My question is:

Prove that the matrix

$A=\begin{bmatrix} 1 & \alpha \\ 0& 1 \end{bmatrix}$

is positive definite in $\Re ^{2}$ for any |$\alpha$|<2.

Here is my reasoning:

Apparently there are several definitions out there regarding if a matrix is positive definite or not. One of the definitions that I read was that it must be a symmetric matrix and have positive eigen values.

There is another definition out there which says that the matrix must be a linear operator on a finite dimensional inner product space. Where the operator is self adjoint and $\left \langle T(x),x \right \rangle>0$.

Now I understand that a matrix is a linear operator, but I do not understand the adjoint aspect.

Also when I attempt to get the eigen values of the matrix, it seems to me that alpha doesn't even matter as the zero element in the matrix will always make alpha zero. And according to the one definition I stated I thought that the original matrix must be a symmetric matrix in the first place an order to be positive definite.

Any help with this?

  • $\begingroup$ A real quadratic form $X^tAX$ is positive definite if and only if all the principal minors of a is positive and also it is positive definite if and only if all the eigenvalues of A are positive. $\endgroup$ Aug 14, 2017 at 19:31
  • $\begingroup$ If $V = \mathbb R^n$, and $T:V\to V$ is a linear operator, then $T$ is a matrix, and self-adjointness corresponds to (conjugate) symmetry. You'll have to use a definition of definiteness that doesn't make use of symmetry. $\endgroup$
    – DominikS
    Aug 14, 2017 at 19:32

1 Answer 1


HINT The common definition is indeed that $<T(x), x> \ge 0 \ \forall x \in \mathbb{R}^n$. So for you, this means $(x,y) A (x,y)^T \ge 0$ for all $x,y \in \mathbb{R}$. Note that $$ (x,y) A (x,y)^T = (x,y) \cdot (x+\alpha y, y) = x(x+ \alpha y) + y^2 = x^2+ y^2 + \alpha xy $$ Can you take it from here?


Then, assuming $x\ne 0$ and $y \ne 0$ (in which case it is true for all $\alpha$), we have $$ 0 \le x^2+ y^2 + \alpha xy \iff -\alpha xy \le x^2 + y^2 \iff \alpha \ge \frac{x^2+y^2}{xy}. $$

Can you take if from here?

  • $\begingroup$ But if I set your result here to greater than zero we have 3 unknowns and 1 equation so I don't see how to get to the $|\alpha|$ <2 part. $\endgroup$
    – Erock Brox
    Aug 14, 2017 at 19:44
  • $\begingroup$ @ErockBrox see the update, can you finish it? $\endgroup$
    – gt6989b
    Aug 15, 2017 at 2:40
  • $\begingroup$ Yes, I got that far on my own, but could not figure out how to get the value of $\alpha$ < 2 from that result. $\endgroup$
    – Erock Brox
    Aug 15, 2017 at 21:26
  • $\begingroup$ Well I graphed the function (x^2+y^2)/(xy) and it said that it's range its (-infinity, -2) U (2, infinity) $\endgroup$
    – Erock Brox
    Aug 15, 2017 at 21:37
  • 1
    $\begingroup$ You could also use $x^2+y^2+\alpha xy = \frac{1}{4}\left((2+\alpha)(x+y)^2+(2-\alpha)(x-y)^2\right).$ $\endgroup$ Aug 16, 2017 at 20:43

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