Linear Algebra Flashcards
When is a matrix in Upper Echelon from?
All non-zero rows are below zero rows. The first entry of a non zero row is 1. The first entry of each non zero row is strictly to the right of the row above.
What are the properties of a vector space? (5)
i) a(U+V) = aU + aV ii) (a+b)V = aV + bV iii) (ab)V = a(bV) iv) 1V = V v) addition properties of a field
What is a linear combination of v(1), v(2), … , v(n) ?
a vector of the form v = av(1) + bv(2) + …. + nv(n) for integers a,b,…,n
When are vectors linearly independent?
When the only combination of the vectors that makes zero is the trivial one
Lemma; The vectors v(1),…,v(n) are linearly dependent if and only either v(1)=0 or, for some r, the vector v(r) is a linearly combination of v(1),…,v(r-1)
If v(1)=0 then for any a(1) = 1 and a(i) = 0 for all i not equal to one, the linear combination is equal to zero so the vectors are linearly independent. If v(r) is a linear combination then a(1)v(1)+ … + a(r-1)v(r-1) - v(r) = 0 so the vectors are linearly dependent. Conversely, if they’re linearly independent then a(1)v(1)+ … + a(r)v(r)=0 with the coefficients not all zero. Let r be the maximal with a(r) not zero, then if r=1 v(1)=0. Otherwise, we get v(r) = a(1)v(1)/a(r)+ …. + a(r-1)v(r-1)/a(r) in which case v(r) is a linear combination.
When do vectors span a set V
When every vector in V is a linear combination of a subset S
When are a subset S in V a basis of V
When they span V and the vectors in S are linearly independent
Lemma; The vectors v(1),…..,v(n) form a basis of V if and only if every vector in V can be written uniquely as a linear combination of v(1),…..,v(n)
Suppose v(1),…..,v(n) form a basis of V. Then each vector in V can be written as a(1)v(1)+ … + a(n)v(n). Suppose also that v= b(1)v(1) + …. + b(n)v(n). Then v-v = (a(1)-b(1))v(1) + ….. + (a(n)-b(n))v(n) = 0 This implies by linear independence that a(i) = b(i) for all i. So each vector can be written uniquely Conversely, suppose each vector in V can be written uniquely. Then v(1),….,v(n) span V. Specifically a(1)v(1)+….+a(n)v(n)=0 and since its unique all a(i) = 0, so the vectors a linearly independent and therefore form a basis.
What is the dimension of V?
The number of vectors in the basis of V
Lemma; Suppose that vectors v(1),….,v(n),w span V and that w is a linear combination of v(1),….,v(n). Then v(1),….,v(n) span V
Since v(1),….,v(n),w span V then every vector v can be written as a linear combination v= a(1)v(1)+….+a(n)v(n)+a(w)w. Since w is a linear combination w=b(1)v(1)+….+b(n)v(n) this can be rearranged to get v=(a(1)+a(w)b(1))v(1)+….+(a(n)+a(w)b(n))v(n). so the vectors v(1),….,v(n) span V
Theorem; Suppose that the vectors v(1),….v(r) span the vector space V. Then there is a subsequence of v(1),…,v(r) than form a basis of V
We sift the vectors v(1),….,v(r). The vectors that are sifted are linear combinations of the remaining vectors, and the remaining vectors still span. The remaining vectors are linearly independent, and hence form a basis of V
Theorem; Let V be a vector space over K which has a finite spanning set, and suppose that the vectors v(1),….,v(r) are linearly independent. Then we can extend the sequence to a basis v(1),….,v(n) where n >= r
Suppose that w(1),…..,w(q) is a spanning set. We sift the sequence v(1),….,v(r),w(1),….,w(q). Since w(1),…..w(q) spans V, the whole sequence spans V. Furthermore, v(1),….,v(r) are linearly independent none of these are sifted so the basis contains v(1),….v(r)
What is the row reduced matrix of vectors that form a basis
Identity
Proposition; Suppose that vectors v(1),….,v(n) span V and that vectors w(1),…..w(m) are linearly independent. Then m<=n
Since v(1),….,v(n) span V then w(1),v(1),…..,v(n) are linearly independent, so we sift and at least one v(j) is deleted. We add w(2) to the front and sift again, and keep repeating. None of the w(i) are deleted since they are linearly independent. In total, we add m vectors, and a v(j) is deleted each time, so we must have m<=n
Can n-1 vectors span a vector space of dimension n?
No
Can n+1 vectors span a vector space of dimension n?
Yes
Can n-1 vectors be linearly independent in a vector space of dimension n?
Yes
Can n+1 vectors be linearly independent in a vector space of dimension n?
No
What are the 3 conditions of a subset W in a vector space V for it to be a subspace?
i) W is non-empty ii) Closed under addition iii) Closed under scalar multiplication
Proposition; If W(1) and W(2) are subspaces then W(1)nW(2) is also a subspace
Take u,v in W(1)nW(2). Then u+v is in W(1) , and u+v is in W(2), so its in the intersection group. Similarly au is in the intersection, so its a subspace
Define the subset W(1) + W(2)
w(1) + w(2), where w(1) is in W(1) and w(2) is in W(2)
Proposition; W(1) + W(2) is a subspace where W(1) and W(2) are subspaces, in fact it’s the smallest subspace containing both
Take u,v in W(1) + W(2), u= u(1) + u(2), v= v(1)+v(2). Then u+v = (u(1)+u(2)) + (v(1)+v(2)) is in the subset. a(u) = a(u(1)+u(2)) is in the subset. So its a subspace. Any other subspace will contain this subspace, so its the smallest
What is the equation for dim(W(1)+W(2))
dim(W(1)) + dim(W(2)) - dim(W(1)nW(2))
What are the two properties of a linear transformation T from U to V
T(u(1)+u(2)) = T(u(1)) + T(u(2)) T(au) = aT(u)
Proposition; Let U,V be vector spaces, let S be a basis of U and let f:S -> V be a function assigning to each vector in S an element of V. Then there is a unique linear map T from U to V such that for every s in S T(s) = f(s)
Since S is a basis of U each element in U is uniquely determined. u = a(1)s(1) + …. + a(n)s(n). so T(u) = T(a(1)s(1)+…..+T(a(n)s(n) = a(1)f(s(1)) + a(n)f(s(n)). So T if it exists is uniquely determined
Let T(1), T(2) map U to V, with non matrices A,B. What is the matrix of the transformation i) T(1) + T(2) ii) T(1)(T(2))
i) A+B ii) AB