# Construction of inner product from given $3\times 3$ matrix

Can the following matrix $$A=\begin{pmatrix} 3 & 1 & -5\\ 0 & 2 & 3\\ 0 & 0 & -1 \end{pmatrix}$$ be a matrix of self-adjoint operator (for some euclidean inner product space in $$\mathbb{R}^3$$)? If yes then find corresponding Gram matrix.

My solution: It is easy to compute that this matrix has $$3$$ eigenvalues, namely $$\{-1,2,3\}$$. We can find each corresponding eigenspace $$V_{3}=\langle (1,0,0)\rangle,V_{2}=\langle (-1,1,0)\rangle, V_{-1}=\langle (3,-2,2)\rangle$$. Hence this matrix is diagonalizable. And we have to find such inner product in $$\mathbb{R}^3$$ such that those eigenvectors are orthogonal. More precisely, the function $$\langle \cdot,\cdot\rangle:\mathbb{R}^3\times\mathbb{R}^3\to \mathbb{R}, \quad \langle x,y\rangle=(x_1,x_2,x_3)A\begin{pmatrix} y_1 \\ y_2\\ y_3 \end{pmatrix}$$ where $$A$$ should be symmetric and positive definite, i.e. $$A=\begin{pmatrix} a & b & c\\ b & d & e\\ c & e & f \end{pmatrix}$$. Let''s denote vectors $$(1,0,0), (-1,1,0), (3,-2,2)$$ by $$a_1,a_2,a_3$$, respectively. Since we want $$a_i\perp a_j$$ for $$i\neq j$$.

Then condition $$a_1\perp a_2$$ implies that $$b=a$$. Condition $$a_1\perp a_3$$ implies $$c=-\frac{a}{2}$$. And condition $$a_2\perp a_3$$ implies that $$e=d-\frac{3a}{2}$$. So our matrix looks like $$A=\begin{pmatrix} a & a & -\frac{a}{2}\\ a & d & d-\frac{3a}{2}\\ -\frac{a}{2} & d-\frac{3a}{2} & f \end{pmatrix}.$$ Since $$A$$ is positive definite then all upper-left submatrices should have positive determinants, i.e. it means that $$a>0, ad-a^2>0, \det A=\frac{a}{2}(d-a)(2f-2d+\frac{3a}{2})>0$$ which is equivalent to $$a>0, d>a,f>d-\frac{3a}{4}.$$

In particular If I take $$a=2, d=4$$ then $$f>\frac{5}{2}$$ and one can take $$f=4$$. So It means that Gram matrix can be taken as $$\begin{pmatrix} 2 & 2 & -1\\ 2 & 4 & 1\\ -1 & 1 & 4 \end{pmatrix}$$ which makes our operator to be self-adjoint.

• Are you asking us to verify your solution? Commented Mar 24, 2020 at 14:51
• @Omnomnomnom, yeah, please. Is there some alternative way of the solution I'd be happy to take a look!
– RFZ
Commented Mar 24, 2020 at 15:20
• Once you have an eigenbasis for $A$, let $P$ be the matrix with those vectors as columns and set $G=P^{-T}P^{-1}$.
– amd
Commented Mar 25, 2020 at 6:19
• @amd, Could you explain in detail why such choice of $G=P^{-T}P^{-1}$ works, please?
– RFZ
Commented Mar 25, 2020 at 14:59
• @amd, what do you mean by $P^{-T}$?
– RFZ
Commented Mar 25, 2020 at 15:40

As you’ve written, you’re looking for an inner product for which your eigenvectors are orthogonal. Strengthening this to having them be an orthonormal set, if $$P$$ is the matrix of eigenvectors, then this is equivalent to finding a symmetric matrix $$G$$ such that $$P^TGP$$ is diagonal, with positive diagonal entries. A simple choice for this diagonal matrix is the identity, and multiplying by inverses on both sides yields $$G=P^{-T}P^{-1}$$. Other choices for the diagonal matrix amount to scaling the eigenvectors.

For your choice of eigenvectors, this produces $$G=\begin{bmatrix}1&1&-\frac12\\1&2&\frac12\\-\frac12&\frac12&\frac32\end{bmatrix}.$$ For the $$G$$ that you came up with, we have $$P^TGP=\operatorname{diag}(2,2,6)$$.

• Your $G$ should be positive definite matrix. Buy you did not mention it at all.
– RFZ
Commented Mar 26, 2020 at 0:44
• @ZFR That’s equivalent to the diagonal elements of $P^TGP$ being positive.
– amd
Commented Mar 26, 2020 at 0:45
• I am a bit confused to be honest. Whey are they equivalent?
– RFZ
Commented Mar 26, 2020 at 1:15
• I guess that I got your point. Let me write it in order to be sure that i am right.
– RFZ
Commented Mar 26, 2020 at 1:24
• Since the columns of $P$ are eigenvectors (I denoted them by $a_1,a_2,a_3$) and they form basis of $\mathbb{R}^3$ then for any vector $x$ we have $x=\beta_1 a_1+\beta_2 a_2+\beta_3 a_3$ then $x^TGx=\sum _{i,j}\beta_i\beta_j a_i^TGa_j$ and since $P^TGP$ is diagonal then last sum is equal to $\sum \limits_{i=1}^3 \beta_i^2 \cdot [P^TGP]_{ii}>0$. Is this correct? Would be grateful if check this, please!
– RFZ
Commented Mar 26, 2020 at 1:33