How does this look on a Venn Diagram?

As explained below, System.out.println is the subroutine. This means a command is a subroutine with an inputted data (in this case "Hello World!").

should be p(u,v)

The k's cancel out and the denominator is equal to 1; the numerator is equal to \(\beta\).

The nullity is unknown but we know it is > 0. This assumes that the dimension of the null space is equivalent to the rank r. However, if q = 0, so that \(t - \lambda_1\) is the zero map. But if \(t - \lambda_1\) is the 0 map, then the rank is the empty set, but the the nullity > 0. This is a contradiction. This proves that it cannot be the zero map.

The exact number of y's is not essential for the proof because they all go to \(\overrightarrow{0}\) (see end of proof).

These are cases where lambda is not equal to 0.

The original function was applied to the domain \(\mathbb{C}^n\). Now the transformation (with the same name) is applied to the domain \(W\).

They prove that if the eigenvalues are distinct, the eigenvectors are linearly independent. Does this statement hold in the other direction? That is, if the eigenvectors are linearly independent, are the eigenvalues distinct?

By theorem 3.18 (introduced later), for any distinct set of eigenvalues, there is a linearly independent set of eigenvectors.

Need to prove that every t-string basis gives a matrix similar to a matrix that is all zeros expept for blocks of subdiagonal ones. The examples show this but a formal proof has not been shown.

Note that $$B_{R}^\frown B_N$$ is not a basis for V. (1) The dimension may be equivalent but range(h) + null(h) is not necessarily equal to V unless h is onto and h:V-->V. (2) In addition, the null space and the range space may overlap more than just the zero vector, in which case null(h) + range(h) is not a direct sum and \(B_R\) and \(B_N\) are not linearly independent. See page 435, where the diagram and proof clearly show this.

The sum of the two subspaces is by definition the span of the union of the range space and null space, which is less than V because we are assuming nilpotency > 1 for the inductive step. However, the sum rank(t) + nullity(t) = dim(V). This is because for every vector missing from the domain, there is a redundant vector that is in both the null and range spaces (see diagram above). However, the sum of the null and range spaces does not give V. Only the dimension is equivalent.

The problem with this concatenation is that C hat includes the basis for the intersection of the range and null spaces. I think he meant the concatenation B conc Z. In other words, the null space and range space are not independent. The equation below writes this relationship correctly because they subtract the dimension of the intersection.

We know the left-hand disjoint string basis form is the correct one (see map action graph above), so \(\beta_2\) and \(\beta_4\) are elements of the range. Any two vectors that satisfy the map action graph will do.

This is because the powers between 0 and n are not necessarily one to one so more than one vector may be sent to the zero vector and overlap with the range space (see example below).

They do not have to shrink. There is the case where k = 1.

Note that they prove that if \(t(\beta_i) = \lambda_i\beta_i \), then it is diagonalizable. But they do not prove that if it is diagonalizable, then \(t(\beta_i) = \lambda_i\beta_i \). It is trivial to prove this because it is true by the definition of diagonalizability of transformations.

if \(d(x) = x - \lambda\), then \(d(\lambda) = 0 \) and \( \ p(\lambda) = r(\lambda).\) Because we know that r(x) must be constant (based on division theorem for polynomials), then \(r(x) = c_0\). Thus, \(r(\lambda) = r(x)\).

Theorem 4.3 says that "a matrix is invertible if and only if it is nonsingular".

See the definition for a nonsingular linear map in Definition 2.7 and also the requirements for an inverse function in the appendix.

zero is considered "even".

This proof assumes that a function with the properties in Definition 2.1 exists. However, they have yet not proven that a function with these properties exists.

check proof later

Note that if D has less than n elements, it is not a basis for the enclosing set, but it is for a subset. The theorem proves that D cannot have more than n elements but not that it cannot have less. Thus, spanning the set is not required for D to be linearly independent.

He finally defines what a "spanning set" is.

They used polynomial division to get this result. However, multiplying by \(\frac{ 1/x}{1/x}\) is simpler. $$\lim_{x\to\infty} \frac{x}{x+1} = \lim_{x\to\infty} \frac{1}{1+ \frac{1}{x}} = 1$$

1 | a. Also, a | 1 if and only if a = 1. Therefore, pk = 1. However, by definition, every prime is divisible by one, so 1 is not a prime number.

I have not seen a theorem that shows that if ab | cd then a | cd (where they are all primes). However, they assume that r < s, so it is feasible that r = 1 with this assumption. Why can they assume this without "loss of generality"?

Refer to theorem 1.2.3. We know \(m_1\ | \ m_1\) (a). Thus, since \(m_1 \ | \ m_1\) and \(m_2 \ | \ n_1\), \(m_1m_2 \ | \ m_1n_1\) (f).

They prove using only the natural numbers and the multiplicative property. I think it could also be proven using the rational numbers and the multiplicative property by multiplying 1/c on both sides of ac = bc. However, they did not do this probably because they want to only use the natural numbers, and it is not necessary. The contrapositive statement, if a = b, then ac = bc is clear using the multiplicative property of equality.

Chain rule on the left side. We do not know y explicitly, so we factor out y'. However, we can easily solve for y' by implicit differentiation to check our solution.

What do they mean by sensible? I think they had commented earlier.

Let $$r = \begin{pmatrix} r_1 \ r_2 \end{pmatrix}$$. According to Lemma 2.3, when multiplying a linear combination of linear combinations by a factor, in this case r,

$$r_1s_1 + r_2s_2$$

the variables can be rearranged into a singular linear combination by regrouping the shared variables. In this case, the variables are the shared vectors.

$$\begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix}(r_1y_1 + r_2y_2) + \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}(r_1z_1 + r_2z_2) = \begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix}r(y_1 + y_2) + \begin{pmatrix} -1 \ 0 \ 1 \end{pmatrix}r(z_1 + z_2) $$

Because \(y,z \in \mathbb{R}\), \(y_1 + y_2, z_1 + z_2 \in \mathbb{R}\). Also, \(r(z_1 + z_2), r(y_1 + y_2) \in \mathbb{R}\). Therefore, the elements of the set of vectors above are equivalent to the elements of S; i.e., the two sets are equal. QED

See example 2.8. Notice the wording in the next sentence:

we cannot ... be confident the result is an element of T.

Many of the vectors in T will satisfy the conditions, but not all.

Stated in other words, they are proving that any system's solution set differs from the corresponding homogenous system's solution set only by the particular solution.

It is a bit confusing, because in the Lemma, \(\overrightarrow{h}\) is assumed to be a set of 1. I am not using \(\overrightarrow{s}\) and \(\overrightarrow{h}\) in the same way as the proof. \(\overrightarrow{s}\) and \(\overrightarrow{h}\) represent sets with \(i\) solutions.

Part I. From Appendix A-5, if two sets have the same members, they are equivalent. For set inclusion the first way (\(\overrightarrow{s}_i \in \overrightarrow{p} + \overrightarrow{h}\)). In this case, for any solution in \(\overrightarrow{s}\), \(\overrightarrow{s}_i - \overrightarrow{p}\) gives \(\overrightarrow{0}\), just as any solution of the associated homogenous system \(\overrightarrow{h_i}\) gives \(\overrightarrow{0}\) (i.e. the particular solution of any homogenous system is \(\overrightarrow{0}\)).

Part II. Likewise, for set inclusion the second way (\( \overrightarrow{p} + \overrightarrow{h}_i \in \overrightarrow{s}\)): any solution of the solution set of the associated homogenous system plus the particular solution of the system is sufficient to solve any system. By definition \(\overrightarrow{h}_i\) always gives \(\overrightarrow{0}\), but adding \(\overrightarrow{p}\) will give the complete solution set because in the first part of the proof they showed that \(\overrightarrow{s}_i - \overrightarrow{p} = \overrightarrow{h}_i.\).

Thus, we can use the solution set of an associated homogenous system to get the solution set of any system (if we have the particular solution). The other lemma proved that any homogenous system could be expressed in the pattern: particular + unrestricted combination. Combining that lemma with this one shows that any system can be expressed in that pattern because you only need to add the particular solution to get an equivalent solution set. However, they do not show this proof.

They prove that if the first condition on addition is satisfied for the subset, that all other conditions on addition are satisfied.

For if P, then Q statements, the statement is true only if Q is true, independent of P. For equivalent statements, both if P then Q and if Q then P need to be true. (3) --> (1) is the hardest, so the "intermediate" (2) statement was made to help prove (3) --> (1). In addition, (2) is more practical when trying to prove that S is a subspace of V.

The following were treated as self-evident: If (3) is true, then (2) is true because a pair of vectors in S is a subset of any vectors in S (3). If (1) is true, then (3) must be true by the definition of a vector space.

Vector spaces are "safe" spaces in which vector addition and multiplication can be performed. Although the name is "vector space", it encompasses all collections of linear combinations, not just vector structures or linear systems.

(2) They are proving that \(\overrightarrow{w}\) = \(\overrightarrow{v}\cdot-1\). Thus, we can use \(\overrightarrow{-v} = \overrightarrow{w}\) without worrying about confusing the term with \({\overrightarrow{v}\cdot-1}\).

\(a_n = (n-1)^2 + 1\) or \(a_{n-1} + 2n - 3, a_0 = 2\)

Was this an exercise?

Here they are using a homogenous system. They proved in the ealier chapter that an associated homogenous system differs from a system only by the particular solution.

By theorem here, he means the Lemma.

He has not stated the theorem yet. It's possible the author wrote the proofs in one coherent argument, but then split the theorem into Lemmas and did not bother to patch them up properly.

Proof by contradiction. Assume there is a row in R that is a linear combination of the others and see if the Lemma 2.5 holds.

(From the previous section) Linear surfaces are defined by a set of vectors. This means that there exists a vector that connects any two points in the surface. This definition unifies the properties of the vector with the properties of the linear surface, which allows us to explain the properties of the surface with the theorem. Thus, if the theorem does not hold in higher dimensions, the lines in higher dimensional spaces would not be flat.

The theorem says (in other words): take any two vectors (\(\overrightarrow{u}\) and \(\overrightarrow{v}\)) connecting two points in the plane they define, and no matter how much they approximate linearity, \(|\overrightarrow{u} + \overrightarrow{v}|\) will always have less length than the sum of their lengths. If they are multiples of each other, they are parallel and thus they do not form a plane.

In other words, any set of linear combinations of a variable can be regrouped into a single linear combination of the variable. Furthermore, any scalar multiplied to each element of the set of linear combinations becomes part of the new coefficients of the new linear combination of the variable.

This is important because as they showed in the examples above, Gaussian row operations essentially create sets of linear combinations of the variable within a row, but Lemma 2.3 shows that the variables can be regrouped so that set of linear combinations collapses to singular linear combination of the variable.

The Lemma is a bit confusing because they are using "linear combination" in a general and a specific sense. What they mean is:

A linear combination of linear combinations of x is a linear combination of x.

They use "linear combination" in the general sense because they create a linear combination of the scalar "d" with the set of linear combinations of x. Then they regroup to create a singular linear combination of x.

They are proving the theorem by lack of contradiction to inequality theorems. See Equivalent statements in appendix A-2. This seems to be a proof by "lack of contradiction". Assume that P is true, and see if Q holds because P is true if and only if Q is true.

I think the author may have initially used \(p_j \) and \(s_j\) instead of \(p_n \) and \(s_n\).

According to the definition of a partition, overlapping parts are equal. So all of the \(S_x\) parts (there are infinite) that share at least one \(y\) will collapse into one part. The result is two subsets that compose \(\Omega.\) From what I can tell, in the context of partitions, \(S = \Omega.\)

Earlier (A-4), they defined \(\hat{n}\) using \(n = 2^{k_n} \cdot \hat{n}\), where \(k_n\) refers to the number of 2's in \(n\). This definition is extended to \(\hat{d}\): \(d = 2^{k_d} \cdot \hat{d}\). However, in this context, \(k\) refers to the number of 2's that satisfy the following condition: \(\hat{n}d = n\hat{d}\).

This must be a typo because -1 \(\neq\) 1.