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Let $D=\text{diag}(d_1,\dots,d_n)$ be a real diagonal matrix, where $0\le d_1 \le d_2 \le \dots \le d_n$. Let $a_1 < a_2 < \dots < a_m$ be its distinct eigenvalues (counted without multiplicities).

Now, let $A$ be a real symmetric $n \times n$ matrix, satisfying $A^2=D$. Is it true that $A$ must be of the form

$A = \begin{pmatrix} \sqrt{a_1} B_1 & & & & 0 \\ & \sqrt{a_2} B_2 & & & \\ & & \sqrt{a_3} B_3 & & \\ & & & \ddots & \\ 0 & & & & \sqrt{a_m} B_m \end{pmatrix} \;\;,\;\; $ where $B_i$ are symmetric and $B_i^2 = I$?

The size of $B_i$ should be the multiplicity of the value $a_i$ as an eigenvalue of $D$.

I basically ask whether $A$ should have a block-structure corresponding to the different eigenvalues.

I tried to orthogonally diagonalize $A$, but couldn't proceed.

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If one writes $D$ in the form of $a_1I_{k_1}\oplus\cdots\oplus a_mI_{k_m}$, from $AD=DA$ one can derive that $A$ has a block-diagonal structure conforming to the partitioning of $D$. Since each diagonal sub-block must squares to some $a_jI_{k_j}$, the result follows.

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  • $\begingroup$ Thanks. How do you deduce that $AD=DA$ from the assumption $A^2=D$? $\endgroup$ Commented Jan 1, 2020 at 16:30
  • $\begingroup$ @AsafShachar $AD=A^3=DA$. More generally, $D$ is a polynomial in $A$. Hence it commutes with $A$. $\endgroup$
    – user1551
    Commented Jan 1, 2020 at 16:31
  • $\begingroup$ Thanks, this is clear now. $\endgroup$ Commented Jan 1, 2020 at 16:33

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