10.25 Linear combination

Tutorials and Workshops.pdf

Consider the vectors \(u = (1, 2, 3)\) and \(v = (2, 3, 1)\) in the real vector space \(\mathbb{R}^3\).
- (a) Find \(k \in \mathbb{R}\) so that \(w = (1, k, 4)\) is a linear combination of \(u\) and \(v\).
\(a.\text{ We need }w=\lambda\cdot u+\alpha\cdot v\text{ with }\lambda,\alpha\in\mathbb{R}\;\iff\;(1,k,4)=\lambda\cdot(1,2,3)+\alpha\cdot(2,3,1)\;\iff\;\begin{cases}1=\lambda+2\alpha\\ k=2\lambda+3\alpha\\ 4=3\lambda+\alpha\end{cases}\;\iff\;\begin{cases}\lambda=1-2\alpha\\ k=2-4\alpha+3\alpha=2-\alpha\\ 4=3-6\alpha+\alpha=3-5\alpha\end{cases}\;\text{then }k=\frac{11}{5},\lambda=\frac75,\alpha=-\frac15\text{ then }w=\frac75u-\frac15v.\) * (b) Find conditions on \(a, b, c\) so that \(w = (a, b, c)\) is a linear combination of \(u\) and \(v\).

\(b.\text{We need to write }(a,b,c)=\lambda(1,2,3)+\alpha(2,3,1),\lambda,\alpha\in\mathbb{R}\;\iff\;\begin{cases}a=\lambda+2\alpha\\ b=2\lambda+3\alpha\\ c=3\lambda+\alpha\end{cases}\;\iff\;\begin{cases}\lambda=a-2\alpha\\ b=2a-4\alpha+3\alpha=2a-\alpha\\ c=3a-6\alpha+\alpha=3a-5\alpha\end{cases}\;\iff\;\begin{cases}\lambda=a-4\alpha+2b=-3a+2b\\ \alpha=2a-b\\ c=3a-10a+5b=-7a+5b\end{cases}\Rightarrow\text{from here we get }7a-5b+c=0.\text{ So when we have }7a-5b+c=0\text{ we get }(a,b,c)=(-3a+2b)u+(2a-b)v.\)
Consider the real vector space \(V = \{f : \mathbb{R} \to \mathbb{R} : f \text{ is a function}\}\) and the functions \(\cos\), \(\sin\), \(\text{id}_{\mathbb{R}} \in V\). Show that if we have a linear combination of these elements in \(V\) equal to the zero function \(0 \in V\) as in: \(0 = a \cdot \cos + b \cdot \sin + c \cdot \text{id}_{\mathbb{R}}\) then \(a = b = c = 0\) necessarily.

We have \(0 = a \cdot \cos(x) + b \cdot \sin(x) + c \cdot x \quad \forall x \in \mathbb{R}.\)

Let's consider: \(x = 0: \quad 0 = a \cdot 1 + b \cdot 0 + c \cdot 0 \implies a = 0.\)

\(x = \pi: \quad 0 = b \cdot 0 + c \cdot \pi \implies c = 0.\)

\(x = \frac{\pi}{2}: \quad 0 = b \cdot 1 \implies b = 0.\)
Consider the complex vector space \(\mathbb{C}^3\) and \(v_1 = (1, 1 + i, i)\), \(v_2 = (i, -i, 1 - i)\) and \(v_3 = (0, 1 - 2i, 2 - i)\) vectors in \(\mathbb{C}^3\).
- (a) Is it possible to write one of these three vectors as a linear combination with coefficients in \(\mathbb{R}\) of the other two? In other words, is it possible to write \(0 = a \cdot v_1 + b \cdot v_2 + c \cdot v_3\), with \(a, b, c \in \mathbb{R}\) not all zero?
a. We need \(0 = a(1, 1+i, i) + b(i, -i, 1-i) + c(0, 1-2i, 2-i).\)

We have \(\begin{cases}0=a+bi,\\ 0=a(1+i)-b(i)+c(1-2i),\\ 0=ai+b(1-i)+c(2-i).\end{cases}\)

Since \(a, b \in \mathbb{R},\) the only way to have \(a = -b\) is if \(a = b = 0.\)

And when \(a = b = 0 \implies c = 0.\)

It is not possible. * (b) Is it possible to write one of these three vectors as a linear combination with coefficients in \(\mathbb{C}\) of the other two? In other words, is it possible to write \(0 = a \cdot v_1 + b \cdot v_2 + c \cdot v_3\), with \(a, b, c \in \mathbb{C}\) not all zero?

We'll continue from above

We have \(\begin{cases} a = -bi, \\ 0 = -bi(1 + i) - bi + c(1 - 2i) = b(1 - 2i) + c(1 - 2i), \\ 0 = b + b(1 - i) + c(2 - i) = b(2 - i) + c(2 - i). \end{cases}\)

\(\iff \begin{cases} a = -bi = ci, \\ b = -c, \\ b = -c. \end{cases}\)

So \(a = ci\) and \(b = -c.\)

So we can take, for example, \(c = 1,\) \(a = i,\) and \(b = -1.\)

And we have \(0 = ar_1 + br_2 + cr_3.\)
Show that the following vector spaces share the same structure (i.e. their operations behave "the same"):
- (a) \(M_2(\mathbb{R})\) and \(\mathbb{R}^4\) as real vector spaces with the natural addition and scalar multiplication operations.
Let's consider \(M_2(\mathbb{R})\) with its operations:

For \(A, B \in M_2(\mathbb{R})\) and \(\alpha \in \mathbb{R}\), we have that

\(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \quad B = \begin{pmatrix} \tilde{a} & \tilde{b} \\ \tilde{c} & \tilde{d} \end{pmatrix}\) \(A+B=\begin{pmatrix}a+\tilde{a} & b+\tilde{b}\\ c+\tilde{c} & d+\tilde{d}\end{pmatrix}\) \(\alpha \cdot A = \begin{pmatrix} \alpha a & \alpha b \\ \alpha c & \alpha d \end{pmatrix}\)

Let's consider \(\mathbb{R}^4\) with its operations: let \(r = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}\), \(w = \begin{pmatrix} \tilde{a} \\ \tilde{b} \\ \tilde{c} \\ \tilde{d} \end{pmatrix} \in \mathbb{R}^4\) and \(\alpha \in \mathbb{R}\).

So we have: \(r+w=\begin{pmatrix}a+\tilde{a}\\ b+\tilde{b}\\ c+\tilde{c}\\ d+\tilde{d}\end{pmatrix}\), \(\alpha \cdot r = \begin{pmatrix} \alpha a \\ \alpha b \\ \alpha c \\ \alpha d \end{pmatrix}\)

Note that, we can consider each matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) as a vector \(r = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}\) and the operations are unchanged. * (b) \(\mathbb{R}^4\) and \(P_3 = \{p \in \mathbb{R}[x] : \deg(p) \leq 3\}\) as real vector spaces with the natural addition and scalar multiplication operations.

We've already written the operations for \(\mathbb{R}^4\). Let's see for \(P_3(\mathbb{R})\):

Let \(p(x) = a + bx + cx^2 + dx^3\) and \(q(x) = \tilde{a} + \tilde{b}x + \tilde{c}x^2 + \tilde{d}x^3\), \(\alpha \in \mathbb{R}\).

\(p(x)+q(x)=(a+\tilde{a})+(b+\tilde{b})x+(c+\tilde{c})x^2+(d+\tilde{d})x^3\)

\(\alpha \cdot p(x) = (\alpha a) + (\alpha b)x + (\alpha c)x^2 + (\alpha d)x^3\)

We notice that we can consider each \(r = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix} \in \mathbb{R}^3\) as the polynomial \(p(x) = a + bx + cx^2 + dx^3\) and we get the "same" operations.

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