A subspace the a vector space V is a subset H that V that has actually the following properties

**(0) V includes H(1) The zero vector of V is in H(2) H is closeup of the door under vector addition. That is, because that each u and v in H, the sum u + v is in H(3) H is closeup of the door under multiplication by scalars. That is, for each u in H and each scalar c, the vector cu is in H.if v1,...vp room in a vector room V, climate Spanv1,...,vp is a subspace of V.**

You are watching: Explain why the columns of an nxn matrix a are linearly independent when a is invertible

An n through n procession A is said to it is in invertible if over there is an n by n matrix c such the CA=I and also AC = I whereby I equates to In the n through n identity matrix. C is an inverse of A, A^-1. C is unique determined by A

For n better than or equal to zero, the set Pn the polynomials of level at most n is composed of all polynomials the the form p(t)=a0 +a1t+a2t^2+...+ant^n whereby the coefficients a0,...,an and the change t are genuine numbers. The level of p is the highest power of t whose coefficient is no zero. If p(t)=a0 does not equal 0, degree of ns is zero. If all coefficients are 0, ns is called the zero polynomial. The zero polynomial is had in Pn even though, for technical reasons, 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 for this reason A^-1b is a distinctive solution

If A is an n by n matrix and the equation Ax=b has more then one equipment for part b, climate the transformation x|->Ax is not one to one, What else can be said about this transformation?

Suppose a linear change T: Rn |-> Rn has actually the home that T(u) = T(v) for part pair of distinct vectors u and also v in Rn. Can not map Rn top top Rn?

Assume A is the typical matrix the T and T is not onto through hypothesis. A is no invertible by the IMT, A must have actually linearly dependence columns, because A is square, the columns that A perform not span Rn, therefore T can not map Rn top top Rn.

When the entries along its main diagonal space nonzero => usage pivots come row reduce to echelon form => row indistinguishable to In => have to be invertible

When the entries follow me its key diagonal space nonzero => in echelon type => row tantamount to In => should be invertible

No, should not have actually a pivot in every row, due to the fact that it is square that can"t be row tantamount to In so that can"t it is in invertible.

If A is invertible, climate A^t is invertible so A^t must have linearly live independence columns by the IMT.

No because if the matrix has actually identical columns climate its columns should be linearly dependent so by the IMT the matrix must not be invertible.

No since then the matrix can be heat reduced, by individually the identical two rows, come a form where one row has actually all zeros for this reason the matrix can"t be row tantamount to the identity matrix therefore the procession can"t be invertible.

If the columns that a 7 by 7 identification matrix D space linearly elevation what can be said about the services of Dx=b?

If the columns of D are linearly independent and also since D is square, then D should be invertible. If D is invertible, over there exists a distinct solution come Dx=b because that every b in R^7.Dx=bD(D^-1b)=b(DD^-1)b=bInb=b for this reason D^-1b is a solutionDu=bD^-1(Du)=D^-1b(D^-1D)u=D^-1bInu=D^-1bu=D^-1b so D^-1b is a distinctive solution

If A is a 5 by 5 matrix and also the equation Ax=b is continuous for every b in R^5, is it feasible that for some b, the equation Ax=b has an ext than one solution?

No, if Ax=b is consistent for every b and square then A is invertible, for this reason A must have a unique solution for every b in R^5.

Explain why the columns of A^2 span R^n whenever the columns of one n by n matrix A space linearly independent.

If the columns of A space linearly independent and A is square then A is invertible. A^2 = AA = the product of invertible matrices which must be invertible.

If A is invertible, then the equation Ax=0 has actually a distinctive solution, the trivial solution, so the columns that A need to be linearly independent.

If A is invertible climate the equation Ax=b has a distinct solution because that every b in Rn, so the columns of A must span Rn.

If Ax=b has a equipment for every b in Rn then A must have actually a pivot in every column because A is square A has a pivot in every row together well. For this reason A should be row equivalent to In and also A must as such be invertible.

Suppose the last tower of abdominal is totally zeros but B itself has actually no columns the zeros, what can be said about the columns of A?

AB=A = = Abp = 0 however bp does not equal zero; because of this the columns the A should be linearly dependent.

Bx=0 whereby x walk not always = 0A(Bx)=0, (AB)x=0where x go not always equal zero, for this reason the columns of ab must it is in linearly dependent.

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Let A it is in a square matrix, if a multiple of one row of A is included to another row to produce a matrix B, climate 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 it is in square matrices. Present that even though abdominal may no equal BA, detAB will 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

You are watching: Explain why the columns of an nxn matrix a are linearly independent when a is invertible

An n through n procession A is said to it is in invertible if over there is an n by n matrix c such the CA=I and also AC = I whereby I equates to In the n through n identity matrix. C is an inverse of A, A^-1. C is unique determined by A

For n better than or equal to zero, the set Pn the polynomials of level at most n is composed of all polynomials the the form p(t)=a0 +a1t+a2t^2+...+ant^n whereby the coefficients a0,...,an and the change t are genuine numbers. The level of p is the highest power of t whose coefficient is no zero. If p(t)=a0 does not equal 0, degree of ns is zero. If all coefficients are 0, ns is called the zero polynomial. The zero polynomial is had in Pn even though, for technical reasons, 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 for this reason A^-1b is a distinctive solution

If A is an n by n matrix and the equation Ax=b has more then one equipment for part b, climate the transformation x|->Ax is not one to one, What else can be said about this transformation?

Suppose a linear change T: Rn |-> Rn has actually the home that T(u) = T(v) for part pair of distinct vectors u and also v in Rn. Can not map Rn top top Rn?

Assume A is the typical matrix the T and T is not onto through hypothesis. A is no invertible by the IMT, A must have actually linearly dependence columns, because A is square, the columns that A perform not span Rn, therefore T can not map Rn top top Rn.

When the entries along its main diagonal space nonzero => usage pivots come row reduce to echelon form => row indistinguishable to In => have to be invertible

When the entries follow me its key diagonal space nonzero => in echelon type => row tantamount to In => should be invertible

No, should not have actually a pivot in every row, due to the fact that it is square that can"t be row tantamount to In so that can"t it is in invertible.

If A is invertible, climate A^t is invertible so A^t must have linearly live independence columns by the IMT.

No because if the matrix has actually identical columns climate its columns should be linearly dependent so by the IMT the matrix must not be invertible.

No since then the matrix can be heat reduced, by individually the identical two rows, come a form where one row has actually all zeros for this reason the matrix can"t be row tantamount to the identity matrix therefore the procession can"t be invertible.

If the columns that a 7 by 7 identification matrix D space linearly elevation what can be said about the services of Dx=b?

If the columns of D are linearly independent and also since D is square, then D should be invertible. If D is invertible, over there exists a distinct solution come Dx=b because that every b in R^7.Dx=bD(D^-1b)=b(DD^-1)b=bInb=b for this reason D^-1b is a solutionDu=bD^-1(Du)=D^-1b(D^-1D)u=D^-1bInu=D^-1bu=D^-1b so D^-1b is a distinctive solution

If A is a 5 by 5 matrix and also the equation Ax=b is continuous for every b in R^5, is it feasible that for some b, the equation Ax=b has an ext than one solution?

No, if Ax=b is consistent for every b and square then A is invertible, for this reason A must have a unique solution for every b in R^5.

Explain why the columns of A^2 span R^n whenever the columns of one n by n matrix A space linearly independent.

If the columns of A space linearly independent and A is square then A is invertible. A^2 = AA = the product of invertible matrices which must be invertible.

If A is invertible, then the equation Ax=0 has actually a distinctive solution, the trivial solution, so the columns that A need to be linearly independent.

If A is invertible climate the equation Ax=b has a distinct solution because that every b in Rn, so the columns of A must span Rn.

If Ax=b has a equipment for every b in Rn then A must have actually a pivot in every column because A is square A has a pivot in every row together well. For this reason A should be row equivalent to In and also A must as such be invertible.

Suppose the last tower of abdominal is totally zeros but B itself has actually no columns the zeros, what can be said about the columns of A?

AB=A

Bx=0 whereby x walk not always = 0A(Bx)=0, (AB)x=0where x go not always equal zero, for this reason the columns of ab must it is in linearly dependent.

See more: How Did John D Rockefeller Treat His Workers, Please Wait

Let A it is in a square matrix, if a multiple of one row of A is included to another row to produce a matrix B, climate 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 it is in square matrices. Present that even though abdominal may no equal BA, detAB will 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