### map1 edits first pass

parent d53bc95c
This diff is collapsed.
 ... ... @@ -10787,7 +10787,7 @@ \end{ans} \subsection{Subsection Three.I.2: Dimension Characterizes Isomorphism} \begin{ans}{Three.I.2.8} \begin{ans}{Three.I.2.9} Each pair of spaces is isomorphic if and only if the two have the same dimension. We can, when there is an isomorphism, state ... ... @@ -10819,7 +10819,7 @@ \end{exparts} \end{ans} \begin{ans}{Three.I.2.9} \begin{ans}{Three.I.2.10} \begin{exparts*} \partsitem $$\rep{3-2x}{B}=\colvec[r]{5 \\ -2}$$ \partsitem $$\colvec[r]{0 \\ 2}$$ ... ... @@ -10827,25 +10827,25 @@ \end{exparts*} \end{ans} \begin{ans}{Three.I.2.10} \begin{ans}{Three.I.2.11} They have different dimensions. \end{ans} \begin{ans}{Three.I.2.11} \begin{ans}{Three.I.2.12} Yes, both are $$mn$$-dimensional. \end{ans} \begin{ans}{Three.I.2.12} \begin{ans}{Three.I.2.13} Yes, any two (nondegenerate) planes are both two-dimensional vector spaces. \end{ans} \begin{ans}{Three.I.2.13} \begin{ans}{Three.I.2.14} There are many answers, one is the set of $$\polyspace_k$$ (taking $$\polyspace_{-1}$$ to be the trivial vector space). \end{ans} \begin{ans}{Three.I.2.14} \begin{ans}{Three.I.2.15} False (except when $$n=0$$). For instance, if $$\map{f}{V}{\Re^n}$$ is an isomorphism then multiplying by any ... ... @@ -10854,7 +10854,7 @@ possible is $\zero_V\mapsto 0_W$.) \end{ans} \begin{ans}{Three.I.2.15} \begin{ans}{Three.I.2.16} No. A proper subspace has a strictly lower dimension than it's superspace; if $U$ is a proper subspace of $V$ then any linearly independent subset ... ... @@ -10862,7 +10862,7 @@ be a basis for $V$, and $U$ wouldn't be proper. \end{ans} \begin{ans}{Three.I.2.16} \begin{ans}{Three.I.2.17} Where $$B=\sequence{\vec{\beta}_1,\ldots,\vec{\beta}_n}$$, the inverse is this. \begin{equation*} ... ... @@ -10871,11 +10871,11 @@ \end{equation*} \end{ans} \begin{ans}{Three.I.2.17} \begin{ans}{Three.I.2.18} All three spaces have dimension equal to the rank of the matrix. \end{ans} \begin{ans}{Three.I.2.18} \begin{ans}{Three.I.2.19} We must show that if $$\vec{a}=\vec{b}$$ then $$f(\vec{a})=f(\vec{b})$$. So suppose that ... ... @@ -10894,11 +10894,11 @@ and so the function is well-defined. \end{ans} \begin{ans}{Three.I.2.19} \begin{ans}{Three.I.2.20} Yes, because a zero-dimensional space is a trivial space. \end{ans} \begin{ans}{Three.I.2.20} \begin{ans}{Three.I.2.21} \begin{exparts} \partsitem No, this collection has no spaces of odd dimension. \partsitem Yes, because $\polyspace_{k}\isomorphicto\Re^{k+1}$. ... ... @@ -10907,7 +10907,7 @@ \end{exparts} \end{ans} \begin{ans}{Three.I.2.21} \begin{ans}{Three.I.2.22} One direction is easy:~if the two are isomorphic via $$f$$ then for any basis $$B\subseteq V$$, the set $$D=f(B)$$ is also a basis (this is shown in ... ... @@ -10935,7 +10935,7 @@ and the scalar multiplication calculation is similar. \end{ans} \begin{ans}{Three.I.2.22} \begin{ans}{Three.I.2.23} \begin{exparts} \partsitem Pulling the definition back from $$\Re^4$$ to $$\polyspace_3$$ ... ... @@ -10950,7 +10950,7 @@ \end{exparts} \end{ans} \begin{ans}{Three.I.2.23} \begin{ans}{Three.I.2.24} Yes. Assume that $$V$$ is a vector space with basis ... ... @@ -10990,7 +10990,7 @@ Preservation of scalar multiplication is similar. \end{ans} \begin{ans}{Three.I.2.24} \begin{ans}{Three.I.2.25} Because $$V_1\intersection V_2=\set{\zero_V}$$ and $$f$$ is one-to-one we have that $$f(V_1)\intersection f(V_2)=\set{\zero_U}$$. To finish, count the dimensions: ... ... @@ -10998,7 +10998,7 @@ as required. \end{ans} \begin{ans}{Three.I.2.25} \begin{ans}{Three.I.2.26} Rational numbers have many representations, e.g., $$1/2=3/6$$, and the numerators can vary among representations.
This diff is collapsed.
This diff is collapsed.
Markdown is supported
0% or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!