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We have the matrix $$\begin{pmatrix}1&1&0\\1&1&1\\0&1&1\end{pmatrix}.$$ I learned the terms positive definiteness and signature only recently, so I would really appreciate it if somebody could show me if this matrix if positive definite and what its rank and signature are. Thanks in advance.

Edit: Can somebody show me how to compute the rank and signature of this matrix specifically?

4 Answers4

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By Silvester's criterion a symmetric matrix is positive definite if and only if all of the leading principal minors have positive determinant. Since the $2\times 2$ principal minor has determinant zero, the matrix is not positive definite.

Dietrich Burde
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A matrix is positive definite if it's determinant is positive. In this matrix the determinant is negative so the matrix is not positive definite.

Signature: Since the determinant is negative, so this matrix has atmost one negative eigenvalue.

Rank: Rank is equal to the number of non-zero eigenvalues (counting multiplicity)

Kushal Bhuyan
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  • In fact, all determinants of the leading principal minors must be positive, in particular the determinant. But the determinant of a $2\times 2$-matrix is easier to compute. – Dietrich Burde Nov 16 '15 at 16:30
  • A matrix is positive definite if all its eigenvalues are positive. Any matrix with an even number of negative eigenvalues will have a positive determinant. – JessicaK Nov 16 '15 at 16:31
  • You both are correct. @JessicaK – Kushal Bhuyan Nov 16 '15 at 16:33
  • It is obvious that $1$ is an eigenvalue of this matrix. This is not negative definite. – JessicaK Nov 16 '15 at 16:41
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Here is another option how to decide on the definiteness of the matrix simply by definition. Since $[1,-1,1]A[1,-1,1]^T=-1<0$ and $[1,1,1]A[1,1,1]^T=7>0$, the matrix is indefinite.

Algebraic Pavel
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Call your matrix $A$. It has at least one positive eigenvalue because its trace is positive. It follows that $A$ has exactly one negative eigenvalue and two positive eigenvalues because $\det A<0$. Hence $A$ is not positive definite and it has full rank.

user1551
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