Let {U1,U2,...,Un} be linearly independent set of vectors in R^n. Then, the matrix A=[U1,U2,...,Un} is an invertible matrix.
what is this proof?
what is this proof?
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Since R^n is of dimension n, {U1, U2, ..., Un} forms a basis for R^n, which means that it spans R^n. Let E1, E2, ..., En be the standard basis vectors. Then, for each i, Ei can be written as a (unique) linear combination of U1, U2, ..., Un. Let Vi be the vector of coefficients for the linear combination for Ei, for each i. Then, I claim that the matrix:
B = [V1, V2, ..., Vn]
is the inverse of B, thus proving that A is invertible. The way to see this is to consider matrix multiplication. First, you need to notice that A(Vi) = Ei for any i. That's because the way that multiplying a matrix by a column vector works is by taking a linear combination of the column vectors of the matrix, with coefficients read from the column vector. You can see this if you just think about the mechanics of matrix multiplication. Since Vi contains exactly the coefficients necessary to make the linear combination of the column vectors of A (i.e. U1, ..., Un) combine to give Ei, the result of A(Vi) = Ei follows immediately.
Now, to see that AB = I, notice that I = [E1, E2, ..., En], and that, as always:
A[V1, ..., Vn] = [A(V1), ..., A(V2)]
Again, this can be seen by simply considering matrix multiplication. The columns of the right matrix are acted upon independently by the left matrix, and the product will have columns equal to the columns of the right matrix, multiplied by the left matrix. Now, from what we proved before:
AB = A[V1, ..., Vn] = [A(V1), ..., A(Vn)] = [E1, ..., En] = I
So, B is A's inverse. Hence, A is invertible.
B = [V1, V2, ..., Vn]
is the inverse of B, thus proving that A is invertible. The way to see this is to consider matrix multiplication. First, you need to notice that A(Vi) = Ei for any i. That's because the way that multiplying a matrix by a column vector works is by taking a linear combination of the column vectors of the matrix, with coefficients read from the column vector. You can see this if you just think about the mechanics of matrix multiplication. Since Vi contains exactly the coefficients necessary to make the linear combination of the column vectors of A (i.e. U1, ..., Un) combine to give Ei, the result of A(Vi) = Ei follows immediately.
Now, to see that AB = I, notice that I = [E1, E2, ..., En], and that, as always:
A[V1, ..., Vn] = [A(V1), ..., A(V2)]
Again, this can be seen by simply considering matrix multiplication. The columns of the right matrix are acted upon independently by the left matrix, and the product will have columns equal to the columns of the right matrix, multiplied by the left matrix. Now, from what we proved before:
AB = A[V1, ..., Vn] = [A(V1), ..., A(Vn)] = [E1, ..., En] = I
So, B is A's inverse. Hence, A is invertible.