# Diagonalizable by Symmetric Matrices

Can all real diagonalizable matrices with real eigenvalues be diagonalized by a symmetric matrix? That is, if $$A$$ is a real diagonalizable matrix with real eigenvalues, can we write

$$A = S D S^{-1}$$

for some real symmetric matrix $$S$$ and some real diagonal matrix $$D$$?

• An observation: if $S$ and $A$ are symmetric, then $$A^T = [SDS^{-1}]^T = S^{-1}DS = A = SDS^{-1}$$ Which means that we have $S^{-1}DS = SDS^{-1}$, which implies that $$S^4D = DS^4$$ If $A$ has distinct eigenvalues, then the above implies that $S^4$ is a diagonal matrix. – Omnomnomnom Jun 28 '19 at 17:25

Let us consider the more general setting in which $$A$$ is an $$n\times n$$ diagonalisable matrix over a field $$\mathbb F$$. By assumption, $$A$$ admits a diagonalisation $$A=VD_1V^{-1}$$. We are asking whether there exist a symmetric matrix $$S$$ and a diagonal matrix $$D$$ over $$\mathbb F$$ such that $$VD_1V^{-1} = SDS^{-1}$$. Since the two sides have the same spectrum, $$D_1=PDP^T$$ for some permutation $$P$$. Thus the equation can be rewritten as $$(VP)D(VP)^{-1} = SDS^{-1}.\tag{1}$$ We now consider two cases:
1. $$n=2$$. Then $$(1)$$ is always solvable. This is obvious if $$A$$ is a scalar matrix (so that $$D=\lambda I$$). Suppose $$A$$ has two distinct eigenvalues. Then all eigenspaces of $$A$$ are one-dimensional. Therefore, $$(1)$$ is solvable if and only if $$S = VP\Lambda\tag{2}$$ for some nonsingular diagonal matrix $$\Lambda$$. Since $$V$$ is nonsingular, it has at most two zero entries but not any zero row/column. Therefore, there always exists a permutation $$P$$ such that $$VP$$ is in one of the following forms, where $$a,b,c,d\ne0$$: $$\pmatrix{a&b\\ c&d}, \ \pmatrix{0&b\\ c&d}, \ \pmatrix{a&b\\ c&0}, \ \pmatrix{0&b\\ c&0}.$$ It follows that $$VP\Lambda$$ is symmetric when $$\Lambda=\operatorname{diag}\left(1,\frac{c}{b}\right)$$.
2. $$n\ge3$$ and $$\mathbb F$$ has at least $$n$$ elements. Then $$(1)$$ is not always solvable. First, we claim that there exists an invertible and entrywise nonzero matrix of every order $$n$$. We can prove it by mathematical induction. The base case $$n=1$$ is trivial. In the inductive step, suppose $$B\in GL_{n-1}(\mathbb F)$$ is entrywise nonzero. Since $$\mathbb F$$ has at least $$n\ge3$$ elements, some two of them, say $$b_1$$ and $$b_2$$, are nonzero. Therefore $$\det\pmatrix{B&\mathbf1^T\\ \mathbf1&b}=b\det(B)+\text{constant}$$ is nonzero for some $$b\in\{b_1,b_2\}$$ and our claim is proved. Now consider a diagonalisable matrix $$A$$ with distinct eigenvalues (such an $$A$$ always exists because $$\mathbb F$$ has at least $$n$$ elements) and suppose its eigenvectors are given by the columns of $$V=\pmatrix{1&0\\ \mathbf1&B}$$ where $$B$$ is an invertible and entrywise nonzero matrix of order $$n-1$$. As $$A$$ has distinct eigenvalues, the problem again boils down to solving $$(2)$$ for $$S,P$$ and $$\Lambda$$. However, should $$(2)$$ be solvable, $$VP$$ must be sign-symmetric (by a sign, we mean whether an entry is nonzero). Yet, this is impossible, because $$VP$$ always contains an off-diagonal zero entry on the first row but none in any row below.
The remaining case where $$n\ge3$$ and $$\mathbb F$$ has fewer than $$n$$ elements is more intricate. Since $$A$$ must contain some repeated eigenvalues, the above counterexample does not apply.
• But in this case, we can say $$A = \pmatrix{0&1\\1&1} \pmatrix{0&0\\0&1}\pmatrix{0&1\\1&1}^{-1}$$ – Omnomnomnom Jul 1 '19 at 22:35
• I think so. $\phantom{1}$ – Omnomnomnom Jul 3 '19 at 15:20