given an m by n matrix A the transpose of A, A^t, is the n by m matrix whose columns are created by the matching rows of A
A subspace of a vector space V is a subcollection H of V that has actually the adhering to properties(0) V has H(1) The zero vector of V is in H(2) H is closed under vector enhancement. That is, for each u and v in H, the sum u + v is in H(3) H is closed under multiplication by scalars. That is, for each u in H and also each scalar c, the vector cu is in H.if v1,...vp are in a vector area V, then Spanv1,...,vp is a subroom of V.

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An n by n matrix A is shelp to be invertible if there is an n by n matrix c such that CA=I and also AC = I where I equals In the n by n identity matrix. C is an inverse of A, A^-1. C is distinctive figured out by A
For n higher than or equal to zero, the collection Pn of polynomials of degree at most n consists of all polynomials of the form p(t)=a0 +a1t+a2t^2+...+ant^n where the coefficients a0,...,an and also the variable t are genuine numbers. The level of p is the highest possible power of t whose coefficient is not zero. If p(t)=a0 does not equal 0, level of p is zero. If all coefficients are 0, p is called the zero polynomial. The zero polynomial is contained in Pn also though, for technical factors, its level is undefined.(p+q)(t)=p(t)+q(t)=(a0+b0)+(a1+b1)t+...+(an+bn)t^n(cp)(t)=cp(t)=ca0 +(ca1)t+(ca2)t^2+...+(can)t^n
Ax=bA(A^-1b)=b(AA^-1)b=bInb=b so A^-1b is a solutionAu=bA^-1(Au)=A^-1b(A^-1A)u=A^-1bInu=A^-1bu=A^-1b so A^-1b is a distinct solution
If A is an n by n matrix and the equation Ax=b has actually even more then one solution for some b, then the transformation x|->Ax is not one to one, What else have the right to be said about this transformation?
Suppose a straight transdevelopment T: Rn |-> Rn has actually the residential or commercial property that T(u) = T(v) for some pair of distinctive vectors u and also v in Rn. Can T map Rn onto Rn?
Assume A is the traditional matrix of T and T is not onto by hypothesis. A is not invertible by the IMT, A should have actually lialmost dependent columns, given that A is square, the columns of A carry out not expectancy Rn, so T cannot map Rn onto Rn.
When the entries along its main diagonal are nonzero => usage pivots to row alleviate to echelon develop => row tantamount to In => have to be invertible
When the entries along its primary diagonal are nonzero => in echelon form => row indistinguishable to In => must be invertible
No, need to not have actually a pivot in eextremely row, given that it is square it can not be row identical to In so it can not be invertible.
If A is invertible, then A^t is invertible so A^t have to have lipractically independent columns by the IMT.
No bereason if the matrix has the same columns then its columns must be lialmost dependent so by the IMT the matrix should not be invertible.
No because then the matrix have the right to be row decreased, by subtracting the the same 2 rows, to a kind wbelow one row has all zeros so the matrix can"t be row tantamount to the Identity matrix so the matrix can not be invertible.
If the columns of a 7 by 7 identity matrix D are lialmost independent what can be sassist about the options of Dx=b?
If the columns of D are lipractically independent and also given that D is square, then D should be invertible. If D is invertible, there exists a distinct solution to Dx=b for every b in R^7.Dx=bD(D^-1b)=b(DD^-1)b=bInb=b so D^-1b is a solutionDu=bD^-1(Du)=D^-1b(D^-1D)u=D^-1bInu=D^-1bu=D^-1b so D^-1b is a distinct solution
If A is a 5 by 5 matrix and also the equation Ax=b is continual for eincredibly b in R^5, is it feasible that for some b, the equation Ax=b has even more than one solution?
No, if Ax=b is continual for eincredibly b and square then A is invertible, so A must have actually a distinct solution for eincredibly b in R^5.
Exsimple why the columns of A^2 span R^n whenever the columns of an n by n matrix A are linearly independent.
If the columns of A are linearly independent and also A is square then A is invertible. A^2 = AA = the product of invertible matrices which need to be invertible.
If A is invertible, then the equation Ax=0 has actually a distinct solution, the trivial solution, so the columns of A must be livirtually independent.
If A is invertible then the equation Ax=b has a distinctive solution for every b in Rn, so the columns of A have to expectations Rn.
If Ax=b has actually a solution for eextremely b in Rn then A need to have actually a pivot in eextremely column considering that A is square A has a pivot in eextremely row as well. So A must be row indistinguishable to In and also A must therefore be invertible.
Suppose the last column of AB is completely zeros but B itself has actually no columns of zeros, what have the right to be sassist about the columns of A?
AB=A = = Abp = 0 however bp does not equal zero; therefore the columns of A need to be linearly dependent.
Bx=0 wright here x does not constantly = 0A(Bx)=0, (AB)x=0wright here x does not constantly equal zero, so the columns of AB should be lialmost dependent.

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Let A be a square matrix, if a multiple of one row of A is included to one more row to create a matrix B, then detB =?
A(A^-1)=I=(A^-1)Adet(A(A^-1))=detI = det((A^-1)A)det(A)det(A^-1)=1 = det(A^-1)det(A)det(A^-1)=1/detA = det(A^-1)
Let A and B be square matrices. Show that even though AB may not equal BA, detAB will certainly always equal detBA.
det(PAP^-1)=(detP)(detA)(detP^-1)=(detP)(detP^-1)(detA)=(detP^-1)(detP)(detA)=det(PP^-1)(detA)=det(P^-1))(detA)=det(I)det(A)=det(IA)=detA    