# Positive Definite Matrix

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1. Jun 30, 2016

### johnny blaz

1. The problem statement, all variables and given/known data
a positive definite matrix has eigenvalues λ=1 and λ=2. find the matrix

2. Relevant equations

3. The attempt at a solution
I've used a 2x2 matrix with entries a0,a1,a2,a3 as the unknown matrix but no use. (As little as i know a0 and a3 should be 1 and 2 respectively; corresponding to the eigenvalues) Any ideas on how to solve this?

Last edited: Jun 30, 2016
2. Jun 30, 2016

### Staff: Mentor

The fact, that the matrix is $2\times 2$ should be in the list of conditions I guess.
Start at the beginning: What does it mean for $\begin{pmatrix}a && b \\ c && d \end{pmatrix}$ to be positive definite?
(I find this notation easier in this case for I'll have to type less indices.)
Can you really assume already $a=1$ and $d = 2$? What do you really know about $λ_1 \cdot λ_2 = 2$ and $λ_1 + λ_2=3$ with respect to the characteristic polynomial?

3. Jun 30, 2016

### johnny blaz

I've assumed it to be a 2x2 matrix as a positive 2x2 matrix would have only two eigenvalues. No other conditions are provided for this problem......

4. Jun 30, 2016

### Staff: Mentor

Yes, but you have not mentioned whether $1$ and $2$ are the only eigenvalues nor their multiplicity. Anyway, can you answer my questions?

5. Jun 30, 2016

### Ray Vickson

The concepts of "definiteness" (positive, negative, semi-, etc) apply to symmetric matrices, so your matrix has three unknown parameters in it:
$$A = \pmatrix{a & b \\b & c}$$
Its characteristic polynomial is a quadratic function of $\lambda$ having roots 1 and 2. What does that tell you about $a,b,c$?

Why would you think that your $a_0$ and $a_3$ should be 1 and 2 (or 2 and 1)?

6. Jun 30, 2016

### Staff: Mentor

This is not necessary to be positive definite. The definition applies to all bilinear forms and every quadratic matrix defines one.
(I'm not quite sure whether I have made some mistake or not, but my solution isn't symmetric. The elements on the second diagonal may cancel out.)

Last edited: Jun 30, 2016
7. Jun 30, 2016

### johnny blaz

The product of eigenvalues is equal to the determinant of a matrix, and the sum of eigenvalues is equal to the trace. and the determinant of a 2 × 2 matrix A − λI can be written as:
λ^2 − (trace)λ + determinant = 0

8. Jun 30, 2016

### Staff: Mentor

Correct. So the characteristic polynomial has to be $char(t) = t^2-3t+2 = \det\begin{pmatrix}a-t && b \\ c && d-t\end{pmatrix}.$
(I've changed the variable to $t$ because you already used $λ$ for the eigenvalues, i.e. the solutions of $char(t)=0.$)
Furthermore the definition of positive definiteness means, that $(x,y)\begin{pmatrix}a && b \\ c && d\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} > 0$ for all possible values of $x,y$ where not both are $0.$
These conditions, if written out, can be used to calculate possible values of $a,b,c,d$. And there is an obvious solution to it and another one.

9. Jun 30, 2016

### Ray Vickson

Whether or not the 'definiteness' concept applies generally, or only to symmetric matrices, is a point of disagreement in the literature.

Anyway, one can reduce the general form for matrix $B$ to the equivalent symmetric form $A = (1/2) B + (1/2) B^T$, get the general symmetric solution, then deal with the multitude of solutions to $b_{12} + b_{21} = 2 a_{12}$. There are still infinitely many symmetric solutions, but all satisfying some specific relationships between $a_{11} , a_{12} = a_{21}$ and $a_{22}$..

10. Jun 30, 2016

### johnny blaz

Thank you all for helping me out