# Positive Definite Matrix

## Homework Statement

a positive definite matrix has eigenvalues λ=1 and λ=2. find the matrix

## 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?

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cdsandstuff

fresh_42
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## Homework Statement

a positive definite matrix has eigenvalues λ=1 and λ=2. find the matrix

## 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?
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?

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?
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......

fresh_42
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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......
Yes, but you have not mentioned whether ##1## and ##2## are the only eigenvalues nor their multiplicity. Anyway, can you answer my questions?

Ray Vickson
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## Homework Statement

a positive definite matrix has eigenvalues λ=1 and λ=2. find the matrix

## 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?

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)?

fresh_42
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The concepts of "definiteness" (positive, negative, semi-, etc) apply to symmetric matrices, ...
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.)

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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?
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

fresh_42
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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
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.

johnny blaz
Ray Vickson
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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.

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}##..

Thank you all for helping me out