# Is it possible to find an orthogonal matrix $V\in M_n(\Bbb R)$ s. t. $A=VDV^T$ with a column not proportional to any column of $U$?

Let $$A\in M_n(\Bbb R)$$ be a symmetric matrix with (strictly) less than $$n$$ distinct eigenvalues. Since $$A$$ is diagonalizable, we can write it as $$A=UDU^T$$ where $$U\in M_n(\Bbb R)$$ is orthogonal and $$D\in M_n(\Bbb R)$$ is diagonal.

Question:

Is it possible to find an orthogonal matrix $$V\in M_n(\Bbb R)$$ s. t. $$A=VDV^T$$ under the condition that at least one column of $$V$$ isn't proportional to any column of $$U$$?

My thoughts:

I think the fact there are less than $$n$$ distinct eigenvalues guarantees it is possible to find such $$V$$, otherwise, it would be impossible.

Since there are less than $$n$$ distinct eigenvalues, there is an eigenspace $$E_{\lambda'}$$ corresponding to the eigenvalue $$\lambda'$$ s. t. $$\dim\left(E_{\lambda'}\right)=k\geqslant 2$$.

Let $$\{e_1,\ldots,e_k\}$$ be an orthonormal basis for the eigenspace $$E_{\lambda'}$$ and let's observe one plane in $$\Bbb R^n$$ spanned by, say, $$M=\operatorname{span}\{e_1,e_2\}$$.

Let $$f_1=\frac{e_1+e_2}{\left\|e_1+e_2\right\|}$$. Then $$f_2\in M$$ is another unit vector (in the same plane) s. t. $$f_1\perp f_2$$.

Actually, we could apply Gramm-Schmidt to an arbitrary basis written as $$\{\alpha e_1+\beta e_2,\gamma e_1+\delta e_2\},\alpha,\beta,\gamma,\delta\in\Bbb R$$.

I thought I could also reach the same result by rotating $$e_1$$ and $$e_2$$ in the plane $$M$$ for some angle $$\varphi\ne k\pi,k\in\Bbb Z$$.

If this part of my statement holds, then, of course, $$\{f_1,f_2,e_3,\ldots,e_k\}$$ is also an orthonormal basis for $$M$$. I believe this could inductively hold for any $$M\leqslant E_{\lambda'}$$, where $$2\leqslant\dim M\leqslant\dim E_{\lambda'}$$.

May I ask for verification of the statement and advice on how to concisely (dis)prove it?

I recommend the following approach. First, note that $$A = VDV^T = UDU^T \implies\\ VDV^T = UDU^T \implies\\ U^TVDV^TU = U^TUDU^TU \implies\\ (U^TV) D(U^TV)^T = D.$$ With that in mind, let $$W$$ denote the orthogonal matrix $$W = U^TV$$. We have $$WDW^T = D \implies WD = DW.$$ In other words, $$W$$ is an orthogonal matrix for which $$WD = DW$$. Keep in mind that once we have $$W$$, we have $$W = U^TV \implies V = UW$$.
Now, $$A$$ has a repeated eigenvalue; call this eigenvalue $$\lambda$$. Without loss of generality, suppose that $$\lambda$$ comes first among the diagonal entries of $$D$$, and write $$D = \pmatrix{\lambda I_k & 0\\0 & D'}$$ where $$I_k$$ is a size $$k$$ identity matrix (with $$k \geq 2$$) and $$D'$$ is also diagonal. I claim that if $$W_1$$ is any $$k \times k$$ orthogonal matrix an $$W_2$$ is diagonal with $$\pm1$$'s, then the block matrix $$W = \pmatrix{W_1 & 0\\0 & W_2}$$ will be orthogonal and satisfy $$WD = DW$$. Let's stipulate that for our choice of $$W$$, $$W_1$$ has no zero-entries.
Now, note that the entries of $$W$$ are the dot-products of columns of $$U$$ with columns of $$V$$. With that in mind, conclude that because the first column of $$W$$ has $$k \geq 2$$ non-zero entries, the first column of $$V$$ is not a multiple of any of the columns of $$U$$.
• First, thank you for the answer! I' ll ask, just in case, to revise: $W=U^TV$, so rows of $U^T$ are columns of $U$, as you mentioned in the last paragraph. If the first column of $V$ were a multiple of any column of $U$, the dot product of the first column of $V$ with any other column of $U$ would be $0$, right? To show $W$ and $D$ commute, we can mulitply the block matrices ($D$ and $W$) as if their blocks were elements, since the sizes of blocks are appropriate? It's clear now. Could we interpret $W_1$ as a rotation in the eigensace corresponding to the eigenvalue $\lambda$? – Invisible Aug 20 '20 at 5:35
• @Invisible Yes to all of those, with the caveat that $W_1$ would generally be a combination of rotations and possibly a reflection. – Ben Grossmann Aug 20 '20 at 5:42