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Does the concept of matrix similarity apply to the condition < Xv,v >?

In other words, if a real square non-symmetric matrix X is similar to a symmetric positive definite matrix, do we have < Xv,v > > 0 for all nonzero vector v?

I feel this is a trivial question, but I am a bit confused with the concept of matrix similarity.

Thank you very much.

Lio
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  • update my answer with references on general transformatioons which preserve positive definiteness, if that is what the question asks (apart from counter examples) – Nikos M. Jun 18 '14 at 19:52

3 Answers3

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Nope. If a matrix $A$ is similar to an SPD matrix $B$, this does not generally mean that $x^TAx>0$ for nonzero all $x\neq 0$ (well, obviously yes for some, namely the eigenvectors).

Consider a simple example: $$ A=\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}^{-1} =\begin{bmatrix} 1 & 0 \\ -5 & 2 \end{bmatrix}. $$ The matrix $A$ is clearly similar to an SPD matrix (a diagonal one with positive diagonal entries), but $$ x^TAx=-2<0 \quad\text{for}\quad x=[1,1]^T. $$

Algebraic Pavel
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  • Thanks. And can we find conditions on matrix A such that <Ax,x> > 0 for all nonzero x? For instance, what if A is nonnegative? – Lio Jun 15 '14 at 22:15
  • It's actually the congruence transformations ($A\mapsto P^TAP$, $P$ nonsingular), which preserve this. In particular, if $P$ is orthogonal, you get also the similarity. – Algebraic Pavel Jun 16 '14 at 09:07
  • What if $A = D B D^{-1}$, where $B$ is semi-positive definite, and $D$ is a diagonal matrix? Is A semi-positive definite? – Elias Strehle Feb 07 '17 at 16:04
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Matrix similarity is an equivalence relation between matrices in the same space.

A similarity transformation (one way to see it) transforms a vector $\vec{x}$ represented in the basis vectors of matrix $A$ to a vector $\vec{y}$ represented to the basis vectors of matrix $B$ in such a way so that the inner product (or norm) remains equal.

i.e $$<\vec{x}, A\vec{x}> = <B\vec{y}, \vec{y}>$$ (for a similarity transformation by an orthogonal matrix)

Let Q be an orthogonal matrix (i.e $Q^{-1} = Q^T$) and let $\vec{x} = Q\vec{y}$, then:

$$ x^TAx = (Qy)^T A (Qy) = y^T Q^TAQy = y^TQ^{-1}AQy = y^TBy$$

What this means intuitively is that in general for any similarity transform that is an isometry both maintain the inner product of the space (in a normed space).

For an investigation and results on general (similarity) transformations that preserve positive-definiteness check this arxiv paper

Nikos M.
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  • How come that this is true? Could you elaborate? – Algebraic Pavel Jun 15 '14 at 21:18
  • @AlgebraicPavel, for similarity transformations that transform an SPD matrix into diagonal form (eigenvalues) is clearly true for every x > 0 – Nikos M. Jun 16 '14 at 02:07
  • @AlgebraicPavel, however i just saw your answer and of course it is a clear counter-example, i am not sure this is what the OP asks, if so, i can delete this answer – Nikos M. Jun 16 '14 at 02:09
  • @AlgebraicPavel, elaborated on similarity transformations by an orthogonal matrix – Nikos M. Jun 16 '14 at 02:28
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The answer by Nikos M. can be generalized as follows:

Let $A\geq 0$, i.e. $\langle x|A|x\rangle \geq 0$ for all $|x\rangle$, and let $B$ be arbitrary. Then we have $B^\dagger A B\geq 0$:

Let $|x\rangle$ be arbitrary, and define $|y\rangle = B|x\rangle$. Then

$$\langle x| B^\dagger A B |x\rangle = (B|x\rangle)^\dagger A (B|x\rangle) = \langle y|A|y\rangle \geq 0$$

by the assumption that $A\geq 0$.

In particular, if $A$ is positive and $H$ is Hermitian, then $HAH$ is positive.

(Hope you're fine with bra-ket notation...)

Felix L.
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