\mathrm{r}}_{\mathrm{ji}} \\ ⇔\mathrm{R}\mathrm{is}\mathrm{a}\mathrm{symmetric}\mathrm{matrix}. \end{gathered}$$

Therefore, equation implies that R is symmetric.

Since D is diagonal, using the argument in the reverse way, we get $\mathrm{D}={\mathrm{R}}^{\mathrm{T}}\mathrm{DR}$. That is, $\mathrm{D}={\left({\mathrm{QP}}^{\mathrm{T}}\right)}^{\mathrm{T}}\mathrm{E}\left({\mathrm{QP}}^{\mathrm{T}}\right)$(since D = E).

This gives ${\mathrm{P}}^{\mathrm{T}}\mathrm{DP}={\mathrm{Q}}^{\mathrm{T}}\mathrm{EQ}$. which means A = B.

The given statement is TRUE.