《线性代数应该这样学》（Linear Algebra Done Right）中说：You cannot read mathematics the way you read a novel. If you zip through a page in less than an hour, you are probably going too fast. When you encounter the phrase “as you should verify”, you should indeed do the verification, which will usually require some writing on your part. 这个笔记也由此而生。
笔记将会记录一部分书内要求自行验证的题目。出于某些原因，我将试着用英语来进行记录。这是我第一次尝试着用英语写数学证明，因此在格式、单词、语法等地方出错应该是难以避免的，望谅解。
CHAPTER 1 Vector Spaces
- With the usual operations of addition and scalar multiplication, is a vector space over , as you should verify.
Proof 1.13 states the commutativity. As for associativity, suppose , and . Then
Suppose . Then
For additive identity, let . Obviously, for all , suppose
For additive inverse, suppose . There exists such that
For multiplicative identity, suppose . Then
For distributive properties, suppose , , . Then
Thus is a vector space over .
- You should verify all three bullet points in the next example.
- If is a nonempty set, then (with the operations of addition and scalar multiplication as defined above) is a vector space over .
- The additive identity of is the function defined by for all .
- For , the additive inverse of is the function defined by for all .
Proof Commutativity. Suppose , then
Associativity. Suppose , , then
Additive identity. Suppose and defined by for all , then
Additive inverse. Suppose and defined by for all , then
Multiplicative identity. Suppose , then
Distributive properties. Suppose and , then
-
You should verify all the assertions in the next example.
(a) If , then is a subspace of if and only if .
(b) The set of continuous real-valued functions on the interval is a subspace of .
(c) The set of differentiable real-valued functions on is a subspace of .
(d) The set of differentiable real-valued functions on the interval such that is a subspace of if and only if .
(e) The set of all sequences of complex numbers with limit is a subspace of .
Proof (a) Denote the set by .
If it is a subspace of , then in it. Hence . This happens if .
If , then belongs to the set. For any and , suppose , . So we have that and . Then
i.e. is closed under addition. Similarly,
i.e. is closed under scalar multiplication.
Hence if , is a subspace of .
(b) Denote the set of continuous real-valued functions on the interval by .
The additive identity of is the constant function on , and it obviously belongs to the set.
The sum of tow continuous functions is continuous, i.e. is closed under addition.
The product of constant with continuous function is continuous, i.e. is closed under scalar multiplication.
Thus is a subspace of .
(c) Omitted
(d) Denote the set by .
If is a subspace of , then the additive identity . Hence .
If , it is clear that the additive identity is contained in .
Closed under addition: suppose , then are differentiable real-valued functions. So is . Moreover,
Closed under scalar multiplication: suppose and , then is differentiable read-valued function. So is . Moreover,
(e) Denote the set by .
Additive identity: it is clear that .
Closed under addition: suppose and , so and .
Hence, .
Closed under scalar multiplication: suppose and , then .
Hence, .
- Suppose is the set of all elements of whose second and third coordinates equal , and is the set of all elements of whose first and third coordinates equal :
Then
as you should verify.
Proof Suppose and , then
Hence .
Every vector in , can be written as
where the first vector on the right side is in , the second vector is in .
Hence .
Thus .
- Suppose is the subspace of of those vectors whose last coordinate equals , and is the subspace of of those vectors whose first two coordinates equal :
Then , as you should verify.
Proof , hence is direct sum.
Clearly , because every vector can be written as
where the first vector on the right side is in , the second vector is in .
Thus .