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I need to show two parts of the implication are true. First: if $A$ is $2\times 2$ and is symmetric positive definite then $trace(A)>0$ and $\det(A)>0$.

Second: if $trace(A)>0$ and $\det(A)>0$ then $A$ is symmetric positive definite.

For the first part I was thinking: $A$ is symmetric and positive definite then $A$ has its eigenvalues positive. If $A$ is $2\times 2$ then characteristic polynomial of $A$ is $x^2−x.trace(A)+\det(A)=0$ If we compute the discriminant we get $(tr(A))^2 ≥ 4.\det(A)$ Now $tr(A)$ is squared so it is positive. How do I know that $\det(A)$ is also positive?

After copper.hat's response we argue that the eigenvalues of A are all positive because $A$ is spsd and the $\det(A)$ is the product of its eigenvalues. Thus $\det(A)$ is strictly positive.

Now, how do I verify the second part?

Thanks.

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    $\begingroup$ The determinant is the product of the eigenvalues. $\endgroup$
    – copper.hat
    Mar 3, 2015 at 18:17
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    $\begingroup$ Look at the possible five canonical forms of $2\times2$ symmetric matrices. The sign of the determinant is an invariant, hence if $\det\ne0$ must be definite or indefinite. Indefinite have $\det<0$. Thus if $\det>0$ is definite and both entries in the diagonal have the same sign, which is the sign of the trace. $\endgroup$
    – Jesus RS
    Mar 3, 2015 at 18:29
  • $\begingroup$ @Jesus RS What do you mean by canonical forms? $\endgroup$
    – AVP
    Mar 4, 2015 at 18:11
  • $\begingroup$ I am interested in proving the second part of the implication. Meaning, if $A$ is symmetric and $trace(A)>0$ and $det(A)>0$ then $A$ is symmetric positive definite. $\endgroup$
    – AVP
    Mar 4, 2015 at 18:18
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    $\begingroup$ The canonical forms are $\begin{pmatrix}1&0\\0&1\end{pmatrix}$, $\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$, $\begin{pmatrix}-1&0\\0&0\end{pmatrix}$, $\begin{pmatrix}1&0\\0&-1\end{pmatrix}$. Any symmetric $2\times 2$ matrix is congruent-similar to one of these. $\endgroup$
    – Jesus RS
    Mar 4, 2015 at 20:07

3 Answers 3

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The classification of symmetric $2\times 2$ real matrices (or bilinear symmetric $2$-forms, or quadratic $2$-forms) through trace and determinant can be obtained in different ways, depending on the machinery one accepts. From more to less:

1) Spectral theorem. Then one knows the classification is done through eigenvalues. For instance positive semidefinite means two positive eigenvalues $\lambda>0,\mu>0$, which is equivalent to $\lambda\cdot\mu>0,\,\lambda+\mu>0$, that is determinant and trace both positive. Honestly, I think that for $2\times 2$ matrices this is too heavy.

2) Canonical forms. Any symmetric $2\times 2$ real matrix $A$ is equivalent to one of the following five canonical forms $$ \begin{pmatrix}1&0\\0&1\end{pmatrix},\, \begin{pmatrix}-1&0\\0&-1\end{pmatrix},\, \begin{pmatrix}1&0\\0&0\end{pmatrix},\, \begin{pmatrix}-1&0\\0&0\end{pmatrix},\, \begin{pmatrix}1&0\\0&-1\end{pmatrix}. $$ The matrix $A$ shares with its canonical form the sign of the determinant (including being $0$). Thus we see that $\det>0$ immediately gives $A$ definite, and it remains to distinguish whether $A$ is positive or negative. In any case, the two entries in the diagonal of $A$ have the same sign, hence the sign of their sum, which is the trace of $A$. Thus $\det(A)>0$, tr$(A)>0$ means positive definite.

3) Nothing. In other words just from the definition. Let $A= \begin{pmatrix}a&b\\b&c\end{pmatrix}$.

Then the corresponding quadratic form is $q(x,y)=ax^2+2bxy+cy^2$, and we have to study the sign variations of this function for $(x,y)\ne(0,0)$. For instance $f(x,0)=ax^2>0$ if and only if $a>0$. Then if $y\ne0$ we can write: $$ \frac{1}{y^2}q(x,y)=at^2+2bt+c=P(t),\quad t=\frac{x}{y}, $$ and we discuss the signs of $P(t)$. For $t$ big enough, $P(t)>0$, since $a>0$. Then $P(t)>0$ for all $t$ means the polinomial has no zero, that is, its discriminant is negative, which gives $$ 0>\varDelta=b^2-ac=-\det(A). $$ And we get the condition $\det(A)>0$. Thinking this over one realizes this characterizes being positive semidefinite (that is, is a back and forth argument). And trace? Since $0<\det=ac-b^2$, and $a>0$, necessarily $c>0$ and trace$=a+c>0$.

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    $\begingroup$ For part 3): in other words, geometrically, fix $y$ at any non-zero value, taking a cross-section of the 2D quadratic form surface, look at the resulting 1D quadratic as $x$ varies, find its minimum, and make sure that it's zero or greater. (A more visual way of saying the same thing about the discriminant.) $\endgroup$ Sep 30, 2016 at 2:13
  • $\begingroup$ What isn't quite proved here is 2nd part in number 3 of Jesus RS' answer, I think? Added an answer. $\endgroup$
    – BCLC
    Nov 13, 2020 at 15:37
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What isn't quite proved here is 2nd part in number 3 of Jesus RS' answer, I think?

Anyhoo:

Proof:

  1. Real symmetric is orthogonally diagonalisable. (I think we use just diagonalisable._ Let $A=QDQ^T$.

  2. We want to show $x^TAx=v^TDv > 0$ if $v=Q^Tx \ne 0$ (iff $x \ne0$ since $Q^T$ is invertible (since $Q^T$ is orthogonal. again, I think just diagonalisable is fine))'.

  3. For $D=diag[d_1,d_2]$, we have $v^TDv=d_1v_1^2+d_2v_2^2$.

  4. We have $0 < tr(A)=tr(D) = d_1 + d_2$ and $0 < \det(A)=\det(D) = d_1 d_2$, i.e. $d_1,d_2>0$.

  5. By (3) and (4), we have $v^TDv > 0$ if $v \ne 0$. QED

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  • $\begingroup$ I don’t really understand what’s missing... and the point in this part was to use nothing. I show first that PD implies det>0 & tr>0 (just discussing a second degree equation). For the converse: det>0 implies ac>0, plus tr>0 gives a>0, and then read backwards the discussion of P>0. $\endgroup$
    – Jesus RS
    Nov 13, 2020 at 18:22
  • $\begingroup$ @JesusRS You mean something like all the steps in your part 3 are reversible? $\endgroup$
    – BCLC
    Nov 14, 2020 at 9:52
  • $\begingroup$ That’s it (I believe...) $\endgroup$
    – Jesus RS
    Nov 15, 2020 at 11:53
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Well, this question is old but it seems to me that nobody answered the original question even if it has been seen 34 thousand times. Above answers use methods different from the question. However, it is quite simple to complete @AmirHosein reasoning without any quadratic form nor linear algebra theory :

Denote $d = \det(A), t = trace(A), \Delta = t^2-4d$. The equation roots are $\lambda_1 = (t+\sqrt{\Delta})/2, \lambda_2 = (t-\sqrt{\Delta})/2$.

We want $\lambda_1>0$ and $\lambda_2>0$, which is equivalent to $\lambda_1 \lambda_2>0$ and $\lambda_1 +\lambda_2>0$.

But $\lambda_1 \lambda_2 = d$ therefore $d>0$.

And $\lambda_1 + \lambda_2 = t$ therefore $t>0$.

We proceeded by equivalence therefore:

$\lambda_1$ and $\lambda_2>0$ is equivalent to $d>0$ and $t>0$.

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