Linear Algebra Assignment | Mathematics Assignment

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Linear AlgebraA32: Matricesa)R23Assuming that v is a vector space then a linearly independent spanning set for V iscalled the basis.For a matrix inRnthen the standard basis isei=(1,0,0,0,0,0,0),ise2=(0,1,0,0,0,0,0,0),isen=(0,0,0,0,0,0,0,0,1)but theRnis either ann1foracolumn¿1nforarow¿.Then theR23will be a 2*3. For a 2*3 the basis ise1=(1,0,0),e2=(0,1,0)b)A=(121)B=(311212)C=[1010]calculatingallthepossible¿ABCA*ABACBBA*BCCCACB*The areas marked by * have a matrix appearing twice hence are ignored as per therequirements.Now applying the matrix multiplication rule that is; to multiply two matrices, then thenumber of columns in the first matrix should be equal to the number of rows in thesecond matrix.We now check the matrices for compatibility.1st; AB¿(121)(311212)the matrices are compatible hence the multiplicationgives[43]2ndAC=(121)[1010]in this matrices matrix A has 3 columns while C has 2rows from the guiding rule these matrices are not compatible hence multiplication isnot possible.
Linear Algebra3rdBA;Bhas 2 columns while A has 1 row. Matrices are therefore not compatible.4th;BC=(311212)[1010]=[201010]5th; CA; C has 2 columns while A has 1 row hence not compatible.6th; CB, C has 2 columns while B 3 rows therefore not compatible.Afterwards we now check matricesABCBCAABC=[43][1010]=[10]BCA=[201010](121)In this case BC has 2 columns while a has only one row hence incompatibility.The obtained multiplications are the only possible matrix product from matrix A, Band Cc)CalculatingAnA=[4124251251]First, we begin by computingA1,A2,A3¿A1as for any number gives the same number so in our case we obtain[4124251251]now moving toA2=AA¿[4124251251][4124251251]=[088044044]A3=AAA¿[4124251251][4124251251][4124251251]=0¿herewe can conclude thatAn=0forallthen>2To prove this, we apply the diagonalization formulaSince A is 3 *3 matrix we try to put A into a diagonal matrix
Linear AlgebraSupposeD=P1APTheDn=[P1AP]n=P1APP1AP¿P1AP=P1AnPFrom here we can deriveAn=P1DnPThe aim is to put A in a diagonal format therefore there is need to obtain a basis ofeigenvectors.Obtaining the eigenvalues from the characteristic polynomial[4λ12425λ1251λ]¿λ3=(1)(λ)(λ)(λ)This givesλ1=λ2=λ3=0Now obtaining the eigenvectors from the eigenvalueAλE=[4124251251]AλE=0which gives a homogenous system of linear equations. Solving bygaussian elimination we get[4124251251]gives[102011000]From herex1+2x3=0x2x3=0Obtaining the variablesx2the system 2 givesx2=x3The vectorX=[2x3x3x3]Nowx3[211]which is the basis for the eigenvector for AIf this gives usPwhich means thatP1does not exist.The number of eigenvectors is not equal to the matrix dimensions hence matrix A isnot diagonalizable as it only has one eigenvector which is less than 3.
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