Math 2270 - Lecture 33 : Positive Definite Matrices

1 Math 2270 - Lecture 33 : Positive DefiniteMatricesDylan ZwickFall 2012 This Lecture coverssection the we re going to talk about a special type of symmetric matrix,called apositive definitematrix. A Positive definite matrix is a symmetricmatrix with all Positive eigenvalues. Note that as it s a symmetric matrixall the eigenvalues are real, so it makes sense to talk about them beingpositive or , it s not always easy to tell if a matrix is Positive definite. Quick,is this matrix? 1 22 1 Hard to tell just by looking at way to tell if a matrix is Positive definite is to calculateall theeigenvalues and just check to see if they re all Positive . The only problemwith this is, if you ve learned nothing else in this class, you ve probablylearned that calculating eigenvalues can be a real pain.

2 Especially for largematrices. So, today, we re going to learn some easier ways totell if a matrixis Positive assigned problems for this section are:Section - 2, 3, 7, 11, 161It s Positive Definite Matrices - What Are They, andWhat Do They Want?I ve already told you what a Positive definite matrix matrix is Positive definite if it s symmetric and all its eigenvalues are thing is, there are a lot of other equivalent ways to definea positivedefinite matrix. One equivalent definition can be derived using the factthat for a symmetric matrix the signs of the pivots are the signs of theeigenvalues. So, for example, if a4 4matrix has three Positive pivotsand one negative pivot, it will have three Positive eigenvalues and onenegative eigenvalue.

3 This is proven in section of the textbook. We canapply this fact to Positive definite Matrices to derive the next matrix is Positive definite if it s symmetric and all its pivots are are, in general,wayeasier to calculate than eigenvalues. Justperform elimination and examine the diagonal terms. No problem. Inpractice this is usually the way you d like to do it. For example, in thatmatrix from the introduction 1 22 1 If we perform elimination (subtract2 row 1 from row 2) we get 1 20 3 The pivots are1and 3. In particular, one of the pivots is 3, and sothe matrix is not Positive definite. Were we to calculate the eigenvalueswe d see they are3and det(Ak)k det(Ak_l) (Ak)> ,ifallupperleftkxkdeterminantsofasymmetr icmatrixarepositive, 10 12 1\0 123 \-L-/L1707jcsiveIfxisaneigenvectorofAthe nx0andAx= >0,thenasxTx>0wemusthaveXTAX> ,foranynon-zerovectorx, , > (orXTAx) +B.

4 I)dIiCfifl/-,Ourfinaldefinitionofpositiv edefiniteisthatamatrixAispositivedefinit eifandonlyifitcanbewrittenasA=RTR,whereR isamatrix,possiblyrectangular, ,usingourenergy-baseddefinition,it seasytoprovethatifA= , tproventhisyet,butthat (Rx)T(Rx) ,andsoxTAx> 1b 12 1b 12bb (-)( 25