Uniqueness of a $2\times 2$ real symmetric matrix under certain conditions.

Question

Let $a$ and $b$ be real numbers. We are to show that there exists a unique $2\times 2$ real symmetric matrix $A$ with $\operatorname{tr}(A)=a$ and $\det(A)=b$ if and only if $a^2=4b$.

Now I know that if $\lambda_1$ and $\lambda_2$ are eigenvalues of a real $2\times 2$ symmetric matrix $A$ with $\operatorname{tr}(A)=a$ and $\det(A)=b$ then $\lambda_1+\lambda_2=a$ and $\lambda_1\lambda_2=b$. But I am absolutely lost while proving uniqueness when $a^2=4b$. I am also unable to use the property that $A$ is symmetric. Please help. This is a Masters level entrance examination question.

Answer 1

We can use Cayley-Hamilton equation for checking uniqueness (is it really unique?) when $a^2=b/4$.

$A^2-\text{tr}(A)A+\det(A)I=0$

With $a^2=b/4$ we have

$A^2-a A+(a^2/4)I=0$

$4A^2- 4aA +a^2I=(2A-aI)^2=0$

Now we know that $A$ is symmetric so $2A-aI$.

Which symmetric matrix squared would give $0$ ?
Only $A=aI/2$ i.e. multiply of identity which is also constrained with determinant.

Solution is not unique.. there are plenty of such symmetric matrices type $cI$.

But if $a$ is specified there is of course only one matrix of type $cI$.

Answer 2

Hint: Since $A$ is symmetric, it is diagonalizable. Write $$A=PDP^{-1}$$ with $P$ invertible. Show that an unique $A$ exists if and only if $D$ commutes with all the invertible matrices.

Answer 3

Here is a simpler but not enlightening approach. Let $A=\pmatrix{x&y\\ y&z}$. For the on "only if" part, note that $B=\pmatrix{z&-y\\ -y&x}$ has the same trace and determinant as $A$. For the "if" part, when $\operatorname{tr}(A)^2=4\det(A)$, what are the implications for the values of $x,y,z$?

Answer 4

The "only if" part (That is, we are considering that $A$ is unique)

To Show: $a^2=4b$ (under the conditions given)

Consider $A$ to be a real symmetric matrix such that $$ A = \left[ \begin{array}{cccc} \alpha_1 & \alpha_2 \\ \alpha_2 & \alpha_3 \\ \end{array} \right] $$
($\alpha_1,\alpha_2,\alpha_3,\alpha_4 $ are all real)

Now, $trace(A)=\lambda_1+\lambda_3=a$
$|A|=\alpha_1\times\alpha_3-{\alpha_2}^2=b$

If, $\lambda_1,\lambda_2$ be the characteristic root of $A$ then the characteristic equation of $A$ is of the form

$(\lambda-\lambda_1)(\lambda-\lambda_1)=0$
$\Rightarrow \lambda^2-(\lambda_1+\lambda_2)\lambda+\lambda_1\lambda_2=0$
$\Rightarrow \lambda^2-tr(A)\lambda+|A|=0$
$\Rightarrow \lambda^2-a\lambda+ b=0$

By Cayley-Hamilton Theorem,
$A^2-tr(A)\lambda+|A|=0$
$\Rightarrow A^2-(\lambda_1+\lambda_2)A+|A|=0$
$\Rightarrow (A-\lambda_1I)(A-\lambda_2I)=0$

Since $A$ is unique we must have $\lambda_1=\lambda_2$

That is, the characteristic equation should have equal roots.
By applying theory of equations, we have,
the characteristic equation has equal roots if
$a^2-4b=0$
$\Rightarrow a^2=4b$

The if part:

Suppose $a^2=4b$
To show, $A$ is unique

Now, solving the characteristic equation, we have

$\lambda^2-tr(A)\lambda+|A|=0$

$\lambda=\frac{a \pm \sqrt{a^2-4b}}{2}$

So, $\lambda_1= \frac{a + \sqrt{a^2-4b}}{2}$
$\lambda_2= \frac{a - \sqrt{a^2-4b}}{2}$

But, since $a^2=4b$, so
$\lambda_1=\frac{a}{2}=\lambda_2$

Again, $A=\lambda_1I$, or $A=\lambda_2I$
Since, $\lambda_1=\lambda_2=\frac{a}{2}$,

$A=\frac{a}{2}I$
$\Rightarrow$ $A$ is unique