OSA: Linear Algebra

Question 1

Let \(F\) be a field and \(U_1, U_2 \subseteq V\) two linear subspaces of the \(F\)-vector space \(V\). Which of the following are equivalent to the assertion that \(V\) is the direct sum of \(U_1\) and \(U_2\)?

property	equivalent?	explanation
Every \(v \in V\) has a unique representation as a sum \(v = u_1 + u_2\) with \(u_1 \in U_1, u_2 \in U_2\).	yesno	This is the usual definition of direct sums.
Every affine subspace of \(V\) has non-empty intersections with both \(U_1\) and \(U_2\).	yesno	For a counterexample note that translating \(U_1\) by a non-zero vector from \(U_2\) gives an affine subspace which does not intersect \(U_1\) if \(V\) is the direct sum of \(U_1\) and \(U_2\).
There is a projection \(\pi:V\to V\) such that \(\mathrm{ker}(\pi) = U_1\) and \(\mathrm{image}(\pi) = U_2\).	yesno	If \(V\) is the direct sum of \(U_1\) and \(U_2\) we may choose \(\pi\) to be the identity on \(U_2\) and zero on \(U_1\). For the other direction, note that a projection is by definition idempotent, so \(\pi\) is the identity on \(U_2\).
For every pair of linear maps \(\varphi_1 \in \mathrm{Hom}(U_1,V)\), \(\varphi_2 \in \mathrm{Hom}(U_2,V)\), there is a \(\varphi \in \mathrm{Hom}(V,V)\) that extends both \(\varphi_1\) and \(\varphi_2\) in the sense that \(\varphi(u_i) = \varphi_i(u_i)\) for \(i = 1, 2\) and \(u_i \in U_i\).	yesno	Note that we do not ask for \(\varphi\) to be unique in the question, so any two subspaces \(U_1\) and \(U_2\) of \(V\) with \(U_1 \cap U_2 = \emptyset\) have the stated property.
Every basis \(B_1\) of \(U_1\) can be extended to a basis \(B\) for \(V\) using only vectors from \(U_2\) in \(B \setminus B_1\).	yesno	Note that the statement is always true for \(U_2 = V\).
For every basis \(B\) of \(V\), \(B \cap U_i\) forms a basis of \(U_i\) for \(i = 1,2\).	yesno	Assume \(V\) is the direct sum of \(U_1\) and \(U_2\), both of which have dimension \(\geq 1\). Take a basis \(B_1\) of \(U_1\) and add an arbitrary non-zero element of \(U_2\) to each basis element. Together with a basis \(B_2\) of \(U_2\), the resulting vectors form a basis of \(B\), but its intersection with \(U_1\) is empty and therefore not a basis of \(U_1\).
Every union \(B = B_1 \cup B_2\) of bases \(B_i\) of \(U_i\) for \(i = 1,2\) is a basis for \(V\).	yesno	This is not true if \(V = F_2\), the one-dimension vector space over the field \(F_2\): Since \(V\) has exactly one basis we may choose \(U_1 = U_2 = V\). Note that if \(F\) does not have characteristic 2 and \(U_1 \cap U_2\) contains a non-zero vector \(v\), we may choose a basis of \(U_1\) containing \(v\) and a basis of \(U_2\) containing \(2v\), but the union of these bases is not linearly independent. So \(U_1 \cap U_2 = \{0\}\) in this case.
For every \(v \in V\) there is a \(u \in U_1\) such that \((v-u) \in U_2\).	yesno	Take \(U_1 = V\).

Question 2

Consider the standard \(3\)-dimensional vector space \(V = (F_5)^3\) over the 5-element field \(F_5\) with addition and multiplication modulo 5. Specify

question	answer	result	explanation
the number of points in an affine subspace of dimension 1			Any affine subspace is a translation of a linear subspace, which has \(5^d\) many points if it is of dimension \(d\).
the number of points in a linear subspace of dimension 2			as above
the number of points in the quotient space \(V/U\) for a linear subspace of dimension 2			This is the same as the number of points in a linear complement of a linear subspace of dimension 2.
the number of affine subspaces associated to a fixed linear subspace of dimension 1			as above
the number of distinct linear complements of a fixed linear subspace of dimension 2			A linear subspace of dimension 2 leaves 125-25 = 100 points in its complement (as a set). Any four of these (namely its non-zero multiples) generate the same linear complement, leaving a total of 25 complements.

Question 3

Classify the following matrices up to similarity in \(R^{3\times 3}\). Hint: You can use invariants such as the trace, the determinant, the characteristic polynomial and the minimal polynomial. \begin{align*} A_1 &:= \begin{pmatrix} 2 & 1 & 0\\ 0 & 4 & -2\\ 0 & 0 & 2 \end{pmatrix} & A_2 &= \begin{pmatrix} 4 & 6 & 7\\ 0 & 1 & 5\\ 0 & 0 & 3 \end{pmatrix} & A_3 &= \begin{pmatrix} 1 & 1 & 0\\ -4 & 5 & 3\\ 0 & 0 & 1 \end{pmatrix} \\ A_4 &= \begin{pmatrix} 3 & 1 & 0\\ 0 & 3 & 1\\ 0 & 0 & 1 \end{pmatrix} & A_5 &= \begin{pmatrix} 20 & -30 & -15\\ 5 & -10 & -5\\ 6 & -6 & -4 \end{pmatrix} & A_6 &= \begin{pmatrix} 2 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 2 \end{pmatrix} \end{align*}

question	answer	explanation
\(A_1\) and \(A_2\)	yesno	Note that \(\det(A_1) = 16\) while \(\det(A_2) = 12\).
\(A_1\) and \(A_3\)	yesno	Note that \(\det(A_1) = 16\) while \(\det(A_3) = 9\).
\(A_1\) and \(A_4\)	yesno	Note that \(\det(A_1) = 16\) while \(\det(A_4) = 9\).
\(A_1\) and \(A_5\)	yesno	Note that \(\mathrm{tr}(A_1) = 8\) while \(\mathrm{tr}(A_5) = 6\).
\(A_1\) and \(A_6\)	yesno	Note that the eigenspace of \(A_6\) corresponding to eigenvalue 2 has dimension 2, whereas the eigenspace of \(A_1\) corresponding to eigenvalue 2 has dimension 1.
\(A_2\) and \(A_3\)	yesno	Note that \(\det(A_2) = 12\) while \(\det(A_3) = 9\).
\(A_2\) and \(A_4\)	yesno	Note that \(\det(A_2) = 12\) while \(\det(A_4) = 9\).
\(A_2\) and \(A_5\)	yesno	Note that \(\mathrm{tr}(A_2) = 8\) while \(\mathrm{tr}(A_5) = 6\).
\(A_2\) and \(A_6\)	yesno	Note that \(\det(A_2) = 12\) while \(\det(A_3) = 16\).
\(A_3\) and \(A_4\)	yesno	Both \(A_3\) and \(A_4\) have characteristic polynomial \((1-\lambda)(3-\lambda)^2\), and the eigenspace correspnding to eigenvalue \(3\) has dimension \(1\) is both cases.
\(A_3\) and \(A_5\)	yesno	Note that \(\mathrm{tr}(A_3) = 7\) while \(\mathrm{tr}(A_5) = 6\).
\(A_3\) and \(A_6\)	yesno	Note that \(\det(A_3) = 9\) while \(\det(A_3) = 16\).
\(A_4\) and \(A_5\)	yesno	Note that \(\mathrm{tr}(A_4) = 7\) while \(\mathrm{tr}(A_5) = 6\).
\(A_4\) and \(A_6\)	yesno	Note that \(\det(A_4) = 9\) while \(\det(A_3) = 16\).
\(A_5\) and \(A_6\)	yesno	Note that \(\mathrm{tr}(A_5) = 6\) while \(\mathrm{tr}(A_6) = 8\).