Not all matrix with positive eigenvalues is positive definite, i.e. $\mathbf{x}^\mathsf{T}A\mathbf{x}>0$ for all non zero vector $\mathbf{x}$. For example consider matrix

$$A = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix}.$$

How to prove that if we add symmetry into hypothesis then the assertion is true? That is, a symmetric matrix with positive eigenvalues is positive definite.


1 Answer 1


The fact builds upon the facts on eigenvalue and eigenvectors of symmetruc matrix. The one directly leads to the fact you asked is that: a symmetric matrix A can decomposed as $$ A = Q^TDQ$$ where Q is an orthogonal matrix and D is diagonal matrix and entries of diagonal of D is eigenvalues $\lambda_i$ of A: $d_{ii} = \lambda_i > 0$

Now $$x^TAx = x^TQ^TDQx = (Qx)^TD(Qx) = y^TDy = \sum \lambda_i y_i^2 > 0$$ Here $ y =(y_1, y_2, ..., y_n) = Qx \ne 0$ if $x \ne 0$. $0$ here is $0$ vector.

More facts like all eigenvalues of A are real value, etc.

  • $\begingroup$ Just to add information: $A$ can be decomposed as $A = Q^TDQ$ due to the Spectral Theorem, that can be obtained from the Schur Decomposition. $\endgroup$
    – Figurinha
    Commented Aug 28, 2020 at 19:41

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