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I'm having a bit of trouble solving a homework problem. I've given the specific problem below, but general answers are 100% welcome:

Consider the following three bases for a vector space , which is a subspace of $C^\infty(-\infty,\infty):$ $$B_1 = \{ 1, e^x, e^{2x}\}$$ $$B_2 = \{ 1, 1+e^{2x}, e^x -e^{2x}\}$$ $$B_3 = \{ 1+e^x, 1+e^{2x}, 1+e^x+e^{2x}\}$$

Obtain the following transition matrices: $$P_{B_1 \to B_2}, P_{B_2 \to B_1}, P_{B_2 \to B_3}$$

I've read about change of basis and done problems for questions like this in finite vector spaces ($\mathbb{R}^n$ and $M_{nm}$, etc). But the way we're given to solve equations involves utilizing the inverse of the basis, which I'm not certain on calculating. At first observation, I can see that I could pull out $\vec{x} = \{1,e^x,e^{2x}\}$ in order to at least simplify what I'm looking at:

$$B_1 = A_1\vec{x} = \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} \begin{bmatrix} 1\\e^x\\e^{2x} \end{bmatrix}$$ $$B_2 = A_2\vec{x} = \begin{bmatrix} 1&0&0\\1&0&1\\0&1&-1 \end{bmatrix} \begin{bmatrix} 1\\e^x\\e^{2x} \end{bmatrix}$$ $$B_3 = A_3\vec{x} = \begin{bmatrix} 1&1&0\\1&0&1\\1&1&1 \end{bmatrix} \begin{bmatrix} 1\\e^x\\e^{2x} \end{bmatrix}$$

This allows me to view each basis at least in the sense of something which isn't one dimensional. I know from my text that a transition vector can be expressed as $P_{A \to B} = [B]^{-1}[A]$ (where A and B are bases), as well as:

$$AA^{-1} = I\\ IA = A\\ (AB)^{-1} = B^{-1}A^{-1}\\ [B] = [\vec{u}_1| \vec{u}_2| ...| \vec{u}_n]\\ A\vec{x}=\vec{b}\\ \vec{x}=A^{-1}\vec{b}$$

With the final two holding true if A is invertible. I've expressed all the bases in the form $\vec{b} = A\vec{x}$, so following the above logic (since every A from our bases is invertible), I believe the following should hold true: $$ P_{B_1 \to B_2} = B_2^{-1}B_1\\ ... = (A_2\vec{x})^{-1}(A_1\vec{x})\\ ... = (\vec{x})^{-1}A_2^{-1}A_1\vec{x}\\ \vec{x}P_{B_1 \to B_2} = A_2^{-1}A_1\vec{x}$$

However, I'm having trouble getting rid of or simplifying that $\vec{x}$ around. I'm also not certain if it's even appropriate to denote a non-square $\vec{x}^{-1}$. But I feel like there must be a way to simplify $(\vec{x})^{-1}M\vec{x} \to M$.

Failing that, I also tried:

$$rank(B_{1...3}) = 3\\ \therefore dim(V) = 3\\ \therefore \forall \vec{v} \in V, \vec{v} \in \mathbb{R}^3$$

So by this, I believe we can determine that while $C^\infty$ is an infinite-dimensional vector space, $V$ is actually a 3-dimensional subspace described by our bases.

But I suppose this also leaves me a littlle confused, as it would mean e.g. $1 + e^x$ (from $B_{3,1}$) is actually a third-dimensional vector, which is what lead me to breaking down the bases given based on the variable inputs over addition.

In short:

  • Is my approach flawed or incorrect? If so, how?
  • I'm a bit short on terminology since I'm learning, is there a name for what I did in that first step?
  • Most importantly: How do I generate the transition vector for this subspace?
Rogue
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  • Your first set of assertions that $$B_1 = A_1\vec{x} = \begin{bmatrix} 1&0&0\0&1&0\0&0&1 \end{bmatrix} \begin{bmatrix} 1\e^x\e^{2x} \end{bmatrix}$$ etc, are not correct. Here you're saying that the basis $B_1$ is the $3$-dimensional vector $(1, e^{x}, e^{2x})$, but a basis is not a vector it is a set of vectors, in this case, the set of the $1$-dimensional vectors ${1, e^x, e^{2x}}$. You can't manipulate a basis in the same way you manipulate a vector or a matrix. – SeraPhim Jul 18 '20 at 14:59
  • @SeraPhim okay, so the initial premise was wrong. Though isn't $A_1\vec{x}$ still a $3x1$ vector ($3x3 * 3x1$)? I'm aware a basis is a set of vectors (see the end paragraph), which is why it's confusing to me to assert that ${1, e^x, e^{2x}}$ is a basis, as in order to calculate the transition vector with our given equations I would need to calculate $[B]^{-1}$. But as far as I'm aware, you can't inverse a non-square matrix / vector. Plus in order for $[B]^{-1}(\vec{v})_B = \vec{v}$ to hold for an $m x 1$ matrix $\vec{v}$, I'd assume $B$ would have to be an $m x m$ matrix. – Rogue Jul 18 '20 at 15:08

1 Answers1

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Let us take $$b_1=1$$ $$b_2=e^x$$ $$b_3=e^{2x}$$ and $$c_1=1$$ $$c_2=1+e^{2x}$$ $$c_3=e^x-e^{2x}$$ which translate in $$c_1=b_1$$ $$c_2=b_1+b_3$$ $$c_3=b_2-b_3$$ the second has associated matrix $C=\left[\begin{array}{ccc}1&0&0\\1&0&1\\0&1&-1\end{array}\right]$.

But also if $$d_1=1+e^x$$ $$d_2=1+e^{2x}$$ $$d_3=1+e^x+e^{2x}$$ then $$d_1=b_1+b_2$$ $$d_2=b_1+b_3$$ $$d_3=b_1+b_2+b_3$$ with associated matrix $D=\left[\begin{array}{ccc}1&1&1\\1&0&1\\0&1&1\end{array}\right]$.

So if you want to see how to express $d_1,d_2,d_3$ in terms of the $c_1,c_2,c_3$ you ought to solve and get $$b_1=c_1$$ $$b_2=-c_1+c_2+c_3$$ $$b_3=-c_1+c_2$$ that has associated matrix $C^{-1}=\left[\begin{array}{ccc}1&-1&-1\\0&1&1\\0&1&0\end{array}\right]$.

And subbing we have \begin{eqnarray*} d_1&=&c_1+(-c_1+c_2+c_3)=c_2+c_3\\ d_2&=&c_1+(-c_1+c_2)=c_2\\ d_3&=&c_1+(-c_1+c_2+c_3)+(-c_1+c_2)=-c_1+2c_2+c_3 \end{eqnarray*} that has associated matrix $A=\left[\begin{array}{ccc}0&0&-1\\1&1&2\\1&0&1\end{array}\right]$.

This, is the multiplication $$C^{-1}D=\left[\begin{array}{ccc}1&-1&-1\\0&1&1\\0&1&0\end{array}\right] \left[\begin{array}{ccc}1&1&1\\1&0&1\\0&1&1\end{array}\right]. $$

janmarqz
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