# Finding the symmetric square roots of diagonal matrices

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.

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