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The matrix $A\in Mat_n(\mathbb{R})$ is given by:

$$ A=\begin{pmatrix} 2 & -1 & -1 & 0\\ -1 & 3 & -1 & -1\\ -1 & -1 & 3 & -1\\ 0 & -1 & -1 & 2 \end{pmatrix}$$

and let $q: \mathbb{R^4} \to \mathbb{R}$ be the corresponding quadratic form defined by: $q(x)=x^TAx$ for $x\in\mathbb{R}^4$

Show that $q$ is positive definite on the orthogonal complement (with regard to the scalar product) to the null space $N(A)$ for A.


I am really uncertain if this is the right approach to the problem, but I just wanted to give it my attempt and hope that someone can guide me in the right direction.


I have calculated the null space $N(A) = span(\begin{pmatrix} 1\\ 1 \\ 1 \\ 1 \end{pmatrix})=span(\begin{pmatrix} \alpha\\ \alpha \\ \alpha \\ \alpha \end{pmatrix}) \in W$. Then I believe it is possible to form a quadratic form of $W$ by the symmetric matrix A such that: $q_{A}(W)= 2\alpha^2+3\alpha^2+3\alpha^2+2\alpha^2-10\alpha^2=0$. Which means that $q$ is negative semidefinte on $W$.

Furthermore, the positive index of inertia for $ı_+q=3$, then $q_{A}$ will be positive definite on the subspace $V$ on $\mathbb{R^4}$ and the dimension of $V$ is given by $dim⁡(V)=ı_+q=3$

$Dim⁡(N(A))=1$, in order to get the $dim⁡(V)=4$ it must be true that $dim⁡(N(A))^⊥=3$. As $q$ is negative semidefinite on $N(A)$, $q$ must be positive definite on the orthogonal complement $N(A)^⊥$

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Not necessary to go as far as orthogonal matrices for this. I am referring to Sylvester's Law of Inertia. The first column of $R$ is a basis for the kernel of $A$ as a linear transformation; the other three columns are evidently a basis for the orthogonal complement, and we see how the quadratic form acts on the orthogonal complement in the final diagonal matrix.

Let the columns of $R$ be the new basis, with $$ R = \left( \begin{array}{cccc} 1&-1&0&1 \\ 1&0&-1&-1 \\ 1&0&1&-1 \\ 1&1&0&1 \\ \end{array} \right) $$ so that $$ R^T R = \left( \begin{array}{cccc} 1&1&1&1 \\ -1&0&0&1 \\ 0&-1&1&0 \\ 1&-1&-1&1 \\ \end{array} \right) \left( \begin{array}{cccc} 1&-1&0&1 \\ 1&0&-1&-1 \\ 1&0&1&-1 \\ 1&1&0&1 \\ \end{array} \right) = \left( \begin{array}{cccc} 4&0&0&0 \\ 0&2&0&0 \\ 0&0&2&0 \\ 0&0&0&4 \\ \end{array} \right) $$ and $$ R^T A R = \left( \begin{array}{cccc} 1&1&1&1 \\ -1&0&0&1 \\ 0&-1&1&0 \\ 1&-1&-1&1 \\ \end{array} \right) \left( \begin{array}{cccc} 2&-1&-1&0 \\ -1&3&-1&-1 \\ -1&-1&3&-1 \\ 0&-1&-1&2 \\ \end{array} \right) \left( \begin{array}{cccc} 1&-1&0&1 \\ 1&0&-1&-1 \\ 1&0&1&-1 \\ 1&1&0&1 \\ \end{array} \right) = \left( \begin{array}{cccc} 0&0&0&0 \\ 0&4&0&0 \\ 0&0&8&0 \\ 0&0&0&16 \\ \end{array} \right) $$

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  • $\begingroup$ I am not sure I quite follow your method: So you form the matrix $R$ by the null spaces of A, but why are the columns a basis for the orthogonal complement? The Sylvesters Law of Inertia in my textbook is as follows: "Let $q=q_A$ be a quadratic form on $\mathbb{R^n}$, which is represented by the real symmetric matrix A. Additionally let $V = (v_1, ... v_n)$ be a basis for $\mathbb{R^n}$, and let $c_1,...,c_n \in {-1,0,1}$ be integers such that $q(x_1\cdot v_1 + ... x_n \cdot v_n) = c_1\cdot {x_1^2} + ... + c_n \cdot {x_n^2}$ so $ı_+q,ı_-q = c_i$ is equal to 1 and -1 respectively" $\endgroup$ – Mads Jeppesen May 13 '18 at 12:22
  • $\begingroup$ Wouldn't the last column be (1,-2,0,1) if the columns are the eigenvectors for the $\mathbb{R^4}$ ? $\endgroup$ – Mads Jeppesen May 13 '18 at 15:50
  • $\begingroup$ @MadsJeppesen you should calculate my $AR$ by hand. It is just a multiplication of integer matrices. $\endgroup$ – Will Jagy May 13 '18 at 16:08
  • $\begingroup$ Sorry I was unclear in that comment, I mean shouldn't the last column in $R$ be $(1,-2,0,1)^T$ as I assume $R$ is a basis for $\mathbb{R^4}$ consisting of the eigenvectors - which was found by calculating the Null space of A's eigenvalues. $\endgroup$ – Mads Jeppesen May 13 '18 at 16:18
  • $\begingroup$ @MadsJeppesen the 4 eigenspace is two dimensional. I picked a basis I liked. You can confirm that I picked eigenvectors very easily. In particular, the column vector you want is the sum of my third and fourth columns $\endgroup$ – Will Jagy May 13 '18 at 16:20

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