# Alternative definition of positive definite matrix [duplicate]

In a text book I am following, there is a definition of positive definite matrices, which I did not see before:

$$x^TAx \geq \alpha x^Tx$$

where $$\alpha$$ is a positive scalar and $$x \in \mathbb{R^n}$$.

The usual definition I know is for every non-zero vector $$x \in \mathbb{R^n}$$, it is $$x^TAx > 0$$.

These definitions must be equivalent. It is trivial to show that the first definition implies $$x^TAx > 0$$. But I failed to see a way to show the other way around of the equivalence. I thought of applying SVD to the matrix $$A$$ and tried to find a way to show that $$x^TAx$$ is always lower bounded by a $$\alpha x^Tx$$ with $$\alpha$$ being positive, using singular values and vectors of $$A$$ but couldn't move forward. What is the correct way of approach here?

(The other question in the board assumes symmetric matrices. This one does not).

## marked as duplicate by Martin R, Hans Lundmark, Paul Frost, Jyrki Lahtonen, ChristopherNov 26 '18 at 14:15

• I think we can prove that, assuming that $A$ is PD, there exists $a$ such that $A-aI$ is PSD, that is because the eigenvalues of $A-aI$ will be bigger than those of $A$ minus $a$ (this needs to be proven) and since they were strictly positive in the first place then you can find $a$ such that they still are positive. – P. Quinton Nov 26 '18 at 7:25

Let $$f(x)=x^TAx$$. Furthermore let $$f(x)>0$$ for all non-zero $$x$$.

Let $$C:=\{x \in \mathbb R^n: ||x||_2=1\}$$. Then $$C$$ is compact and $$f$$ is continuous on $$C$$. Thus there is $$x_0 \in C$$ such that

$$f(x_0) \le f(x)$$ for all $$x \in C$$.

Now put $$\alpha =f(x_0)$$. Then $$\alpha >0$$.

It is your turn to show that $$x^TAx \geq \alpha x^Tx$$ for all $$x \in \mathbb R^n$$.

Hint: for $$t \in \mathbb R^n$$ and $$t \in \mathbb R$$ we have $$f(tx)=t^2f(x).$$

• Thank you for the answer. I think we have the following: $C$ is compact and $f(x)$ is a continuous mapping, which implies $f(C)$ is also compact, which additionally means $f(C)$ is closed and bounded. This provides us with a $x_0 \in C$ such that $\alpha = f(x_0) = \inf f(C)$ and $f(x_0) \in f(C)$. For any $x \in \mathbb{R}^n$, we have $\left(\dfrac{x}{||x||_2} \right)^T A \left(\dfrac{x}{||x||_2} \right) \geq \alpha$. Seeing that $x^Tx = ||x||_2^2$ we multiply both sides with $x^Tx$ and then we are good to go: $x^T Ax \geq \alpha x^T x$. – Ufuk Can Bicici Nov 26 '18 at 14:44
• Well done ! In $\left(\dfrac{x}{||x||_2} \right)^T A \left(\dfrac{x}{||x||_2} \right) \geq \alpha$ you should mention that $x \ne 0$. – Fred Nov 26 '18 at 14:57

The following is valid only if the matrix $$A$$ doesn't have complex values, which is at least true if it is also symmetric.

Since $$A\succ 0$$, we can use it's diagonalisation as $$A=U\Sigma U^{\dagger}$$ such that $$U U^\dagger = U^\dagger U = I$$ and $$\Sigma$$ is the diagonal matrix containing the eigenvalues of $$A$$, then for any $$a>0$$, $$A-aI = U(\Sigma - aI) U^\dagger$$ and so the eigenvalues of $$A-aI$$ are those of $$A$$ minus $$a$$.

Being positive definite means that the eigenvalues are strictly positive and so we can find $$0 where $$\lbrace\lambda_i\rbrace_{i\in[1{:}n]}$$ are the eigen values such that the eigen values of $$A-aI$$ are positive which is $$A\succ aI$$ and hence $$x^\dagger A x \geq a x^\dagger x$$

• $A$ can be nonsymmetric, so its eigenvalues do not have to be real; then how can we say that positive definite means the eigenvalues are strictly positive? – Ufuk Can Bicici Nov 26 '18 at 9:11
• Well, I guess this is one of the properties of the positive definite matrix en.wikipedia.org/wiki/… Also if an eigenvalue $\lambda$ is complex, then if $v$ is the associated eigenvector, then $v^\dagger A v = \lambda v^\dagger v$ which is a complex value and cannot be compared to $av^\dagger v$ so the problem is not solvable. – P. Quinton Nov 26 '18 at 9:17
• Interestingly the question here: math.stackexchange.com/questions/83134/… shows complex eigenvalues of a positive definite matrix. I think being symmetric is implicitly assumed in the text then. – Ufuk Can Bicici Nov 26 '18 at 9:19
• Then I guess you would need to compare the real values in the question, otherwise it doesn't make sense. And yes you are right, in the wikipedia page they assume symetric. I think Fred's way is the way to go, but you should add real values everywhere. – P. Quinton Nov 26 '18 at 9:21